MTH 4201LN 2014-15(1).pdf

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Chapter 1

1. Relations and equivalence relations

Let A and B be any sets. The Cartesian productof A and B is defined in the following way:

A´B={(a,b): aÎA and bÎB}.In this set two elements (a1,b1) and (a2,b2) are equal

if and only if îíì

==

21

21

bb

aa.

Examples. Let A={1,2,3} and B={b}. Thena) A´B={(1,b), (2,b), (3,b)}, b) A´A={(1,1), (2,1), (3,1), (1,2), (2,2), (3,2), (1,3),(2,3), (3,3)},c) B´A={(b,1), (b,2), (b,3)},d) B´B={(b,b)ú bÎB}.2. If A=B=R the set of all real numbers thengeometrically R´R can be considered as the set of all points of a plane.

1.1.1. Definition. Let S be a set. A subset R of S´S is called an equivalence relation on S if

(i) for all aÎS , (a,a)ÎR;(ii) for all a,bÎS , if (a,b)ÎR then (b,a)ÎR;(iii) for all a,b,cÎS, if (a,b)ÎR and (b,c)ÎR,then (a,c)ÎR.We will write a~b to denote the fact that (a,b)ÎR.

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Using the above definition and notation, it isclear that R is an equivalence relation if and only if for all a,b,cÎS we have(i) [ Reflexive law] a~a;(ii) [Symmetric law] if a~b, then b~a;(iii) [Transitive law] if a~b and b~c, then a~c.

Example. Let n be any fixed positive integer,Z={…, -4, -3, -2, -1, 0, 1, 2, 3, 4, …}

be the set of all integer numbers. Consider the

followingR n={(a,b): a,bÎZ and n½(a-b)}

subset of Z´Z. We use n½(a-b) to say that (a-b)¸n

is an integer number. Let us show that R n is aequivalence relation on Z. Indeed, for any a,b,cÎZ

we have:(i) a~a for any aÎS, because n½(a-a), 0¸n=0,(ii)if a~b, then b~a, because if (a-b)¸n is anyinteger then (b-a)¸n=-(a-b)¸n is also integer,(iii) (iii) if a~b and b~c, then a~c, because if

(a-b)¸nÎZ and (b-c)¸nÎZ, then (a-b)¸n+(b-c)¸n=(a-c)¸nÎZ. 1.1.2. Definition. Let ~ be an equivalencerelation on the set S. For a given element aÎS, wedefine the equivalence class of a to be the set of all

elements of S that are equivalent toa

. We will usethe notation [a]. In symbols, [a]={ xÎS| x~a}.

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The notation S/~ will be used for the collection of allequivalence classes of S under ~. The set S/~ is said to be the factor set of S with respect to theequivalence relation ~.

Example. For the above equivalence relation R non Z the set Z/~ consists of the following classes:

[0]={0, ±n, ±2n, ±3n,…},[1]={1, ±n+1, ±2n+1, ±3n+1,…},[2]={2, ±n+2, ±2n+2, ±3n+2,…},

…[n-1]={n-1, ±n+(n-1), ±2n+ (n-1), ±3n+(n-1),…}.

Example 1.1.2 Let f : S®T be any function.For x1, x2ÎS we define x1~ x2 if f ( x1)= f ( x2). Then for all x1, x2, x3ÎS we have

(i) f ( x1)= f ( x1);(ii) if f ( x1) = f ( x2), then f ( x2) = f ( x1); and (iii) if f ( x1) = f ( x2) and f ( x2) = f ( x3), then f ( x1) = f ( x3).This shows that we have defined an equivalencerelation on the set S. The proof of this is easy

because the equivalence relation is defined in termsof equality of the images f ( x), and equality is themost elementary equivalence relation. The collectionof all equivalence classes of S under the abovedefined ~ will be denoted by S/ f .

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Example. Let f : S®S, where S={1,2,3,4}, be

function given by its table (graph)1

4

3

3

2

2

2

1

)( x f

x. In

this case S/ f ={{1,2},{3},{4}}.Example. Let f : R®R, where R is the set of

real numbers and f ( x)= x2. In this case

R/ f = {{-a, a}: a ³0}.1.1.3. Proposition. Let S be a set, and let ~ be an

equivalence relation on S. Then each element of S belongs to exactly one of the equivalence classes of S determined by the relation ~.

Proof: Let a be any element of the set S. ThenaÎ[a]. If also aÎ[b] for some bÎS, then a~b, henceb~a, bÎ[a] and therefore [a]=[b].

1.1.4. Definition. Let S be any set. A family P of subsets of S is called a partition of S if eachelement of S belongs to exactly one of the membersof P.

1.1.5. Proposition. Any partition P of a set S

determines an equivalence relation. Proof: Let P={Si} be a partition of a set S. Wedefine an equivalence relation on S as follows:

a~b if and only if there exists a set S0i ÎP such that

a,bÎS0i. It is easy to see that all the axioms for

equivalence relation (1.1.1) are satisfied.

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1.1.6. Theorem. If f : S®T is any function, and ~ is the equivalence relation defined on S by letting x1~ x2 if f ( x1) = f ( x2), for all x1, x2ÎS, then there is a

one-to-one correspondence between the elements of the image f (S) of S under f and the equivalenceclasses S/ f of the relation ~ .

Proof: We define function f ~

: S/ f ® f (S) as

follows: f ~

([ x])= f ( x). It is easy to see that f ~

is well

defined and onto. If f ~

([ x1])= f ~

([ x2]) then by the

definition of the function f ~

, f ( x1)= f ( x2), hence x1~ x2.

Therefore [ x1]=[ x2] and function f ~

is one-to-one.

This completes the proof.If f : S®T is a function and y belongs to the

image f (S), then { xÎS | f ( x)= y } is usually called the inverse image of y, denoted by f

-1( y). Theinverse images of elements of f (S) are the

equivalence classes in S/ f .

2. Binary operation and induced (factored)

binary operation

A binary operation * on a set S is a rule that

assigns to each ordered pair (a,b) of elements of S aunique element a*b of S. For example, the ordinary

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operations of addition, subtraction, and multiplication are binary operations on the set of realnumbers. The operation of division is not a binaryoperation on the real numbers because it is notdefined for all ordered pairs of real numbers.Although subtraction is a binary operation on the setof all real numbers, it is not a binary operation onthe set of natural numbers, since, for example, 1-2 isnot in the set of natural numbers.

Definition 1.2.1 A binary operation * on a set Sis a function *:S´S®S from the set S´S of allordered pairs of elements in S into S.

The operation * is said to be commutative if a * b = b * a

for all a,bÎS.The operation * is said to be associative if

a * (b * c)=(a * b) * c

for all a, b, c Î S.

An element eÎS is called an identity element for *

if a * e = a and e * a = a

for all aÎ S.If * has an identity element e, and aÎ S , then bÎ S

is said to be an inverse for a if

a * b = e and b * a = e.

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Let us assume that on a set S an equivalencerelation ~ and a binary operation *: S´ S®S aregiven. Consider the set of equivalence classes S/~. Can we define a binary operation on this set bythe use of * ? If the equivalence relation ~ is in concordancewith the binary operation * then we can introduce a

binary operation on S/~. Definition 1.2.2. If for any a, b, c, d ÎS wehave a* c ~ b* d whenever a ~ b and c ~ d then it issaid that the equivalence relation ~ is in concordancewith the binary operation * (or the equivalencerelation ~ respects the binary operation *)

Examples 1. Let S=Z, n be fixed positive integer

and consider on Z the equivalence relation ~ given by

R n={(a,b): a,b ÎZ and n½(a-b)}.For a binary operation * on Z consider ordinaryaddition +. Let a, b, c, d ÎZ be integers. If a ~ b

( i.e. n½(a-b)) and c ~ d ( i.e. n½(c-d )) thena+c ~ b+d, because (a+c)- (b+d )= (a-b)+(c-d ) and n½[(a-b)+(c-d )]. So “~“ respects “+”.

Example 2. Now in the above example change

the binary operation + to multiplication ×. If a ~ b(i.e. n½(a-b)) and c ~ d (i.e. n½(c-d )) then

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a×c ~ b×d, because (a×c)- (b×d )= c×(a-b)+b×(c-d ) and n½[c×(a-b)+b×(c-d )]. So “~“ respects “×” as well.

If an equivalence relation ~ on S respects a binary operation * on S we can introduce thefollowing binary operation [*] on S/~ : Bydefinition

[a][*][b]= [a*b]for any [a], [b] ÎS/~. Pay attention that here we define the result of

binary operation [*] over the equivalence classes [a]and [b] by the use of their representatives a, b and the binary operation *. The binary operation [*] is“well defined” because of concordance of ~ with *:

If [a]=[b] and [c]=[d ] then a*c~b*d i.e. [a*c]=[b*d ].In future we use * instead of [*] and say that it is theinduced (factored) operation on S/~ by the binaryoperation * on S. It should be emphasized that wecan speak about induced(factored) operation on S/~

by the binary operation * on S if only ~ respects *.Proposition 1.2.3. a) If e is the identity element

with respect to the binary operation * on S then [e]is the identity element for the induced operation * onS/~.

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b) If the binary operation * on Sholds commutativity then the induced operation * on S/~ does as well.

c) If the binary operation * on S holdsassociativity then the induced operation * on S/~does as well.

The proof is evident.

3. Fields; Roots of Polynomials

Definition 1.3.1. Let F be a set on which two binary operations are defined, called addition and multiplication, and denoted by + and × respectively.Then F is called a field with respect to theseoperations if the following properties hold:

(i) Closure: For all a,bÎF the sum a+b and the product a×b are uniquely defined and belong toF.

(ii) Associative laws: For all a,b,cÎF(a+b)+c=a+(b+c) and a × (b× c) = (a×b) × c

(iii) Commutative laws: For all a,bÎF ,a + b = b + a and a×b = b×a .

(iv) Distributive laws: For all a, b, cÎF ,a× (b + c) = (a× b) + (a× c).

(v) Identity elements: The set F contains an

additive

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identity element, denoted by 0, such that for allaÎF , a + 0 = a.

The set F also contains a multiplicative identityelement, denoted by 1 (and assumed to be differentfrom 0) such that for all aÎF , a×1 =a .(vi) Inverse elements: For each aÎF , the

equationsa + x = 0

has a solution xÎF, called an additive inverse of a,and denoted by -a. For each nonzero element aÎF , the equation

a× x = 1has a solution xÎF , called a multiplicative inverse of a, and denoted by a

-1.

We will shortly show that both additive inversesand multiplicative inverses are unique, and this will

justify the notation in the preceding definition.The basic examples you should keep in mind in

this section are (Q;+,×), (R;+,×), (C;+,×), and (Z p;+,×)

where p is prime. Here is one more example of field. Example. Let

Q ( 2) = {a + b 2: a, bÎ Q}.We note that if a1+b1 2=a¢1+b¢1 2, then

a1 – a¢1=(b1-b¢1) 2 so if b1-b¢1¹ 0

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then we can divide by b1-b¢1, which shows that 2is rational, which is a contradiction. We concludethat b1=b¢1, and a1=a¢1, so there is only one way to

represent an element of {a + b 2: a,bÎQ}. It iseasy to see that {a + b 2: a,bÎQ} is closed under addition and subtraction. It is only slightly moredifficult to check that the same is true for multiplication and division:(a+b 2)(c+d 2) = ac+ad 2+bc 2+2bd

= (ac+2bd ) +(ad+bc) 2;

222

2

2)(2(

)2)(2(

2

22222

d c

ad bc

d c

bd ac

d cd c

d cba

d c

ba

--

+--

=-+-+

=++

In both cases the answer has the correct form: arational number plus a rational number times the

square root of 2. In dividing, we must of courseassume that either c¹0 or d ¹0, and then c

2-2d 2¹0

since 2 is not a rational number, which also showsthat c - d 2¹ 0. Q( 2) is the smallest field thatcontains Q and the one root 2.

We now provide the first example of a finite field different from Z p , for p prime. Example 1.3.2. The following set of matrices,with entries from Z2 , forms a field:

F={

úû

ù

êë

é

úû

ù

êë

é

úû

ù

êë

é

úû

ù

êë

é

11

10,

01

11,

10

01,

00

00}

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Here we have omitted the brackets from thecongruence classes [0] and [1], so that we simplyhave 1+1=0, etc. You should check that F is closed under addition and multiplication. The associativeand distributive laws hold for all matrices. You cancheck that these particular matrices commute under

multiplication. The additive identity is úû

ùêë

é00

00, and

the multiplicative identity is úû

ùêë

é10

01. Each element is

its own additive inverse, and the multiplicative

inverse of úû

ùêë

é01

11is úû

ùêë

é11

10.

Each nonzero element of a field F is invertible; wewill use F* to denote the set of all nonzero elementsof F. When there is no risk of confusion, we willoften write ab instead of a×b. If we assume thatmultiplication is done before addition, we can ofteneliminate parentheses. For example, the distributivelaws can be written as a(b +c)=ab+ac and

(a + b)c=ac+bc. You should also note that the listof properties used to define a field is redundant in anumber of places. For example, since multiplicationis commutative, we really only need to state one of

the distributive laws.

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To simplify matters, we have dealt only with thetwo operations of addition and multiplication. Todefine subtraction we use the additive inverse of anelement:a - b = a + (-b). Similarly, we use the multiplicativeinverse of an element to define division, as follows:

If b¹ 0, then a¸b = ab-1. We will sometimes write

a/b in place of ab-1. There is a problem with these

definitions, since a-b and a¸b should be unique,

whereas we do not as yet even know that -b and b-1

are unique. The next proposition takes care of this problem.

Proposition 1.3.4 Let F be a field, with a,b,cÎF.

(a) Cancellation laws: If a+c=b+c, then a=b and

If c ¹ 0 and ac = bc, then a = b.(b) Uniqueness of identity elements: If a + b = a,

then b = 0. If a c = a and a ¹ 0 , then c=1.(c)Uniqueness of inverses: If a + b = 0, then b=-a.

If a ¹ 0 and ab = 1, then b = a-1.

Proof. Proposition 1.3.4 shows that identityelements and inverses are unique for any binaryoperation. The axioms that hold for the field F showthat F is an abelian group under addition, and so theadditive cancellation law follows from Propositionfor groups.

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Suppose that ac=bc, with c¹0. Since c is nonzero,an inverse c

-1 exists, and we can rewrite the equationin the following way: ac = bc

(ac)c-1 = (bc)c-1

a(c c-1) = b(c c-1)a1 = b1a = b .

This completes the proof. Proposition 1.3.5. Let F be a field.

(a) For all a Î F , a 0 = 0.(b) If a, b Î F with a ¹ 0 and b ¹ 0, then ab ¹ 0.

(c) For all a Î F , -(-a) = a.(d) For all a, b Î F , (a) (-b) = (-a) (b) = -ab.(e) For all a, b Î F , (-a) (-b) = ab.

Proof. (a) We will use the fact that 0 + 0 = 0. Bythe distributive law,

a 0 + a 0 = a (0 + 0) = a 0 = a 0 + 0 ,so the cancellation law for addition shows that a0=0.

(b) If a ¹ 0 and ab = 0, then ab = a 0 and the

cancellation law for multiplication shows that b = 0.(c) In words, the equation -(-a) = a states thatthe additive inverse of -a is a, and this follows fromthe equation -a + a = 0 which defines -a.

(d) Using the distributive law,a b + a (-b) = a (b + (-b)) = a 0 = 0 ,

which shows that (a)(-b) is the additive inverse of ab, and so (a)(-b) = -(ab). Similarly, (-a)(b) = -ab.

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(e) Now consider (-a) (-b). By what we have justshown, (-a)(-b) = -((-a)(b)) = -(-ab) = ab ,and this completes the proof.Having proved some elementary results on fields,we are now ready to discuss polynomials withcoefficients in a field. Definition 1.3.6. Let F be a field.If am, am-1,…, a1, a0Î F , then any expression of theform

am xm+ am-1 xm-1+ …+ a1 x+ a0

is called a polynomial over F in the indeterminate x

with coefficients am,am-1,…, a1, a0 . The set of all polynomials with coefficients in F is denoted byF[ x]. If n is the largest nonnegative integer such that

an¹ 0, then we say that the polynomial f ( x)=an x

n+...+a0

has degree n, written deg( f ( x)) = n, and an is called the leading coefficient of f ( x). If the leadingcoefficient is 1, then f ( x) is said to be monic.

According to this definition, the zero polynomial(each of whose coefficients is zero) has no degree.For convenience, it is often assigned -¥ as a degree.A constant polynomial a0 has degree 0 when a0¹ 0.Thus a polynomial belongs to the coefficient field if and only if it has degree 0 or -¥.

Two polynomials are equal by definition if theyhave the same degree and all corresponding

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coefficients are equal. It is important to distinguish between the polynomial f ( x) as an element of F[ x]and the corresponding polynomial function from Finto F defined by substituting elements of F in placeof x.

If f ( x) = an xn + .. . + a0 and c Î F, then f (c) = an c

n + .. . + a0.

In fact, if F is a finite field, it is possible to have twodifferent polynomials that define the same

polynomial function. For example, let F be the field Z5 and consider the polynomials x

5 -2 x+l and 4 x +1. For any cÎZ5 , by Fermat's theorem (Theorem1.4.12) we have c

5ºc(mod5), and so c5-2c+1º-

c+1º4c+1(mod5), which shows that x5 - 2 x + 1 and

4x + 1 are identical, as functions. For the polynomials

f ( x)=am xm+ am-1 x

m-1+ …+ a1 x+ a0

and g( x)=bn x

n+ an-1 xn-1+ …+ b1 x+ b0

the sum of f

( x

) and g( x

) is defined by just addingcorresponding coefficients. The product f ( x)g( x) isdefined to bebnam x

m+n+(bnam-1+bn-1am) xn+m-1+…+(b0a1+a0

b1) x+a0b0

The coefficient ck of xk in f ( x)g( x) can be described

by the formula

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ck = å å== =+

-

k

i k ji jiik i baba

0

This definition of the product is consistent with whatwe would expect to obtain using a naive approach:Expand the product using the distributive lawrepeatedly (this amounts to multiplying each term beevery other) and then collect similar terms. With thisaddition and multiplication, F[ x] has propertiessimilar to those of the integers. Checking that the

following properties hold is tedious, though notdifficult. The necessary proofs use the definitions,and depend on the properties of the coefficient field.We have omitted all details.(i) Associative laws: For any polynomials f ( x) ,

g( x) , h( x) over F, f ( x) + (g( x) + h( x)) = ( f ( x) + g( x)) + h( x) ,(ii) Commutative laws: For any polynomials f ( x) ,

g( x) over F, f ( x) (g( x) h( x)) = ( f ( x) g( x)) h( x)

f ( x) + g( x) = g( x) + f ( x) , f ( x) g( x) = g( x) f ( x).

(iii) Distributive laws: For any polynomials f ( x),g( x), h( x) over F,

f ( x)×(g( x)+h( x)) = ( f ( x)g( x))+( f ( x)h( x)) ,(iv) Identity elements: The additive and

multiplicative identities of F, considered as

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constant polynomials, serve as identityelements.

(v) Additive inverses: For each polynomial f ( x)over F, the polynomial -f (x) serves as an additiveinverse.In the formula above, two polynomials aremultiplied by multiplying each term of the first byeach term of the second, and then collecting similar terms.

Proposition 1.3.7. If f ( x) and g( x) are nonzero polynomials in F[ x] , then f ( x)g( x) is nonzero and

deg( f ( x)g( x)) = deg( f ( x)) + deg(g( x)) .Proof. Suppose that

f ( x)=am xm+am-1 x

m-1+ …+ a1 x+ a0

and g( x)=bn xn+ an-1 x

n-1+ …+ b1 x+b0

with deg( f ( x)) = m and deg(g( x)) = n. Thus am¹ 0and bn¹ 0. It follows from the general formula for multiplication of polynomials that the leadingcoefficient of f ( x)g( x) must be ambn, which must benonzero. Thus the degree of f ( x)g( x) is m + n, sincein f ( x)g( x) the coefficient of xm+n is ambn . Note that we could extend the statement of

Proposition 1.3.7. to include the zero polynomial, provided we would use the convention that assigns

to the zero polynomial the degree -¥.

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Corollary 1.3.8 If f ( x) , g( x), h( x) Î F[ x], and f ( x)is not the zero polynomial, then f ( x)g( x) = f ( x)h( x)implies g( x) = h( x).

Proof. If f ( x)g( x) = f ( x)h( x), then we can use thedistributive law to rewrite the equation as f ( x)(g( x)-h( x))=0, and since f ( x)¹ 0, the previous propositionimplies that g( x)-h( x)=0, or simply g( x)= h(x).

Having proved Proposition 4.1.5, we can makesome further observations about F[ x]. If f ( x)g( x)=1,

then both f ( x) and g( x) must be constant polynomials, since the sum of their degrees must be0. This shows that the only polynomials that havemultiplicative inverses are those of degree 0, whichcorrespond to the nonzero elements of F. In this

sense F[ x] is very far away from being a field itself,although all of the other properties of a field aresatisfied.

Definition 1.3.9 Let f ( x), g( x) Î F[ x]. If f ( x) =q( x)g( x) for some q( x)Î F[ x] , then we say that g( x) is

a factor or divisor of f ( x) , and we write g( x)½ f ( x).The set of all polynomials divisible by g( x) will bedenoted by ág( x)ñ. Lemma 1.3.10 For any element cÎ F, and any

positive integer k , ( x-c)½( xk -c

k )Proof. A direct computation shows that we have

the following factorization:

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( xk -c

k )= ( x-c)( xk -1+cxk -2+…+c

k -2 x+c

k -1) Note that the quotient xk -1+cx

k -2+…+ck -2

x+ck -1 has

coefficients in F. The next theorem shows that for any polynomial f ( x) , the remainder when f ( x) is divided by x - c is

f (c). That is,c x

c f xq

c x

x f

-+=

-)(

)()(

or simply, f ( x) = q( x)( x - c) + f (c) .

The remainder f (c) and quotient q( x) are unique. Theorem 1.3.11 (Remainder Theorem) Let f ( x)ÎF[ x] be a nonzero polynomial, and let cÎ F.

Then there exists a polynomial q( x)Î F[ x] such that f ( x) = q( x)( x - c) + f (c) .

Moreover , if f ( x)=q1( x) ( x - c) + k, where q1( x)ÎF[ x] and k ÎF , then q1( x)=q( x) and k = f (c).Proof. If f ( x)= f ( x)=am x

m+ am-1 xm-1+ …+ a1 x+ a0,

then f ( x)-f (c)= am ( xm

- cm )+am-1( x

m-1- cm-1)+…+ a1( x-c)

But, x - c is a divisor of each term on the right-hand equation, and so it must be a divisor of f ( x) - f (c).Thus

f ( x) - f (c) = q( x)( x - c)for some polynomial q( x) Î F[ x], or equivalently,

f ( x) = q( x)( x - c) + f (c) .If f ( x)=q1( x)( x - c)+k , then (q( x) –q1( x))( x-c)= k-f (c) .

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If q( x) – q1( x) ¹ 0, then by Proposition 4.1.5 the left-hand side of the equation has degree ³1, whichcontradicts the fact that the right-hand side of the

equation is a constant. Thus q( x) – q1( x)= 0, whichalso implies that k - f (c) = 0, and so the quotient and remainder are unique.

Example 1.3.12

Let us work through the previous proof in thecase F = Q, for f ( x) =x

2 + 5 x - 2 and c = 5: f ( x) - f (5) = ( x2

+ 5 x - 2) - (52 + 5×5 - 2)= ( x2 - 52) + (5 x - 25)

=( x+5)( x-5)+5( x-5)

= ( x+10)( x-5).Thus we obtain x

2 + 5 x - 2 = ( x +10)( x - 5)+48. Definition 1.3.13 Let

f ( x)=am xm+am-1 x

m-1+ +…+a1 x+a0 Î F[ x].An element cÎ F is said to be a root of the

polynomial f ( x) if f (c) = 0, that is, if c is a solutionof the polynomial equation f ( x)=0. Corollary 1.3.14 Let f ( x)Î F[ x] be a nonzero

polynomial, and let c Î F. Then c is a root of f ( x) if and only if x - c is a factor of f ( x). That is, f (c) = 0if and only if ( x - c)½ f ( x).

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Corollary 1.3.15A polynomial of degree n withcoefficients in the field F has at most n distinct rootsin F.

Proof. The proof will use induction on thedegree of the polynomial f ( x). The result is certainlytrue if f ( x) has degree 0, that is, if f ( x) is a nonzeroconstant. Now suppose that the result is true for all

polynomials of degree n-1. If c is a root of f ( x) , then by Corollary 4.1.11 we can write f ( x) = q( x)( x-c), for

some polynomial q( x). If a is any root of f ( x), thensubstituting shows that q(a) (a - c) = 0, whichimplies that either q(a) = 0 or a = c. By assumption,q( x) has at most n-1 distinct roots and so this showsthat f ( x) can have at most n-1 distinct roots which

are different from c.

1.4 Factors

Our aim in this section is to obtain, for polynomials, results analogous to some of thetheorems for integers. In particular, we are looking

for a division algorithm, an analogue of theEuclidean algorithm, and a prime factorizationtheorem. Many of the arguments in algebrainvolved finding the smallest number in some set of integers, and so the size, in terms of absolute value,

was important. Very similar arguments can be given

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for polynomials, with the notion of degree replacingthat of absolute value. Our first goal is to formulate and prove a divisionalgorithm. The following example is included justto remind you of the procedure that you probablylearned in high-school algebra. Example 1.4.1. In dividing the polynomial

6 x4- x3+ x2+5 x-18

by 2 x2 - 3, the first step is to divide 6 x4 by 2 x2, to get

3 x2 . The next step is to multiply 2 x2-3 by 3 x2 and subtract the result from 6 x4-2 x3+ x2+5 x-18. Thealgorithm for division of polynomials then proceedsmuch like the algorithm for division of integers, asshown in Figure below

…………………3 x2…- x…+5 2 x2 - 3½ 6 x4 - 2 x3 + x

2 + 5 x- 18 6 x4 -9 x2

-2 x3 +10 x2 +5 x -2 x3 +3 x 10 x2+2 x –18 10 x2 -15

2 x -3We finally obtain

6 x4 -2 x3 + x2 +5 x-18=(3 x2- x+5)(2 x2 –3)+(2 x-3),

where the last term is the remainder.

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The proof of Theorem 1.4.2 is merely a formalverification, using induction, that the procedurefollowed in Example 1.4.1 will always work. The

polynomials q( x) and r ( x) given by the theorem,with f ( x)=q( x)g( x)+r ( x) , are called (as expected) thequotient and remainder when f ( x) is divided by g( x).

Notice that if we divide polynomials withcoefficients in a given field, then the quotient and remainder must have coefficients from the same

field.The division algorithm for integers (Theorem

1.1.3) was stated for integers a and b, with b > 0. Itis easily extended to a statement that is parallel tothe next theorem for polynomials: For any a, bÎZ

with b¹0, there exist unique integers q and r suchthat a=bq+r , with 0<r < ïbï. The role played by theabsolute value of an integer is now played by thedegree of a polynomial. Note that assigning thedegree -¥ to the zero polynomial would simplify the

statement of the division algorithm, requiring simplythat the degree of the remainder be less than thedegree of the divisor. Theorem 1.4.2 (Division Algorithm) For any

polynomials f ( x) and g( x) in F[ x] , with g( x)¹ 0, thereexist unique polynomials q( x) , r ( x)Î F[ x] such that

f ( x) = q( x)g( x) + r ( x) , where either deg(r ( x)) < deg(g( x)) or r ( x) º 0.

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Proof. Let f ( x) = am xm+…+ a1 x + a0 and g( x)

= bn xn+…+ b1 x + b0 , where am¹ 0 and bn¹ 0 .

In case f ( x) has lower degree than g( x), we cansimply take q( x) = 0 and r ( x) = f ( x). The proof of the other case will use induction on the degree of f ( x). If f ( x) has degree zero, it is easy to see that thetheorem holds. Now assume that the theorem is truefor all polynomials f ( x) of degree less than m. (Weare assuming that m > n.). The reduction to a

polynomial of lower degree is achieved by using the procedure outlined in Example 1.4.1. We divideam x

m by bn to get ambn-1

xm-n , then multiply by g( x)

and subtract from f ( x). This gives f 1( x) = f ( x) - ambn

-1 x

m-n g( x) ,

where f 1( x) has degree less than m since the leadingterm of f ( x) has been cancelled by ambn-1

xm-n

bn xn .

Now by the induction hypothesis we can write f 1( x) = q1( x)g( x) + r ( x) ,

where the degree of r ( x) is less than n, unless r ( x) º0. Since f ( x) = f 1( x) + ambn

-1 x

m-n g( x)

substitution gives the desired result: f ( x) = (q1( x) + ambn

-1 x

m-n) g( x) + r ( x)The quotient q( x) = q1( x) + ambn

-1 x

m-n hascoefficients in F, since ambn Î F, and q1( x) hascoefficients in F by the induction hypothesis.

Finally, the remainder r ( x) has coefficients in F bythe induction hypothesis.

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To show that the quotient q( x) and remainder r ( x)are unique, suppose that

f ( x) = q1( x)g( x) + r 1( x) and f ( x) =q2( x)g( x) + r 2( x)

Thus (q1( x)- q2( x))g( x) = r 2( x) - r 1( x) and if (q1( x)-q2( x))¹ 0, then the degree of (q1( x)-q2( x))g( x)is greater than or equal to the degree of g( x),whereas the degree of r 2( x)-r 1( x) is less than thedegree of g( x). This is a contradiction, so we can

conclude that q1( x)= q2( x), and this forces r 2( x) -r 1( x)=(q1( x)- q2( x))g( x)= 0 , completing the proof.

Theorem 1.3.11 is a particular case of the generaldivision algorithm. If g( x) is the linear polynomial x-

c then the remainder must be a constant when f ( x) is

divided by x - c. Substituting c into the equation f ( x)=q( x)( x-c)+r ( x) shows that r (c)=f (c), so theremainder on division by x - c is the same element of F as f ( x) evaluated at x = c.

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Example 1.4.3 In this example we illustrate thedivision algorithm for the polynomials in Example1.4.1, over the finite field Z7. Our first step is toreduce coefficients modulo 7, to obtain f ( x)=6 x4+5 x3+x2+5 x + 3 and g( x)=2 x2 + 4. Since itis much easier to divide by a monic polynomial, wemultiply g( x) by 4 (the inverse of 2 modulo 7) and work with x

2+2. Proceeding as in Example 4.2.1, we

obtain

6 x4+5 x3+x2+5 x+3=(6 x2+5 x+3))( x2+2)+(2 x+4) ,where the last term is the remainder. To take care of the fact that we divided by g( x)/2 , we need tomultiply the divisor x2

+2 by 2 and divide thequotient 6 x2

+5 x+3 by 2. This finally gives

6 x4

+5 x3

+x

2

+5 x +3=(3 x2

+6 x +5))(2 x2

+4)+(2 x+4) ,with the remainder being unchanged.

The next result is analogues of the theorem,which shows that every subgroup of Z is cyclic.

Theorem 1.4.4 Let I be a subset of F[ x] that

satisfies the following conditions:(i) I contains a nonzero polynomial;(ii) if f ( x) , g( x)Î I, then f ( x)+g( x)Î I;

(iii) if f ( x)Î I and q( x)Î F[ x] , then q( x) f ( x)ÎI.If d ( x) is any nonzero polynomial in I of minimal

degree, thenI = { f ( x)Î F[ x]: f ( x) =q( x)d ( x) for some q( x)ÎF[ x]}.

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Proof. If I contains a nonzero polynomial, thenthe set of all natural numbers n such that I contains a

polynomial of degree n is a nonempty set, so by thewell-ordering principle it must contain a smallestelement, say m. Thus we can find a nonzero

polynomial of minimal degree m in I, say d ( x).Every multiple q( x)d ( x) of d ( x) must be in I by

condition (iii). Next we need to show that d ( x) is a

divisor of any other polynomial h( x)ÎI. One way to proceed is to simply divide h( x) by d ( x), using thedivision algorithm, and then show that the remainder must be zero. We can write h( x)=q( x)d ( x)+r ( x) , where r ( x) is

either zero or has lower degree than d ( x). Solvingfor r ( x), we have r ( x) = h( x)+(-q( x))d ( x) .This shows that r ( x)ÎI, since h( x)ÎI and I is closed

under addition and under multiplication by any polynomial. But then r ( x) must be zero, since Icannot contain a nonzero polynomial of lower degree than the degree of d ( x). This shows that h( x)is a multiple of d ( x). We should note that in any set of polynomials of the form

{ f ( x)Î F[ x]: f ( x)=q( x)d ( x) for some q( x)ÎF[ x]} ,

the degree of d ( x) must be minimal (among degreesof nonzero elements). Multiplying by the inverse of

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the leading coefficient of d ( x) gives a monic polynomial of the same degree that is still in the set.

Definition 1.4.5 A monic polynomial d ( x)Î F[ x]is called the greatest common divisor of f ( x),g( x)ÎF[ x] if

(i) d ( x)½ f ( x) and d ( x)½g( x) , and (ii) if h( x)½ f ( x) and h( x)½g( x) for some h( x)ÎF[ x] ,then h( x)½d ( x).The greatest common divisor of f ( x) and g( x) isdenoted by gcd ( f ( x) , g( x)). If gcd ( f ( x) , g( x)) = 1, thenthe polynomials f ( x) and g( x) are said to be relatively

prime.We can show the uniqueness of the greatest

common divisor as follows. Suppose that c( x) and d ( x) are both greatest common divisors of f ( x) and g( x). Then c( x)½d ( x) and d ( x)½c( x), sayd ( x)=a( x)c( x) and c( x)=b( x)d ( x), and so we haved ( x)=a( x)b( x)d ( x). Therefore a( x)b( x)=1, and so

Proposition 4.1.5 shows that a( x) and b( x) are bothof degree zero. Thus c( x) is a constant multiple of d ( x), and since both are monic, the constant must be1, which shows that c( x)=d ( x).

Theorem 1.4.6 For any nonzero polynomials f ( x), g( x)Î F[ x], the greatest common divisor

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gcd ( f ( x),g( x)) exists and can be expressed as a linear combination of f ( x) and g( x), in the form

gcd ( f ( x), g( x))=a( x) f ( x) + b( x)g( x)for some a( x), b( x)ÎF[ x].

Proof. It is easy to check thatI = {a( x) f ( x)+b( x)g( x): a( x), b( x)ÎF[ x]}

satisfies the conditions of Theorem 1.4.4, and so Iconsists of all polynomial multiples of a monic

polynomial in I of minimal degree, say d ( x). Since f ( x) , g( x)ÎI, we have d ( x)½ f ( x) and d ( x)½g( x). Onthe other hand, since d ( x) is a linear combination of f ( x) and g( x), it follows that if h( x)½ f ( x) and h( x)½g( x), then h( x)½d ( x). Thus d ( x) = gcd ( f ( x),

g( x))ÎI

Example 1.4.7 (Euclidean algorithm for polynomials) Let f ( x), g( x)ÎF[ x] be nonzero polynomials. We can use the division algorithm to

write f ( x)=q( x)g( x)+r ( x), with deg(r ( x)) < deg(g( x))or r ( x)=0. If r ( x) = 0, then g( x) is a divisor of f ( x),and so gcd ( f ( x), g( x)) = cg(x), for some cÎF. If r ( x)¹ 0, then it is easy to check that gcd ( f ( x) ,g( x))=gcd (g( x), r ( x)). This step reduces the degreesof the polynomials involved, and so repeating the

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procedure leads to the greatest common divisor of the two polynomials in a finite number of steps.

The Euclidean algorithm for polynomials issimilar to the Euclidean algorithm for finding thegreatest common divisor of nonzero integers. The

polynomials a( x) and b( x) for whichgcd ( f ( x) ,g( x))=a( x) f ( x)+b( x)g( x)

can be found just as for integers.

Example 1.4.8 To find gcd (2 x4+ x3-6 x2+7 x –2;

2 x3 –7 x2+8 x-4) over Q, divide the polynomial of

higher degree by the one of lower degree, to get thequotient x+4 and remainder 14 x2-21 x+14. The

answer will be unchanged by dividing through by anonzero constant, so we can use the polynomial 2 x2-

3 x+2. Like for integers, we now have gcd (2 x4

+x3-6 x2+7 x-

2, 2 x3 –7 x2+8 x -4)=gcd (2 x3 –7 x2

+8 x-4, 2 x2-3 x+2).Dividing as before gives the quotient x-2, with

remainder zero. This shows that the greatestcommon divisor that we are looking for is x

2 –

(3/2) x+1 (we divided through by 2 to obtain a monic polynomial).

Proposition 1.4.9 Let p( x), f ( x) , g( x)Î F[ x]. If gcd ( p( x), f ( x))=1 and p( x)½ f ( x)g( x) , then p( x)½g( x).

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Proof. If gcd ( p( x), f ( x))=1, then 1 = a( x) p( x) +b( x) f ( x)

for some a( x), b( x)ÎF[ x]. Thusg( x) = a( x)g( x) p( x) + b( x) f ( x)g( x),

which shows that if p( x)½ f ( x)g( x), then p( x)½g( x).

Definition 1.4.10 A nonconstant polynomial is

said to be irreducible over the field F if it cannot be

factored in F[ x] into a product of polynomials of lower degree. It is said to be reducible over F if sucha factorization exists.

All polynomials of degree 1 are irreducible. Onthe other hand, any polynomial of greater degree

that has a root in F is reducible over F, since by theremainder theorem it can be factored into polynomials of lower degree. To check that a polynomial is irreducible over a field F, in general itis not sufficient to merely check that it has no rootsin F. For example, x4+4 x2+4=( x2+2)2 is reducibleover Q, but it certainly has no rational roots.However, a polynomial of degree 2 or 3 can befactored into a product of polynomials of lower degree if and only if one of the factors is linear,which then gives a root. This remark proves the

next proposition.

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Proposition 1.4.11 A polynomial of degree 2 or 3 is irreducible over the field F if and only if it hasno roots in F. The field F is crucial in determiningirreducibility. The polynomial x2+1 is irreducibleover R, since it has no real roots, but considered as a

polynomial over C, it factors as x2+1=( x+i)( x-i).

Over the field Z2 , the polynomial x2+x+1 is

irreducible, since it has no roots in Z2. But on the

other hand, it is reducible over the field Z3, since x

2+x+1=( x+2)2 when the coefficients are viewed as

representing congruence classes in Z3.

Lemma 1.4.12 The nonconstant polynomial

p( x)ÎF[ x] is irreducible over F if and only if for all f ( x) , g( x)ÎF[ x] , p( x)½( f ( x)g( x)) implies p( x)½ f ( x) or p( x)½g( x)

Proof. First assume that p( x)½ f ( x)g( x). If p( x) is

irreducible and p( x) ∤ f ( x) , then gcd ( p( x), f ( x)) = 1,and so p( x)½g( x) by Proposition 1.4.9.

Conversely, if the given condition holds, then p( x) ¹ f ( x)g( x) for polynomials of lower degree,since p( x)½ f ( x) and p( x)½g( x).

Because of the similarity between Z and F[ x], itis evident that irreducible polynomials should play a

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role analogous to that of prime numbers, and one of the results we should look for is a uniquefactorization theorem. The corresponding proof from Z can be carried over to polynomials by usingirreducible polynomials in place of prime numbersand by using the degree of a polynomial in place of the absolute value of a number. For this reason wehave chosen to omit the proof of the next theorem,even though it is extremely important.

Theorem 1.4.13 (Unique Factorization) Anynonconstant polynomial with coefficients in the field F can be expressed as an element of F times a

product of monic polynomials, each of which is

irreducible over the field F. This expression isunique except for the order in which the factorsoccur.

Example 1.4.14 (Irreducible polynomials over Rand C)

According to the fundamental theorem of algebra,any polynomial over C of positive degree has a rootin C. This implies that the irreducible polynomialsin C[ x] are precisely the linear polynomials. As a consequence of the fundamental theorem of

algebra it can be shown that any polynomial over Rcan be factored into a product of linear and quadratic

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polynomials. Consider the case of a quadratic.Over R, a polynomial ax

2+bx+c with a¹0 has roots

a

acbb x

2

42 -±-=

and these are real numbers if and only if b2-4ac³0.

Since any factors of ax2+bx+c must be linear and

hence correspond to roots, we can see that the polynomial is reducible over R if and only if b2

-

4ac³0. For example, x2

+ 1 is irreducible over R. Insummary, irreducible polynomials in R[ x] must haveone of the forms ax+b, with a¹0 or ax

2+bx+c, with

b2-4ac< 0.Polynomials cannot be factored as easily over the

field of rational numbers as over the field of realnumbers, so the theory of irreducible polynomialsover the field of rational numbers is much richer than the corresponding theory over the real numbers.For example, x2-2 and x4

+x3+x

2+x+1 are irreducible

over the field of rational numbers, the first since 2 is

irrational and the second as a consequence of acriterion we will develop in the next section.

In studying roots and factors of polynomials, it isoften of interest to know whether there are anyrepeated roots or factors. The derivative p'( x) of the

polynomial p( x) can be used to check for repeated roots and factors. It is possible to formally define

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the derivative of a polynomial over any field as incalculus. For the moment we will restrict ourselvesto the case of polynomials with real coefficients, sothat we can feel free to use any formulas we mightneed from calculus.

Definition 1.4.15 Let f ( x)ÎF[ x]. An elementcÎF is said to be a root of multiplicity n > 1 of f ( x) if

( x -c)n½ f ( x) but ( x -c)n+1∤ f ( x).

Proposition 1.4.16 A nonconstant polynomial f ( x) over the field R of real numbers has no repeated factors if and only if gcd ( f ( x) , f' ( x)) = 1.

Proof. We will prove an equivalent statement: anonconstant polynomial f ( x) over R has a repeated factor if and only if gcd ( f ( x) , f ' ( x))¹ 1.

Suppose that gcd ( f ( x) , f ' ( x)) = d ( x) ¹ 1 and that p( x) is an irreducible factor of d ( x). Then

f ( x)=a( x) p( x) and f ' ( x)=b( x) p( x) for some a( x),b( x)ÎF[ x]. Using the product rule to differentiategives

f ' ( x) = a' ( x)p( x) + a( x) p' ( x) = b( x) p( x).This shows that p( x)½a( x) p' ( x), since a( x) p' ( x) =

b( x) p( x) - a' ( x) p( x) , and thus p( x)½a( x) because p( x)

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is irreducible and p( x) ∤ p' ( x). Therefore

f ( x)=c( x) p( x)2 for some c( x)ÎF[ x] , and so f ( x) has arepeated factor.

Conversely, if f ( x) has a repeated factor, say f ( x)=g( x)n

q( x) , with n > 1, then f' ( x) = n g( x)n-1

g' ( x)q( x) + g( x)n q' ( x)

and g( x) is a common divisor of f ( x) and f' ( x).

5 Polynomials with Integer Coefficients In this section we will give several criteria for determining when polynomials with integer coefficients have rational roots or are irreducibleover the field of rational numbers. We will use thenotation Z[ x] for the set of all polynomials with

integer coefficients.

Proposition 1.5.1 Let f ( x)=an x

n+an-1 xn-1 +...+a1 x+a0

be a polynomial with integer coefficients. If r/s a

rational root of f ( x) , with (r,s)=1, then r ½a0 and s½an.

Proof. If f (r/s)=0, then multiplying f (r/s) by sn

gives the equationan r

n + an-1 r

n-1s +...+a1rs

n-1 +a0sn=0

It follows that r ½a0s

n

and s½anr

n

, so r ½a0 and s½ansince (r , s) =1.

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Example 1.5.2 Suppose that we wish to find allintegral roots of f ( x) = x

3 – 3 x2

+2 x - 6.Using Proposition 1.5.1, all rational roots of f ( x) can

be found by testing only a finite number of values.By considering the signs we can see that f ( x) cannothave any negative roots, so we only need to check the positive factors of 6. Substituting, we obtain f (1)= -6, f (2) = -6, and f (3) = 0. Thus 3 is a root of f ( x) ,

and so we can use the division algorithm to showthat

x3 – 3 x2

+ 2 x - 6 = ( x2+ 2)( x - 3)

It is now clear that 6 is not a root, and we aredone.

Example 1.5.3

Let f ( x)Î Z[ x]. If c is an integral root of f ( x) ,then f ( x)=q( x)( x-c) for some polynomial q( x). Ananalysis of the proof of the remainder theoremshows that q( x)ÎZ[ x], since x-c is monic. For anyinteger n, we must have f (n)=q(n)(n -c), and since f (n), q(n)ÎZ, this shows that (c - n) ½ f (n). Thisobservation can be combined with Proposition 1.5.1to find the integer (and thus rational) roots of monicequations such as x

3+15 x2

-3 x - 6 = 0. By

Proposition 1.5.1, the possible rational roots are ±1,±2, ±3, and ±6. Letting f ( x)=x

3+15 x2

-3 x-6 , we find

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that f (1) = 7, so for any root c, we have (c - 1)½7.This eliminates all of the possible values except c =2 and c = -6. We find that f (2) = 56, so 2 is not a

root. This shows, in addition, that (c - 2)½56 for anyroot c, but -6 still passes this test. Finally, f (-6) =336, and so this eliminates -6, and f ( x) has norational roots. We have also shown, by Proposition1.5.1, that the polynomial f ( x)=x

3+15 x2

-3 x –6 isirreducible in Q[ x].

Definition 1.5.4 A polynomial with integer coefficients is called primitive if the greatestcommon divisor of all of its coefficients is 1.

Given any polynomial with integer coefficients,we can obtain a primitive polynomial by factoringout the greatest common divisor of its coefficients(called the content of the polynomial). For example,the polynomial 12 x2

-18 x+30 has content 6, and sowe can write 12 x2

-18 x + 30 = 6(2 x2 -3 x+5), where2 x2 - 3 x + 5 is a primitive polynomial.

Our immediate goal is to prove Gauss's lemma,which states that the product of two primitive

polynomials is again primitive. In the proof, giventwo primitive polynomials we will show that no

prime number can be a divisor of all of thecoefficients of the product. To do this we will use a

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lemma illustrated by the following example. If g( x)= x2

-2 x + 6 and h( x)= x3 –5 x2

+3 x+12, let f ( x)=g( x)h( x)= x5

-7 x4 +19 x3-24 x2

-6 x+72. In the product f ( x) , the coefficients 72, -6, and -24 are eachdivisible by 2. The coefficient of x3 is 19, and this isthe coefficient of least index that is not divisible by2. In g( x) , the coefficient of least index that is notdivisible by 2 is the coefficient 1 of x2

. In h( x), thecoefficient of least index that is not divisible by 2 is

the coefficient 3 of x. The following lemma showsthat it is no coincidence that 19 is the coefficient ing( x)h( x ) of x3

= x2 x.

Lemma 1.5.5 Let p be a prime number, and let f ( x)=g( x)h( x) , where f ( x)=am x

m+ am-1 xm-

1

+...+a1 x+a0, g( x)= bn x

n

+bn-1 x

n-1

+...+b1 x+b0 , and h( x)= ck xk +ck -1 x

k -1+...+c1 x+c0. If bs and ct are thecoefficients of g( x) and h( x) of least index notdivisible by p, then as+t is the coefficient of f ( x) of least index not divisible by p.

Proof. For the coefficient as+t of f ( x) , we have as+t =

b0cs+t +b1cs+t 1+...+bs-1ct+1+bsct +bs+1ct-1+...+bs+t c0 . By assumption,each of the coefficients b0 ,b1 ,...,bs-1 and ct-1 , ct-

2 ,...,c0 is divisible by p. Thus, with the exception of bsct , each term in the above sum is divisible by p .

This implies that as+t is not divisible by p.

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In any coefficient of f ( x) of lower degree, each

term in the sum ak = o

k

i k iib c -=å is divisible by p, and

thus as+t is the coefficient of least degree not

divisible by p.

Theorem 1.5.6 (Gauss's Lemma) The product of two primitive polynomials is itself primitive.

Proof. Let p be any prime number, and let f ( x)=g( x)h( x) be a product of primitive polynomials.Since g( x) and h( x) are primitive, each one has acoefficient not divisible by p, and then by Lemma1.5.5 it follows that f ( x) has at least one coefficientnot divisible by p. Since this is true for every prime,we conclude that f ( x) is primitive.

Theorem 1.5.7 A polynomial with integer coefficients that can be factored into polynomialswith rational coefficients can also be factored into

polynomials of the same degree with integer

coefficients. Proof. Let f ( x)ÎZ[ x], and assume that f ( x)=g( x)h( x) in Q[ x]. By factoring out the appropriateleast common multiples of denominators and greatest common divisors of numerators, we canassume that f ( x)=(m/n)g*( x)h*( x) , where (m,n)= 1and g*( x) , h*( x) are primitive, with the same degreesas g( x) and h( x), respectively. If d i is any coefficient

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of g*( x)h*( x) , then n½md i since f ( x) has integer coefficients, so n½d i since (n,m)=1. By Gauss'slemma, g*( x)h*( x) is primitive, so we have n = 1,

and thus f ( x) has a factorization f ( x)=(mg*( x))(h*( x))into a product of polynomials in Z[ x]. The generalresult, for any number of factors, can be proved byusing an induction argument.

Theorem 1.5.8 (Eisenstein's Irreducibility

Criterion)Let

f ( x)=an xn + an-1 x

n-1 +... + a0

be a polynomial with integer coefficients. If thereexists a prime number p such that

an-1ºan-2º...ºa0º0(mod p) but an≢0(mod p) and

a0≢0(mod p2 ),then f ( x) is irreducible over the field of rationalnumbers.

Proof. Suppose that f ( x) can be factored as

f ( x)=g( x)h( x), where g( x)=bm xm +... + b0 and h( x) =ck x

k +... + c0. By Theorem 1.5.7 we can assume that both factors have integer coefficients. Furthermore,we can assume that either b0 or c0 is not divisible by p, since b0c0=a0 is not divisible by p

2. Let us assume

that p∤b0. If ct is the coefficient of h( x) of leastdegree that is not divisible by p, then by Lemma

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1.5.5 it follows that at is the coefficient of f ( x) of least degree that is not divisible by p. By assumptionai is divisible by p for i<n, so t =n, showing that h( x)and f ( x) have the same degree. Thus f ( x) isirreducible because it cannot be factored into a

product of polynomials of lower degree. In Theorem1.5.8, the conditionan-1ºan-2º...ºa0º0(mod p) can be summed up assaying that p is a divisor of the greatest common

divisor of these coefficients. The theorem cannotalways be applied. For example, if f ( x)= x

3 – 5 x2-

3 x + 6, then gcd (5,3,6) = 1 and no prime can befound for which the necessary conditions aresatisfied. Yet f ( x) is irreducible, since Propositions

1.5.1 can be used to show that f ( x) has no rationalroots. To show that p( x) is irreducible, it is sufficient toshow that p( x +c) is irreducible for some integer c,since if p( x)=f ( x)g( x) , then p( x+c)= f ( x+c)g( x+c).For example, Eisenstein's criterion cannot be applied to x2+1, but substituting x+1 for x gives another

proof that x2+1 is irreducible over the field of rational numbers.

Often the easiest way to make the substitution of

x+c for x is to use Taylor's formula. Recall that for a polynomial p( x) of degree n,

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p( x)= !

)()(

n

c pn

( x-c)n+)!1(

)()1(

-

-

n

c pn

( x-c)n-1+…+ p¢(c)( x-c)

+ p(c) where pk ( x) denotes the k

th derivative of p( x).

Thus p( x+c)=!

)()(

nc p n x

n+)!1(

)()1(

--

nc p n x

n-1+…+ p¢(c) x

+ p(c) ,and it may be possible to apply Eisenstein's criterionto the new set of coefficients.

Corollary 1.5.9 If p is prime, then the polynomial

f ( x) =x p-1

+ x p-2 +...+ x + 1

is irreducible over the field of rational numbers.

Proof. Note that f ( x) =1

1

-

-

x

x p

and consider

f ( x+1)= x

x p 1)1( -+

= x p-1

+ ÷ ø

öçè

æ 1

p x

p-2+ ÷ ø

öçè

æ 2

p x

p-3+...+ p

For 1 < i < p - 1, the prime p is a factor of the

binomial coefficient ÷ ø

ö

çè

æ i

p

, and so Eisenstein's

criterion can now be applied to f ( x+1), proving thatf ( x) is irreducible over Q.

If p is prime, Corollary 1.5.9 shows that x

p - 1 = ( x - 1)( x p-1 + x

p-2 +...+ x + 1)

gives the factorization over Q of x p - 1 into twoirreducible factors. This is not the case when the

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degree is composite. For example, x4-1= ( x-

1)( x+1)( x2+1)

and x15-1=( x-1)( x2

+x+1)( x4+x

3+x

2+x+1)( x8

-x7+x

5-

2 x4+x

3+x+1)

Definition 1.5.10 The roots in C of the polynomial xn -1 are called the complex n

th roots of unity. A complex n

th root of unity is said to be

primitive if it is a root of the polynomial xn -1 , but isnot a root of xm -1 for any positive integer m < n.

For example, the 4th roots of unity are ±1, ±i. Of these, i and -i are primitive 4th roots of unity, and they are roots of the factor x2

+1 of x4-1. Note that if

p is a prime number, then every p

th

root of unityexcept 1 is primitive.

6. Existence of Roots

The polynomial x2+1 has no roots in the field R

of real numbers. However, we can obtain a root by

constructing an element i for which i2 = -1 and adding it (in some way) to the field R. This leads tothe field C, which contains elements of the form a +bi, for a,bÎR. The only problem is to find a way of constructing the root i.

In this section we will show that for any polynomial, over any field, it is possible to construct

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a larger field in which the polynomial has a root. Todo this we will use congruence classes of

polynomials. The construction is similar in manyways to the construction of the field Z p as a set of congruence classes of Z. By iterating the process, itis possible to find a field that contains all of theroots of the polynomial, so that over this field the

polynomial factors into a product of linear polynomials.

Definition 6.1 Let E and F be fields. If F is asubset of E and has the operations of addition and multiplication induced by E, then F is called asubfield of E, and E is called an extension field of F.

Definition 6.2 Let F be a field, and let p( x) be afixed polynomial over F. If a( x), b( x)Î F[ x], then wesay that a( x) and b( x) are congruent modulo p( x),written a( x) º b( x) (mod p( x)) , if p( x)½(a( x) - b( x)).

The set {b( x)Î F[ x]: a( x) º b( x) (mod p( x))} iscalled the congruence class of a( x), and will bedenoted by [a( x)].

The set of all congruence classes modulo p( x) will be denoted by F[ x]/á p( x)ñ.

We first note that congruence of polynomials

defines an equivalence relation. Then since a( x)ºb( x)(mod p( x)) if and only if a( x)-b( x) º q( x) p( x)

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for some q( x)ÎF[ x] , the polynomials in thecongruence class of a( x) modulo p( x) must be

precisely the polynomials of the form b( x)=a( x)+q( x) p( x), for some q( x). We gave a similar description for the congruence classes of Zn. Whenworking with congruence classes modulo n, we haveoften chosen to work with the smallest nonnegativenumber in the class. Similarly, when working withcongruence classes of polynomials, the polynomial

of lowest degree in the congruence class is a naturalrepresentative. The next proposition guarantees thatthis representative is unique.

Proposition 6.3 Let F be a field, and let p( x) be

a nonzero polynomial in F[ x]. For any a( x)ÎF[ x],the congruence class [a( x)] modulo p( x) contains aunique representative r ( x) with deg(r ( x))<deg( p( x))or r ( x)=0. Proof. Given a( x) ÎF[ x] , we can use the divisionalgorithm to write

a( x) = q( x) p( x) + r ( x) ,with deg(r ( x))<deg( p( x)) or r ( x)=0. Solving for r ( x)in the above equation shows it to be in thecongruence class [a( x)]. The polynomial r ( x) is theonly representative with this property, since if b( x) ºa( x) (mod p( x)) and deg(b( x))<deg( p( x)), then b( x) ºr ( x) (mod p( x)) and so p( x)½(b( x) - r ( x)). This is a

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contradiction unless b( x)=r ( x), since either deg(b( x)-r ( x))<deg( p( x)) or b( x)-r ( x)=0.

Proposition 6.4 Let F be a field, and let p( x) bea nonzero polynomial in F[ x]. For any polynomialsa( x), b( x), c( x), and d ( x) in F[ x], the followingconditions hold:

(a)If a( x)ºc( x) (mod p( x)) and b( x) ºd ( x) (mod p( x)), then a( x)+b( x) º c( x)+d ( x) (mod p( x))

and a( x)b( x) ºc( x)d ( x) (mod p( x))(b)If a( x)b( x) º a( x)c( x) (mod p( x)) and

gcd (a( x), p( x)) = 1, then b( x) º c( x) (mod p( x)) .

Proof. The proof is left for students.

Example 6.5 (R[ x]/ á x2+1ñ) Let F=R, the field of real numbers, and let p( x)= x2

+1.Then every congruence class inR[ x]/ á x2+1ñ can be represented by a linear

polynomial of the form a+bx, by Proposition 6.3.

Furthermore, Proposition 6.4 implies that if we add and multiply congruence classes by choosingrepresentatives, just as we did in Zn then additionand multiplication of congruence classes are well-defined. If we multiply the congruence classesrepresented by a+bx and c+dx, we have

ac + (bc + ad ) x + bdx2 .

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Dividing by x2+1 gives the remainder (ac - bd ) +

(bc + ad ) x , which is a representative of the productof the two congruence classes. An easier way tomake this computation is to note that x2+1º 0 (mod x

2+1) , and so x2º-1(mod x2+1) which means that we

can replace x2 by -1 in the product ac + (bc + ad ) x

+ bdx2 .

This multiplication is the same as themultiplication of complex numbers, and gives

another way to define C. Note that the congruenceclass [ x] has the property that its square is thecongruence class [-1], and so if we identify the set of real numbers with the set of congruence classes of the form [a], where aÎR, then the class [ x]

would be identified with i. We can formalize thisidentification after we define the concept of anisomorphism of fields.

Proposition 6.6 Let F be a field, and let p( x) be

a nonzero polynomial in F[ x

]. For anya( x

)Î

F[ x

],the congruence class [a( x)] has a multiplicativeinverse in F[ x]/ á p( x)ñ if and only if gcd (a( x), p( x)) =1.

Proof. To find a multiplicative inverse for [a( x)]we must find a congruence class [b( x)] with

[a( x)][b( x)] = [1]. Since a( x)b( x)º 1(mod p( x)) if and only if there exists t ( x)ÎF[ x] with a( x)b( x)=1+

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t ( x) p( x) , this shows that [a( x)] has a multiplicativeinverse if and only if 1 can be written as a linear combination of a( x) and p( x), which occurs if and only if gcd (a( x), p( x)) = 1.The inverse [b( x)] = [a( x)]-1 can be found by usingthe Euclidean algorithm.

Theorem 6.7 Let F be a field, and let p( x) be anonconstant polynomial over F. Then F[ x]/á p( x)ñ isa field if and only if p( x) is irreducible over F.

Proof. Proposition 6.4 shows that addition and multiplication of congruence classes are well-defined. The associative, commutative, and distributive laws follow easily from the

corresponding laws for addition and multiplicationof polynomials. For example,[a( x)][b( x)] = [a( x)b( x)] = [b( x)a( x)] = [b( x)][a( x)]

for all a( x), b( x)ÎF[ x]. The additive identity is [0]and the multiplicative identity is [1], while theadditive inverse of [a( x)] is [-a( x)]. All that remainsto show that F[ x]/á p( x)ñ is a field is to show thateach nonzero congruence class has a multiplicativeinverse. Since by Proposition 6.3 we can work withrepresentatives of lower degree than deg( p( x)), byProposition 6.5 each nonzero congruence class

[a( x)] has a multipliative inverse if and only if gcd (a( x), p( x))=1 for all nonzero polynomials a( x)

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with deg(a( x))<deg( p( x)). This occurs if and only if p( x) is irreducible, completing the proof. We notethat whenever p( x) is irreducible, the congruenceclass [a( x)] is invertible if a( x)¹0 and deg(a( x))<deg( p( x)). Conversely, if [a( x)] isinvertible, then[a( x)]=[r ( x)] where r ( x)¹0 and deg(r ( x))<deg( p( x)). Since F[ x] is an abelian group under addition and ( p( x)) is a subgroup, addition of congruence classes

is just the operation defined in the correspondingfactor group. Thus results from group theory alsoimply that F[ x]/á p( x)ñ is an abelian group under addition. It is easy to see that in F[ x]/á p( x)ñ amultiplication operation can be introduced and as a

result F[ x]/á p( x)ñ becomes a commutative ring. Definition 6.8 Let F1 and F2 be fields. A

function j :F1®F2 is called an isomorphism of fieldsif it is one-to-one and onto, j (a+b)=j (a)+j (b) , and j (ab)=j (a)j (b) for all a,bÎF1.

Example 6.9 (Construction of the complexnumbers)

We can now give the full story of Example 6.5. Itis known that the set of all complex numbers C isdefined to be the set of all expressions of the forma+bi, where a,bÎR and i2=- 1. Since x2+1 is

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irreducible over R, we know that R[ x]/á x2+1ñ is a

field. Its elements are in one-to-one correspondencewith polynomials of the form a+bx. Furthermore,

the mapping j :R[ x]/á x2

+1ñ ® C defined byj ([a+bx])=a+bi can be shown to be an isomorphism.Since x

2º-1 (mod x2+1), the congruence class [ x] of the

polynomial x satisfies the condition [ x]2=-1. Thus

the construction of R[ x]/á x2+1ñ using congruence

classes allows us to provide a concrete model for theconstruction we gave previously, in which wemerely conjured up an element i for which i

2=-1.

Theorem 6.10 (Kronecker) Let F be a field, and

let f ( x) be any nonconstant polynomial in F[ x]. Thenthere exists an extension field E of F and an elementuÎE such that f (u) = 0.

Proof. The polynomial f ( x) can be written as a product of irreducible polynomials, and so we let p( x)

be one of the irreducible factors of f ( x). It issufficient to find an extension field E containing anelement u such that p(u) = 0. By Proposition 6.7,F[ x]/á p( x)ñ is a field, which we will denote by E. Thefield F is easily seen to be isomorphic to the subfield of E consisting of all congruence classes of the form[a], where aÎF. We make this identification of F

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with the corresponding subfield of E so that we canconsider E to be an extension of F. Let u be thecongruence class [ x]. If p( x)=an x

n+an-1 xn-1+...+a0,

where aiÎF, then we must compute p(u).We obtain

p(u)=an[ x]n+an-1[ x]n1+...+a0=[an xn+an-1 x

n-

1+...+a0]=[0]since p( x)º0(mod p( x)). Thus p(u)=0 and the proof iscomplete.

Corollary 6.11 Let F be a field, and let f ( x) beany nonconstant polynomial in F[ x]. Then thereexists an extension field E over which f ( x) can befactored into a product of linear factors.

Proof. Factor out all linear factors of f ( x) and let

f 1( x) be the remaining factor. We can find anextension E1 in which f 1( x) has a root, say u1. Thenwe can write f 1( x)=( x – u1) f 2( x), and by considering f 2( x) as an element of E1[ x], we can continue thesame procedure for f 2( x). We will finally arrive at anextension E that contains all of the roots of f ( x) , and over this extension, f ( x) can be factored into a

product of linear factors.

Example 6.12 Consider the polynomial x4 – x2-2 ,

with coefficients in F = Q. It factors as ( x2-2)( x2+1),

and as our first step we let E1=Q[ x]/á x2-2ñ , which isisomorphic to Q( 2). Although E1 contains the

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roots ± 2 of the factor x2- 2, it does not contain theroots ±i of the factor x2+1, and so we must obtain afurther extension E2 = E1/á x2+1ñ . In Chapter on

algebraic numbers we will see that E2 is isomorphicto the smallest subfield of C that contains 2 and i,which is denoted by Q( 2,i).

Example 6.13 Let F = Z2 and let p( x)=x2+x+1.

Then p( x) is irreducible over Z 2 since it has no roots

in Z2, and so Z 2[ x] /á x2+x+1ñ is a field. Thecongruence classes modulo x

2+x+1 can be

represented by [0], [1], [ x] and [1+x], since these arethe only polynomials of degree less than 2 over Z2.Addition and multiplication are given in Tables 6.14

and 6.15. To simplify these tables, all brackets have been omitted in listing the congruence classes.

Table 6.14 Addition in Z2[ x] /á x2 + x + 1ñ

0 1 1

0 0 1 1

1 1 0 1

1 0 1

1 1 1 0

x x

x x

x x

x x x

x x x

+ ++

++

+ +

Table 6.15 Multiplication in Z2[ x] /á x2+x+1ñ

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x x x

x x x

x x

x x

1101

110

1101

00000

110

+++

+

+Ä

If q(x) is irreducible over Z p, then Z p[ x]/áq( x)ñhas p

n elements if deg(q( x))=n, since there areexactly p

n-1 polynomials over Z p of degree less than

n (including 0 gives pn elements). It is possible toshow that there exist polynomials of degree n

irreducible over Z p for each integer n > 0. Thisguarantees the existence of a finite field having p

n

elements, for each prime number p and each positive

integer n.

2. COMMUTATIVE RINGS; INTEGRALDOMAINS

1. Definitions and examples

We will now undertake a systematic study of systems in which there are two operations thatgeneralize the familiar operations of addition and

multiplication. The examples you should have inmind are these: the set of integers Z; the set Zn , of

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integers modulo n; any field F (in particular the setQ of rational numbers and the set R of realnumbers); the set F[ x] of all polynomials withcoefficients in a field F. The axioms we will use arethe same as those for a field, with two crucialexceptions. We have dropped the requirement thateach nonzero element has a multiplicative inverse, inorder to include integers and polynomials in theclass of objects we want to study. We have also

dropped the requirement that a multiplicativeidentity element exists, which allows us, for example, to include the set of all even integers in our definition. Definition 1.1 Let R be a set on which two

binary operations are defined, called addition and multiplication, and denoted by + and × . Then R iscalled a commutative ring with respect to theseoperations if the following properties hold:

(i) Closure: If a, b Î R, then the sum a + b

and the product a×b are uniquely defined and belongto R.

(ii) Associative laws: For all a, b, c Î R,

a+ (b+ c)= (a+ b) +c and a× (b× c) = (a × b) ×c.

(iii) Commutative laws: For all a, b Î R,

a + b = b + a and a× b= b× a.

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(iv) Distributive laws: For all a, b, c Î R,a×(b + c) = a× b + a× c and (a + b) × c = a×c+

b×c .

(v) Additive identity: The set R contains anadditive identity element, denoted by 0, such that for all a Î R,

a + 0 = a and 0+a=a. (vi) Additive inverses: For each a Î R, theequations

a+x = 0 and x+a =0have a solution x in R, called the additive inverse of a, and denoted by -a. The commutative ring R is called a commutativering with identity if it contains an element 1,

assumed to be different from 0 such that for all a Î R,

a×1=a and 1×a=a. In this case , 1 is called a multiplicative identityelement, or, more generally, simply an identity

element. Our first observation is that any commutativering R determines an abelian group by justconsidering the set R together with the singleoperation of addition. We call this the underlyingadditive group of R. Although we require thatmultiplication in R is commutative, the set of

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nonzero elements certainly need not define anabelian group under multiplication.

A set with two binary operations that satisfyconditions (i)-(vi), with the exception of thecommutative law for multiplication, is called aring. Although we will not discuss them here,there are many interesting examples of noncommutative rings. From linear algebra, youshould already be familiar with one such example,

the set of all 2´2 matrices over R. The standard rules for matrix arithmetic provide all of theaxioms for a commutative ring, with the exceptionof the commutative law for multiplication.Although this is certainly an important example

worthy of study, we have chosen to work only withcommutative rings, with emphasis on integraldomains, fields, and polynomial rings over them. Before giving some further examples of

commutative rings, it is helpful to have someadditional information about them. Our observationthat any commutative ring is an abelian group under addition implies that the cancellation law holds for addition. This proves part (a) of the next statement. Let R be a commutative ring, with elementsa,b,cÎ R.

(a) If a + c = b + c, then a = b. (b) If a + b = 0, then b = -a.

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(c) If a + b = a for some a Î R, then b = 0.(d) If a×b = a for all a Î R, then R has anidentity 1 and b = 1.

For any commutative ring the following aretrue.(Note that (a) and (c) involve connections between addition and multiplication. Their proofsmake use of the distributive law, since it providesthe only link between the two operations.)

Let R be a commutative ring.

(a) For all a Î R, a× 0 = 0. (b) For all a Î R, -(-a) = a.

(c) For all a,b Î R, (-a)×(-b) = a×b. We will follow the usual convention of

performing multiplications before additions unless

parentheses intervene. Example 1.2 Zn

In Section 1.4 we listed the properties of addition and multiplication of congruence classes,which show that the set Zn of integers modulo n is a

commutative ring with identity. From our study of groups we know that Zn is a factor group of Z (under addition), and so it is an abelian group under addition. To verify that the necessary propertieshold for multiplication, it is necessary to use thecorresponding properties for Z. We checked the

distributive law in Section 1.4. It is worthcommenting on the proof of the associative law, to

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point out the crucial parts of the proof. For example,we can check that the associative law holds for all[a], [b], [c]Î Zn :

[a]([b][c]) = [a][bc] = [a(bc)] and ([a][b])[c] =[ab][c] = [(ab)c] ,

and so these two expressions are equal because theassociative law holds for multiplication in Z. If effect, as soon as we have established thatmultiplication is well-defined (Proposition 1.4.2), it

is easy to show that the necessary properties areinherited by the set of congruence classes. The rings Zn, form a class of commutativerings that is a good source of counterexamples. For instance, itprovides an easy example showing that

the cancella-tion law may fail for multiplication. Inthe commu-tative ring Z6 we have [2][3]=[4][3], but[2]¹ [4].

Definition 1.3 Let S be a commutative ring. Anonempty subset R of S is called a subring of S if itis a commutative ring under the addition and multiplication of S.

Let F and E be fields. If F is a subring of E,according to the above definition, then we usuallysay (more precisely) that F is a subfield of E. Of

course, there may be other subrings of fields that arenot necessarily subfields. Any subring is a subgroup

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of the underlying additive group of the larger ring,so the two commutative rings must have the samezero element. If F is a subfield of E, then more must

be true: The set F* of nonzero elements of F is asubgroup of E*, and so they must share the samemultiplicative identity element. Proposition 1.4 Let S be a commutative ring,

and let R be a nonempty subset of S. Then R is asubring of S if and only if

(i) R is closed under addition and multiplication;and

(ii) if a Î R, then -a Î R.Proof . If R is a subring, then the closure axioms

must certainly hold. Suppose that z is the additive

identity of R. Then z + z = z = z + 0, where 0 is theadditive identity of S, so z = 0, since the cancellationlaw for addition holds in S . Finally, if a Î R and b isthe additive inverse of a in R, then a + b = 0, so b =-a, and this shows that -a Î R.

Conversely, suppose that the given conditionshold. The first condition shows that condition (i) of Definition 2.1 is satisfied. Conditions (ii)-(iv) of Definition 2.1 are inherited from S . Finally, since R

is nonempty, it contains some element, say a Î R.Then -a Î R, so 0 = a + (-a) Î R since R is closed under addition. Thus conditions (v) and (vi) of Definition 2.1 are also satisfied.

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Example 1.5

Let S be the commutative ring Z6 and let R be thesubset {[0], [2], [4]}. Then R is closed under addition and multiplication and contains the additiveinverse of each element in R. Since [4][0] = [0],[4][2] = [2], and [4][4] = [4], the subring R also hasa multiplicative identity, namely [4]. This showsthat a subring may have a different multiplicativeidentity from that of the given commutative ring. If

e is the multiplicative identity element of a subringof S , then e

2 = e.

An element e such that e2 = e is said to be

idempotent. The idempotent elements of Z6 (other than [0]) are [1], [3], and [4]. This can be used to

show that the only subrings of S = Z6 that have anidentity element are S itself, R, and T = {[0], [3]}. Example 1.6 (Gaussian integers) Let Z[i] be the set of complex numbers of the

formm + ni, where m, n ÎZ. Since

(m + ni) + (r + si) = (m + r ) + (n + s)i and (m + ni) (r + si) = (mr - ns) + (nr + ms)i ,

for all m,n,r,sÎZ, the usual sum and product of numbers in Z[i] have the correct form to belong toZ[i]. This shows that Z[i] is closed under addition

and multiplication of complex numbers. Thenegative of any element in Z[i] again has the correct

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form, as does 1 = 1 + 0i, so Z[i] is a commutativering with identity. Example 1.7 Z[ 2]

Z [ 2]

={m + n

2:m,nÎ

Z} is closed under addition. The product of two elements is given by(m1+n1 2)(m2+n2 2)=(m1m2+2n1n2)+(m1n2+m2n1)

2and so the set is also closed under multiplication. Itis easy to show that Z[ 2] is a subring of Q[ 2].

Definition 1.8 Let R be a commutative ring withidentity element 1. An element aÎ R is said to beinvertible if there exists an element bÎ R such that ab = 1. The element a is also called a unit of R,

and its multiplicative inverse is usually denoted by

a-1

.Since 0b = 0 for all bÎ R, it is impossible for 0 to

be invertible. Furthermore, if aÎ R and ab=0 for some nonzero bÎ R, then a cannot be a unit sincemultiplying both sides of the equation by the inverse

of a (if it existed) would show that b=0. An element a such that ab = 0 for some b ¹ 0 iscalled a divisor of zero.

Example 1.9

Let R be the set of all functions from the set of real numbers into the set of real numbers, withordinary addition and multiplication of functions

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(not composition of functions). It is not hard toshow that R is a commutative ring with identity,since addition and multiplication are defined

pointwise, and the addition and multiplication of realnumbers satisfy all of the field axioms. It is easy tofind divisors of zero in this ring: Let f ( x)=0 for x <

0 and f ( x)=1 for x > 0, and let g( x)=0 for x>0 and g( x)=1 for x<0. Then f ( x)g( x)=0 for all x, whichshows that f ( x)g( x) is the zero function.

The multiplicative identity of R is the function f ( x)=1 (for all x).Then a function g( x) has amultiplicative inverse if and only if g( x)¹ 0 for all x.Thus, for example, g( x)=2+sin( x) has amultiplicative inverse, but h( x)=sin( x) does not.

When thinking of the units of a commutativering, here are some good examples to keep in mind.The only units of Z are 1 and - 1. We have showed that the set of units of Zn, consists of the congruenceclasses [a] for which (a, n)=1. We showed that Zn

isa group under multiplication of congruence classes.

We will use the notation R* for the set of units of any commutative ring R with identity.

Proposition 1.10 Let R be a commutative ringwith identity. Then the set R

* of units of R is anabelian group under the multiplication of R.

Proof. If a,bÎ R*, then a-1 and b-1 exist in R, and so abÎ R

* since

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(ab)(b-1a

-1) = 1. We certainly have 1Î R*, and a

-

1Î R* since (a-1)-1=a. Multiplication is assumed to be

associative and commutative.

In the context of commutative rings we can givethe following definition: A field is a commutativering with identity in which every nonzero element isinvertible.

We can say, loosely, that a field is a set on whichthe operations of addition, subtraction,

multiplication, and division can be defined. For example, we showed that Zn, is a field if and only if n is a prime number.

We have already observed that the cancellationlaw for addition follows from the existence of

additive inverses. A similar result holds for multiplication. If ab=ac and a is a unit, thenmultiplying by a

-1 givesa

-1(ab)=a-1(ac), and then by using the associative

law for multiplication, the fact that a-1

a=1, and thefact that 1 is a multiplicative identity element, wesee that b=c.

If the cancellation law for multiplication holds ina commutative ring R, then for any elements a,bÎ R,ab=0 implies that a=0 or b=0. Conversely, if thiscondition holds and ab=ac, then a(b-c)=0, so if a¹0

then b-c=0 and b=c. Thus the cancellation law for

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multiplication holds in R if and only if R has nononzero divisors of zero.

It is precisely this property that is crucial insolving polynomial equations. If ( x-2)( x-3)=0 and xis an integer, then we can conclude that either x-2=0or x-3=0, and so either x=2 or x=3. But if xrepresents an element of Z6 , then we might have x-[2]=[3] and x-[3] =[2], since this still gives ( x-[2])( x-[3]) =[3][2] =[0] in Z6. In addition to the obvious

roots [2] and [3], we can also substitute [5] and [0],respectively, so the polynomial x2

-[5] x+[6] has four distinct roots over Z6.

Definition 1.11 A commutative ring R withidentity is called an integral domain if for all

a,bÎ R, ab=0 implies a=0 or b=0.The ring of integers Z is the most fundamentalexample of an integral domain. The ring of all

polynomials with real coefficients is also an integraldomain, since the product of any two nonzero

polynomials is again nonzero. It can be seen that thelarger ring of all real valued functions is not anintegral domain.

The next theorem gives a condition that is veryuseful in studying integral domains. It showsimmediately, for example, that Z[i] and Z[ 2] are

integral domains. A converse to Theorem 1.12 will be given later, showing that all integral domains can

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essentially be viewed as being subrings of fields.Proving this converse involves constructing a field of fractions in much the same way that the field of rational numbers can be constructed from theintegers.

Theorem 1.12 Let F be a field with identity 1.Any subring of F that contains 1 is an integraldomain.

Proof. Suppose that R is a subring of the field F , with a,bÎ R. If ab=0 in R, then of course the sameequation holds in F . Either a=0 or a¹0, and in thelatter case a has a multiplicative inverse a

-1 in F ,even though the inverse may not be in R.

Multiplying both sides of the equation by a-1

givesb=0.

Example 1.13

We have already seen that Z6 is not an integraldomain. On the other hand, we have seen that if p isa prime number, then Z p is a field, so it is certainlyan integral domain. Which of the rings Zn , areintegral domains? If we use the condition thatabº0(mod n) implies that aº0(mod n) or bº0(mod n),or equivalently, the condition that n|ab implies n|a or n|b, then we can see that n must be prime. Why

should the notions of field and integral domain be

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the same for the rings Zn? The next theorem givesan answer, at least from one point of view.

Theorem 1.14 Any finite integral domain must be a field.

Proof. Let D be a finite integral domain, and let D* be the set of nonzero elements of D. If d Î D and d ¹0, then multiplication by d defines a function from D* into D*, since ad ¹ 0 if a ¹ 0. Let f : D*® D* bedefined by

f ( x)=xd, for all xÎ D*. Then f is a one-to-onefunction, since f ( x) = f ( y) implies xd=yd , and so x=y

since the cancellation law holds in an integraldomain. But then f must map D* onto D*, since anyone-to-one function from a finite set into itself must

be onto, and so 1= f (a) for some aÎ D*. That is, ad =1 for some aÎ D, and so d is invertible. Since wehave shown that each nonzero element of D isinvertible, it follows that D is a field. 2 Ring Homomorphisms

In studying groups in Chapter 3, we found thatgroup homomorphisms played an important role. Now in studying commutative rings we have twooperations to consider. As with groups, we will beinterested in functions which preserve the algebraic

properties that we are studying. We begin the

section with two examples, each of which involvesan isomorphism.

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Example 2.1

We introduce an element i such that i2 = -1, and

letR[i]= {a+ bi : a,b ÎR}.

To look for an alternate description of R[i], wemight try to find such an element i with i

2=-1 insome familiar setting. If we identify real numberswith scalar 2´2 matrices over R, then the matrix

úûùêë

é- 01

10 has the property that its square is equal to

the matrix corresponding to -1. This suggests thatwe should consider the set T of matrices of the form

úû

ùêë

é-=úû

ùêë

é-+úû

ùêë

éab

ba

ba 01

10

10

01

Verify that T is a commutative ring. To show theconnection with R[i], we define j: R[i]®T by

j(a+bi)= úû

ùêë

é

- ab

ba

To add complex numbers we just add thecorresponding real and imaginary parts, and sincematrix addition is componentwise, it is easy to showthat 0 preserves sums. To show that it preserves

products we give the following computations

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j ((a+bi)(c+di)) = j ((ac-bd )+(ad+bc)i)

= úû

ùêë

é-+-+-

bd acbcad

bcad bd ac

)(

j (a+bi) j (c+di) = úû

ùêë

é- ab

baúû

ùêë

é- cd

d c

= úû

ùêë

é-+-+-

bd acbcad

bcad bd ac

)(

Thus j((a + bi)(c + di))=j(a + bi) j (c + di), and since it is clear that j is one-to-one and onto, wecould compute sums and products of complexnumbers by working with the corresponding

matrices. This gives a concrete model of thecomplex numbers. Definition 2.2 Let R and S be commutativerings. A function j: R ® S is called a ringhomomorphism if

j(a+b)=j(a)+j(b) and j(ab)=j(a) j (b) for alla,bÎ R.

A ring homomorphism that is one-to-one and ontois called an isomorphism. If there is anisomorphism from R onto S, we say that R isisomorphic to S, and write

R@S .

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An isomorphism from the commutative ring R ontoitself is called an automorphism of R.

The condition that states that a ringhomomorphism must preserve addition is equivalentto the statement that a ring homomorphism must bea group homomorphism of the underlying additivegroup of the ring. This means that we have at our disposal all of the results that we have obtained for group homomorphisms.

A word of warning similar to that given for grouphomomorphisms is probably in order. If j: R®S is aring homomorphism, with a,bÎ R, then we need tonote that in the equation

j (a + b) = j (a) + j (b)

the sum a+b occurs in R, using the addition of thatring, whereas the sum j(a)+j(b) occurs in S , usingthe appropriate operation of S. A similar remark applies to the respective operations of multiplicationin the equation

j (ab) = j (a) j (b),where a,bÎ R and j(a)j (b)ÎS . We have chosen notto distinguish between the operations in the tworings, since there is generally not much chance for confusion.

Using the terminology of Definition 2.2, the

function defined in Example 2.1 is an isomorphism providing a different way of looking at the set of

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complex numbers. We begin with some basic resultson isomorphisms.

It follows from the next proposition that "isisomorphic to" is reflexive, symmetric, and transitive. Recall that a function is one-to-one and onto if and only if it has an inverse. Thus a ringisomorphism always has an inverse, but it is notevident that this inverse must preserve addition and multiplication.

Proposition 2.3 (a) The inverse of a ring isomorphism is a ring

isomorphism. (b) The composition of two ring isomorphisms

is a ring isomorphism.

Proof. (a) Let j: R®S be an isomorphism of commutative rings. We have shown in modernalgebra that j-1 is an isomorphism of the underlyingadditive groups. To show that j-1 is a ring homo-morphism, let s1 , s2ÎS . Since j is onto, there exist

r 1,r 2Î R such that j(r 1)=s1 and j (r 2)=s2. Then j

-

1(s1s2) must be the unique element r Î R for which

j(r )=s1s2. Since j preserves multiplication,j-

1(s1s2)=

j-1(j(r 1)j(r 2))=j-1(j (r 1r 2))=r 1r 2=j-1(s1)j-1(s2).(b) If j: R®S and q:S ®T are isomorphisms of commutative rings, then

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qj(ab) = q (j (ab)) = q(j (a) j (b)) = q (j(a))q(j(b)) = qj (a)qj (b) .

The remainder of the proof follows immediately

from the corresponding result for grouphomomorphisms.To show that commutative rings R and S are

isomorphic, we usually construct the isomorphism.To show that they are not isomorphic, it is necessaryto show that no isomorphism can possibly be

constructed. Sometimes this can be done by justconsidering the sets involved. For example, if n and m are different positive integers, then Zn is notisomorphic to Zm since no one-to-onecorrespondence can be defined between Zn and Zm.

One way to show that two commutative rings arenot isomorphic is to find an algebraic property thatis preserved by all isomorphisms and that is satisfied

by one commutative ring but not the other. We nowlook at several very elementary examples of such

properties. If j: R®S is an isomorphism, then fromour results on group isomorphisms we know that jmust map 0 to 0 and must preserve additive inverses.If R has an identity element 1, then for any sÎ S ,there exists r Î R such that j (r ) = s, since j is onto.Thus we have

j (1)s = j (1)j (r) = j (1r ) = j (r) =s ,

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showing that j (1) must be an identity element for S.

It is also easy to show that if r is a unit, then so is s,and so j preserves units. This implies that R is a

field if and only if S is a field.As an immediate consequence of the remarks inthe previous paragraph, we can see that Z is notisomorphic to Q, since Q is a field, while Z is not.Another interesting problem is to show that R and Care not isomorphic. Since both are fields, we cannot

use the previous argument. Let us suppose that wecould define an isomorphism j:C®R. Then wewould have

j (i)2 = j (i2 ) = j (-1) = -j (1) =-1and so R would have a square root of -1, which we

know to be impossible. Thus R and C cannot beisomorphic.

In many important cases we will be interested infunctions that preserve addition and multiplicationof commutative rings but are not necessarily one-to-

one and onto. Example 2.4 will be particularlyimportant in later work. Example 2.4

Consider the mapping j:Q[ x] ® R defined byj( f ( x)) = f ( 2), for all polynomials f ( x)ÎQ[ x].That is, the mapping j is defined on a polynomialwith rational coefficients by substituting x= 2. It iseasy to check that adding (or multiplying) two

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polynomials first and then substituting in x= 2 isthe same as substituting first in each polynomial and then adding (or multiplying). Thus j preserves

sums and products and is a ring homomorphism. Note that j is not one-to-one, since j ( x2-2)=0. Wewill show in Chapter 6 that the image of j isQ( 2)={a+b 2: a,bÎQ}, and so j is not onto. Example 2.5 (Evaluation mapping) The previous example can generalized, sincethere is nothing special about the particular fields wechose or the particular element we worked with. LetF and E be fields, with F a subfield of E. For anyelement u Î E we can define a function ju :F [ x]® E

by letting ju( f ( x)) = f (u), for each f ( x)Î F [ x]. Then

ju preserves sums and products sinceju ( f ( x) + g ( x)) = f (u) + g (u) = ju ( f ( x)) + ju (g

( x))

and ju ( f ( x) g( x))= f (u)∙ g (u)= ju ( f ( x))ju (g ( x)) ,

for all f ( x), g ( x) Î F [ x]. Thus ju is a ring

homomorphism. Since the polynomials in F [ x] areevaluated at u, the homomorphism ju is called anevaluation mapping.

Example 2.6 The previous example may become even clearer in a more general context. We

never used the fact that we were working with polynomials over a field, and so we can consider

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polynomials over any commutative ring.Furthermore, one of the essential facts we used wasthat the inclusion mapping from a subfield into thefield containing it is a homomorphism, and so wemight just as well consider the situation in thatgenerality. Let R and S be commutative rings, let q: R® S bea ring homomorphism, and let s be any element of S .Then there exists a unique ring homomorphism

`q: R[ x]®S such that `q(r )= q(r ) for all r Î R and `q( x)=s. We will first show the uniqueness. If j: R[ x]®S is any homomorphism with the required

properties, then for any polynomial f (x) = a0+a1 x+...+am x

m

in R[ x] we must havej (a0+a1 x+...+am x

m ) = j (a0)+j(a1 x)+...+j(am xm)=

j(a0) +j(a1)j( x)+...+j(am)(j( x))m

=j(a0)+j(a1)s+…+j(am)sm.

This shows that the only possible way to define j is

the following:`q(a0+a1 x+...+am xm)=q(a0)+q(a1)s+...+ q(am)sm

Given this definition, we must show that`q is ahomomorphism. Since addition of polynomials isdefined componentwise, and q preserves sums, it is

easy to check that`q preserves sums of polynomials.

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If g( x) = b0 + b1 x +... + bn x

n

then the coefficient ck of the product h( x)=f ( x)g( x)

is given by the formulac

k = å=+ k ji jiba

. Applying q

to both sides gives

q (ck )= q( å=+ k ji

jiba ) since q preserves both sums

and products. This formula is precisely what we

need to check that`q ( f ( x)g( x)) =`q (h( x)) = `q ( f ( x))`q (g( x)) .This finishes the proof that `q is a ring

homomorphism. Let j: R®S be a ring homomorphism. By

elementary results on group theory, we know that jmust map the additive identity of R onto the additiveidentity of S, that j must preserve additive inverses,and that j ( R) must be an additive subgroup of S. Itis convenient to list these results formally in the next

proposition. Part (c) of the proposition followsimmediately from the observation that if 1 is anidentity element for R, then we must have

j (1) j (1)=j(1×1) = j (1). To complete the proof of part (d), the fact that

j( R) is closed under multiplication follows since j preserves multiplication.

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Proposition 2.7 Let j: R®S be a ringhomomorphism. Then (a) j(0) = 0;

(b) j (-a) = -j (a) for all a Î R; (c) If j is onto and 1 is an identity element for R, then j (1) is an identity element of S;

(d) j ( R) is a subring of S.

Example 2.8

Any isomorphism j: R®S of commutative ringswith identity induces a group isomorphism from R*

onto S *. To establish this result, we first need to

show that such an isomorphism maps identity toidentity. Since j is onto, given 1Î S, we have 1 = j(r ) for some r Î R. Then

j (1) = j (1)×1 = j (1) j (r ) =j (1×r ) = j (r ) = 1.

Furthermore, for any aÎ R* we have j(a)j(a-

1)=j(aa-1) = j (1) = 1, and so j maps R

* into S *.

Applying this argument to (j)-1, which is also a ringhomomorphism, shows that it maps S

* into R*. We

can conclude that j is one-to-one and onto whenrestricted to R

* and this shows that R*@S *.

Before giving some additional examples, we need to give a definition analogous to one for groups. Infact, the kernel of a ring homomorphism is just the

kernel of the mapping when viewed as a group

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homomorphism of the underlying additive groups of the rings.

Definition 2.9.4 Let j: R®S be a ringhomomorphism. The set

{aÎ R: j (a) = 0}is called the kernel of j, denoted by ker(j).

Example 2.10

We already know by results on factor groups thatthe mapping p : Z®Zn given by p( x) = [ x]n , for all xÎZ, is a group homomorphism. The formula[ x]n[ y]n=[ xy]n , which defines multiplication of congruence classes, shows immediately that it also

preserves multiplication. Thus p is a ringhomomorphism with ker(p) = nZ.

Example 2.11 (Reduction modulo n)

Let p:Z®Zn , be the natural projectionconsidered in the previous example. In Example 2.6

we discussed the ring of polynomials withcoefficients in a commutative ring, and so in thiscontext we can consider the polynomial rings Z[ x]and Zn[ x]. Define `p : Z[ x]® Zn[ x] as follows: For any polynomial

f ( x) = a0 + a

1 x +... + am x

m

set `p ( f ( x)) = p (a0) + j (a1 x) +... + j (am xm)=

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p (a0) + p (a1) x +... + p (am) xm

That is, `p simply reduces all of the coefficients of f ( x) modulo n. This is actually a special case of the

result obtained in Example 2.6. We can think of p asa homomorphism from Z into Zn[ x], and then wehave extended p to Z[ x] by mapping xÎZ[ x] top( x)ÎZn[ x]. It follows that `p is a homomorphism.Furthermore, it is easy to see that the kernel of ̀ p isthe set of all polynomials for which each coefficientis divisible by n. To illustrate the power of homomorphism,suppose that f ( x) has a nontrivial factorization f ( x) =g( x)h( x) in Z[ x]. Then

`p ( f ( x)) =`p (g( x)h( x)) =`p (g( x)) `p (h( x)) .If deg(`p( f ( x))) = deg( f ( x)), then this gives anontrivial factorization of

`p ( f ( x)) in Zn[ x]. This means that a polynomial f ( x)

with integer coefficients can be shown to beirreducible over the field Q of rational numbers byfinding a positive integer n such that when thecoefficients of f ( x) are reduced modulo n, the new

polynomial has the same degree and cannot befactored nontrivially in Zn[ x].

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Proposition 2.12 Let j: R®S be a ringhomomorphism.

(a) If a, b Î ker(j) and r Î R, then a + b, a - b, and

ra belong to ker(j).(b)The homomorphism j is an isomorphism if and only if ker(j) = {0} and j ( R) = S.

Proof. (a) If a, b Î ker(j), thenj (a ± b) =j (a) ±j (b) = 0 ± 0 = 0 ,

and so a ±bÎker(j). If r Î R, then j(ra)=j(r )×j(a)=j (r )×0 = 0, showing that ra Î ker(j).(b) This part follows from the fact that j is a grouphomomorphism, since j is one-to-one if and only if

ker(j)=0 and onto if and only if j( R)=S.

Example 2.13

Let j:Z8® Z12 be the ring homomorphismdefined by j([ x]8)=[9 x]12 for all [ x]8ÎZ8 . Thenj(Z8) consists of all multiples of [9]12, and determines the subring {[0]12 , [3]12 , [6]12 , [9]12 } of Z12 . We also have ker(j ) = {[0]8 , [4]8 }.

Let j: R®S be a ring homomorphism. Thefundamental homomorphism theorem for groupsimplies that the abelian group R/ker(j) is isomorphicto the abelian group j( R), which is a subgroup of S.

In order to obtain a homomorphism theorem for

commutative rings we need to consider the cosets of ker(j). Intuitively, the situation may be easiest to

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understand if we consider the cosets to be defined bythe equivalence relation ~ given by a~b iff j(a)=j(b), for all a,bÎ R. The sum of equivalence

classes [a] and [b] in R/ker(j) is well-defined usingthe formula [a]+[b]=[a+b], for all a,bÎ R. The product of equivalence classes [a] and [b] in R/ker(j) is defined by the expected formula[a]×[b]=[ab], for all a,bÎ R. To show that thismultiplication is well-defined, we note that if a ~ cand b ~ d , then ab ~ cd since

j (ab)=j(a)j(b)=j(c)j(d )=j(cd )Since our earlier results on groups imply that

R/ker (j) is an abelian group, to show that R/ker (j)is a commutative ring we only need to verify thedistributive law and the associative and commutativelaws for multiplication. We have

[a] ([b]+[c])=[a][b+c]=[a(b+c)]=[ab+ac]=[ab]+[ac] = [a] [b]+[a]

[c],

showing that the distributive law follows directlyfrom the definitions of addition and multiplicationfor equivalence classes, and the distributive law inR. The proofs that the associative and commutativelaws hold are similar.

Theorem 2.14 (Fundamental HomomorphismTheorem for Rings)

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Let j: R®S be a ring homomorphism. Then R/ker(j)@ j( R).

Proof. Since we have already shown that

R/ker(j) is a commutative ring, we can apply thefundamental homomorphism theorem for groups toshow that the mapping `j given by `j([a])=j(a),for all aÎ R, defines an isomorphism of the abeliangroups R/ker(j) and j( R). We only need to showthat `j preserves multiplication, and this followsfrom the computation`j([a][b])=`j([ab])=j(ab)=j(a)j(b)=j([a])j([b])This completes the proof.The coset notation a + ker(j) is usually used for theequivalence class [a]. With this notation, addition

and multiplication of cosets are expressed by theformulas

(a+ker(j))+(b+ker(j))=(a+b)+ker(j) and (a+ker(j))×(b+ker(j))=(ab)+ker(j).

It is important to remember that these are additive

cosets, not multiplicative cosets.3 Ideals and Factor rings

We have shown that the kernel of any ringhomomorphism is closed under sums, differences,and "scalar" multiples. In other words, the kernel of

a ring homomorphism is an additive subgroup that isclosed under multiplication by any element of the

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ring. For any integer n, the subset nZ of the ring of integers satisfies the same properties. The same istrue for the subset ( f ( x)) of the ring F [ x] of

polynomials over a field F , where f ( x) is any polynomial. In each of these examples we have been able to make the cosets of the given set into acommutative ring, and this provides the motivationfor the next definition and several subsequentresults.

Definition 3.1 Let R be a commutative ring. Anonempty subset I of R is called an ideal of R if

(i) a ± b Î I for all a,bÎ I, and (ii) ra Î I for all aÎ I and r Î R.

For any commutative ring R, it is clear that the set{0} is an ideal, which we will refer to as the trivialideal. The set R is also always an ideal. Amongcommutative rings with identity, fields arecharacterized by the property that these two idealsare the only ideals of the ring.

Proposition 3.2 Let R be a commutative ringwith identity. Then R is a field if and only if it hasno proper ideals.

Proof. First assume that R is a field, and let I be

any ideal of R. Either I = {0}, or else there exists aÎ I such that a¹0. In the second case, since R is a

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field, there exists an inverse a-1 for a, and then for

any r ÎR we have r =r ×1 = r (a-1a) = (ra

-1)a, so by thedefinition of an ideal we have r Î I . We have shown

that either I = {0} or I = R.Conversely, assume that R has no proper nontrivialideals, and let a be a nonzero element of R. We willshow that the set

I ={ xÎ R½ x=ra for some r Î R}is an ideal. First, I is nonempty since a = 1×a Î I . If

r la, r 2aÎ I , then we have r 1a ±r 2a = (r l ± r 2)a,showing that I is an additive subgroup of R. Finally,if x = raÎ I , then for any sÎ R we have sx = (sr )aÎ I ,and so I is an ideal. By assumption we must have I

= R, since I ¹{0}, and since 1Î R, we have 1 = ra for

some r R. This implies that a is invertible, and sowe have shown that R is a field.

Let R be a commutative ring with identity 1. For any aÎ R, let Ra denote the set I={ xÎ R½ x=ra for some r Î R}. The proof of the previous theorem

shows that Ra is an ideal of R that contains a. It isobvious from the definition of an ideal that any idealthat contains a must also contain Ra, and so we are

justified in saying that Ra is the smallest ideal thatcontains a.

Furthermore, R

1 consists of all of R

, since everyelement r Î R can be expressed in the form r 1. Thus

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R is the smallest ideal (in fact, the only ideal) thatcontains 1.

Definition 3.3 Let R be a commutative ring withidentity, and let a Î R.

The ideal Ra={ xÎ R½ x=ra for some r Î R}

is called the principal ideal generated by a. The

notation (a) will also be used. An integral domain in which every ideal is a

principal ideal is called a principal ideal domain.

Example 3.4 Z is a principal ideal domain

In the terminology of the above definition,we see that Theorem 1.1.4 shows that the ring of integers Z is a principal ideal domain. Moreover,given any nonzero ideal I of Z, the smallest

positive integer in I is a generator for the ideal.

Example 3.5 F [ x] is a principal ideal domain We showed in Theorem 4.2.2 that if F is anyfield, then the ring F [ x] of polynomials over F isa principal ideal domain. If I is any nonzero idealof F [ x], then f ( x) is a generator for I if and only if

it has minimal degree among the nonzeroelements of I . Since a generator of I is a divisor

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of every element of I , it is easy to see that there isonly one monic generator for I .

We have also considered the ring of polynomials R[ x] over any commutative ring R.However, if the coefficients do not come from afield, then the proof of the division algorithm is nolonger valid, and so we should not expect R[ x] to bea principal ideal domain. The ring Z[ x] of

polynomials with integer coefficients is an integral

domain, but not every ideal is principal.

Definition 3.6 Let R be a commutative ring withidentity. The smallest positive integer n such thatn×1=0 is called the characteristic of R, denoted by

char( R). If no such positive integer exists, then R issaid to have characteristic zero. The characteristic of a commutative ring R with

identity is just the order of 1 in the underlyingadditive group. If char( R)=n, then it follows fromthe distributive law that n×a=(n×1)×a=0×a=0, and sothe exponent of the underlying abelian group is n.

A more sophisticated way to view the charac-teristic is to define a ring homomorphism j:Z® R byj (n)=n×1. The rules we developed in Chapter 3 for considering multiples of an element in an abelian

group show that j is a homomorphism. The

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characteristic of R is just the (nonnegative) generator of ker(j). Proposition 3.7 An integral domain hascharacteristic 0 or p, for some prime number p.

Proof. Let D be an integral domain, and consider the mapping j:Z®D defined by j(n)=n×1. Thefundamental homomorphism theorem for rings showsthat Z / ker(j) is isomorphic to the subring j(Z). Sinceany subring of an integral domain inherits the

property that it has no nontrivial divisors of zero, thisshows that Z / ker(j) must be an integral domain. Thuseither ker(j)=0, in which case char(D)=0, or ker(j)= pZ for some prime number p, in which casechar(D) = p.

In Chapter1 we used the notion of congruencemodulo n to define the sets Zn , which we nowknow to be commutative rings. Since aºb(mod n) if and only if a-b is a multiple of n, this is preciselythe equivalence relation determined below by the

ideal nZ. In Chapter 3, given an irreducible polynomial p( x), we used the notion of congruencemodulo p( x) to construct a new field from thecongruence classes, in which we could find a rootof p( x). Again in this case, the congruence classesare precisely those determined by the principal ideal( p( x)) of F [ x], consisting of all polynomialmultiples of p( x). We now extend this procedure to

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the equivalence classes defined by ideals incommutative rings. We have chosen to use thenotation of congruence classes rather than additivecosets.

Definition 3.8. Let I be an ideal of thecommutative ring R. For a,bÎ R we define

a º b (mod I )if a-bÎ I. For any aÎ R the congruence classdetermined by a will be denoted by [a]I , or simply

[a] when the context is clear. The set of allcongruence classes modulo I will be denoted by R/I.

Proposition 3.A. Let I be an ideal of thecommutative ring R, and let a, b, c, d be elements of R.

(a)If a º c (mod I) and bº d (mod I) , thena + b º c + d (mod I).

(b) If a º c (mod I) and b º d (mod I), thenab º cd (mod I).

(c) [a]I ={r Î R: r =a +x for some x Î I}.

Proof. Parts (a) and (c) are evident. We includea proof of part (b). Assume that a - cÎ I and b-d Î I.

Multiplying a - c by b and b - d by c gives uselements that still belong to I . Then using thedistributive law and adding gives

(ab - cb)+(cb - cd) = ab - cd,

and this is an element of I , showing that

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ab º cd (mod I ).For an ideal I of a commutative ring R, the nexttheorem provides the justification for calling R/I thefactor ring of R modulo I .

For groups, we were only able to construct factor groups relative to normal subgroups, not allsubgroups. For commutative rings, ideals play ananalogous role. That is, among all subrings of acommutative ring, only those that are ideals can be

used to construct factor rings. Theorem 3.B. If I is an ideal of the commutative

ring R, then R/I is a commutative ring.Proof. The previous proposition shows that

addition and multiplication of congruence classes

are well-defined. To show that the distributive lawholds, let a, b, c Î R. Then[a] ([b] + [c]) = [a] [b + c] = [a(b + c)] =[ab + ac] = [ab] + [ac] = [a] [b] + [a] [c]

Note that we have used the definitions of additionand multiplication of equivalence classes, together with the fact that the distributive law holds in R.Similar computations show that the associative and commutative laws hold for both addition and multiplication, completing the proof that R/I is acommutative ring.

Let R be a commutative ring and let I be an idealof R. For any element aÎ R, the congruence class

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[a] I , is just the additive coset of I determined by a.Using the standard coset notation from group theory,we have [a] I = a+ I . With this notation the formulasfor addition and multiplication in R/I take thefollowing form, where a,b Î R:

(a +I )+(b+I ) = (a +b)+ I and (a +I )× (b +I ) = ab +I .

The most familiar factor ring is Z/nZ , for which wewill continue to use the notation Z n. In this ring,

multiplication can be viewed as repeated addition,and it is easy to show that any subgroup is an ideal.Thus we already know the lattice of ideals of Z n, inwhich ideals correspond to the divisors of n. Recallthat in Z we have mêk if and only if mZ Ê kZ , and so

the lattice of ideals of Z n corresponds to the lattice of all ideals of Z that contain nZ . As shown by the next

proposition, the analogous result holds in any factor ring.

Proposition 3.9. Let I be an ideal of thecommutative ring R.

(a) The natural projection mapping p : R® R/I

defined by p (a)=[a] I for all aÎ R is a ringhomomorphism, and ker (p ) = I.

(b) There is a one-to-one correspondence between the ideals of R/I and ideals of R that contain I .

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Proof . The parts of the proposition that involveaddition can be shown easily. To prove part (a), the

projection mapping must be shown to preservemultiplication, and this follows directly from thedefinition of multiplication of congruence classes.

To prove part (b) consider the followingcorrespondence between the ideals of R/I and idealsof R that contain I. If L is any ideal of R/I thenconsider p -1( L) ={a Î R: [a] Î L}- an ideal in R

containing I.We must show that this correspondence preservesideals. If J is an ideal of R that contains I , then itcorresponds to the additive subgroup

p ( J ) = {[a] I : a Î J }

For any element [r ] Î R/I and any element [a] Î p ( J ), we have [r ][a] = [ra], and then [ra] Î p ( J ) sincera Î J. On the other hand, if L is an ideal of R/I, thenit corresponds to the subgroup

p -1( L) = {a Î R: [a] Î L }

For any r Î R and a Î p -1

( L), we have ra Îp -1

( L)since [ra] = [r ][a] Î L.

Example 3.10.

Let R = Q[ x] and let I =( x2-2 x+1). Using the

division algorithm, it is possible to show that thecongruence classes modulo I correspond to the

possible remainders upon division by x2 - 2 x + 1, so

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that we only need to consider congruence classes of the form [a + bx], for all a, b Î Q. Since x

2- 2 x + 1

º0 (mod I ), we can use the formula x2 º2 x-1(mod I )

to simplify products. This gives us the followingformulas:[a + bx] + [c + dx] = [(a + c) + (b + d ) x] and

[a + bx] × [c + dx] =[(ac - bd ) +(bc + ad + 2bd ) x]By the previous proposition, the ideals of R/I

correspond to the ideals of R that contain I. Since

Q[ x] is a principal ideal domain, these ideals aredetermined by the divisors of x2 - 2 x + 1, showingthat there is only one proper nontrivial ideal in R/I,

corresponding to the ideal generated by x - 1.

Example 3.11.

Let f :R® S be an isomorphism of commutativerings, let I be any ideal of R, and let J = f ( I ). Wewill show that R/I is isomorphic to S/J. To do this,let p be the natural projection from S onto S/J, and consider q = pf . Then q is onto since both p and f

are onto, and ker (q ) = {r Î R: f (r ) Î J } = I,

so it follows from the fundamental homomorphismtheorem for rings that R/I is isomorphic to S/J.

To motivate the next definition, consider the ringof integers Z . We know that the proper nontrivialideals of Z correspond to the positive integers, with

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nZ ÊmZ if and only if m|n. Euclid's lemma statesthat an integer p>1 is prime if and only if it satisfiesthe following property: If p|ab for integers a, b, then

either p|a or p|b. In the language of ideals this saysthat p is prime if and only if abÎ pZ implies aÎ pZ or bÎ pZ, for all integers a, b.

Lemma 2.2.7 gives a similar characterization of irreducible polynomials, stating that a polynomial p( x)ÎF [ x] is irreducible if and only if p( x)| f ( x)g( x)implies p( x)| f ( x) or p( x) | g( x). When formulated interms of the principal ideal ( p( x)), this shows that p( x) is irreducible over F if and only if f ( x)g( x)Î( p( x)) implies f ( x)Î( p( x)) or g( x)Î ( p( x))Thus irreducible polynomials play the same role in

F [ x] as do prime numbers in Z . These two examples probably would not provide

sufficient motivation to introduce a generaldefinition of a "prime" ideal. In both Z and F [ x],where F is a field, we have shown that each element

can be expressed as a product of primes or irreducibles, respectively. This is not true in generalfor all commutative rings. In fact, certain subringsof C which turn out to be important in number theory do not have this property. One of the originalmotivations for introducing the notion of an ideal (or "ideal number") was to be able to salvage at least the

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property that every ideal could be expressed as a product of prime ideals. The prime numbers in Z can be characterized inanother way. Since a prime p has no divisors except±p and ±1, there cannot be any ideals properlycontained between pZ and Z . We refer to ideals withthis property as maximal ideals. Definition 3.12. Let I be a proper ideal of the

commutative ring R. Then I is said to be a prime

ideal of R if for all a,bÎ R it is true that abÎ I impliesaÎ I or bÎ I.

The ideal I is said to be a maximal ideal of R if for all ideals J of R such that I Í J Í R, either J = I or J

= R.

Note that if R is a commutative ring withidentity, then R is an integral domain if and only if the zero ideal of R is a prime ideal. This observation

provides an example, namely, the zero ideal of Z , of a prime ideal that is not maximal. On the other hand,

the characterization of prime and maximal idealsgiven in the next proposition shows that in acommutative ring with identity any maximal ideal isalso a prime ideal, since any field is an integraldomain.

Proposition 3.13. Let I be a proper ideal of thecommutative ring R with identity.

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(a) The factor ring R/I is a field if and only if I is a maximal ideal of R.

(b) The factor ring R/I is a integral domain if and only if I is a prime ideal of R.

(c) If I is maximal, then it is a prime ideal.Proof. (a) R/I is a field if and only if it has no

proper nontrivial ideals. Using the one-to-onecorrespondence between ideals given by Proposition5.3.9 (b), this occurs if and only if there are no ideals

properly between I and R, which is precisely thestatement that I is a maximal ideal.

(b) Assume that R/I is an integral domain, and let a, bÎ R with abÎ I. Remember that the zeroelement of R/I is the congruence class (or coset)

consisting of all elements of I. Thus in R/I we have a product [a] [b] of congruence classes that is equal tothe zero congruence class, and so by assumption thisimplies that either [a] or [b] is the zero congruenceclass. This implies that either aÎ I or bÎ I, and so I

must be a prime ideal.Conversely, assume that I is a prime ideal. If

a,bÎ R with [a] [b]=[0] in R/I , then we have abÎ I,

and so by assumption either aÎ I or bÎ I . This showsthat either [a] = [0] or [b] = [0], and so R/I is anintegral domain.

Example 3.14.

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Let f : R® S be an isomorphism of commutativerings. The isomorphism gives a one-to-onecorrespondence between ideals of the respectiverings, and it is not hard to show directly that prime(or maximal) ideals of R correspond to prime (or maximal) ideals of S. This can also be proved as anapplication of the previous proposition, since inExample 5.3.11 we observed that if I is any ideal of R, then f ( I ) is an ideal of S with R/I ® S/ f ( I ). It

then follows immediately from Proposition 5.3.13that I is prime (or maximal) in R if and only if f ( I ) is

prime (or maximal) in S .We have already observed that in the ring Z, prime

numbers determine maximal ideals. One reason

behind this fact is that any finite integral domain is afield, which we proved in Theorem 5.1.8. It is alsotrue that if F is a field, then irreducible polynomialsin F [ x] determine maximal ideals. The next

proposition gives a general reason applicable in bothof these special cases. Theorem 3.15. Every nonzero prime ideal of a

principal ideal domain is maximal. Proof. Let P be a nonzero prime ideal of a

principal ideal domain R, and let J be any ideal withP Í J Í R. Since R is a principal ideal domain, we

can assume that P=Ra and J=Rb for some elementsa,bÎ R. Since aÎP, we have aÎ J, and so there exists

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r Î R such that a=rb. This implies that rbÎP, and soeither bÎ P or r Î P. In the first case, bÎP impliesthat J=P. In the second case , r Î P implies that r=sa

for some sÎ R, since P is generated by a. This givesa=sab, and using the assumption that R is an integraldomain allows us to cancel a to get 1=sb. Thisshows that 1Î J , and so J = R.

Example 3.16. (Ideals of F [ x]) We can now summarize the information we haveregarding polynomials over a field, using a ringtheoretic point of view. Let F be any field. Thenonzero ideals of F [ x] are all principal, of the form( f ( x)), where f ( x) is the unique monic polynomial of minimal degree in the ideal. The ideal is prime (and

hence maximal) if and only if f ( x) is irreducible. If p( x) is irreducible, then the factor ring F [ x]/ ( p( x)) isa field. Example 3.17. (Evaluation mapping) Let F be a subfield of E, and for any element uÎ E

define the evaluation mapping f u : F [ x]® E byf u( f ( x)) = f (u), for all f ( x)Î F [ x]. We have alreadyseen that f u defines a ring homomorphism. Sincef u(F [ x]) is a subring of E that contains 1, it followsthat it must be an integral domain. By the

fundamental homomorphism theorem for rings thisimage is isomorphic to F [ x]/ker (f u), and so by

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Proposition 5.3.13, the kernel of f u must be a primeideal. If ker (f u) is nonzero, then it follows fromTheorem 5.3.15 that it is a maximal ideal. By

Proposition 5.3.13, we know that F [ x]/ker (f u) is afield, so it follows from the fundamentalhomomorphism theorem for rings that the image of f u is in fact a subfield of E . This fact will be veryimportant in our study of fields in Chapter onalgebraic numbers, where we denote f u(F [ x]) byF (u).

4. Field Extensions

4.1 Algebraic ElementsWe recall that a commutative ring with identity 1

(assumed to be different from the element 0) is called a field if every nonzero element is invertible. Thusthe operations of addition, subtraction,

multiplication, and division (by nonzero elements)are all possible within a field. We should also notethat the elements of a field form an abelian groupunder addition, while the nonzero elements form anabelian group under multiplication. This observationallows us to make use of the well known results from

the Theory of groups.

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Our primary interest is to study roots of polynomials. This usually involves the interplay of two fields: one that contains the coefficients of the

polynomial, and another determined by the roots of the polynomial. In many situations we will startwith a known field K, and then construct a larger field F. We now give a restatement of Definition4.4.1 in this context.

Definition 1.1 The field F is said to be an

extension field of the field K if K is a subset of F which is a field under the operations of F .

The above definition is equivalent to sayingthat if K is a subfield of F. That is, the additive and multiplicative groups that determine K are

subgroups of the corresponding additive and multiplicative groups of F. If F is an extension field of K, then the elements of F that are roots of nonzero

polynomials in K [ x] will be called algebraicelements. If uÎF is a root of some nonzero

polynomial f ( x)ÎK [ x], then let p( x) have minimaldegree among all polynomials of which u is a root.Using the division algorithm, we can write f ( x)=q( x) p( x)+r ( x), where either r ( x)=0 or deg(r ( x))<deg( p( x)). Solving for r ( x) and substituting u shows that u is a root of r ( x), which

violates the definition of p( x) unless r ( x)=0, and sowe have shown that p( x) is a divisor of f ( x). With

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the notation above, we next observe that p( x) must be an irreducible polynomial. To show this, let p( x)=g( x)h( x) for polynomials g( x),h( x)ÎK [ x]. Thensubstituting u gives g(u)h(u)=p(u)=0, and so either g(u)=0 or h(u)=0 since these are elements of F,

which is a field. From the definition of p( x) as a polynomial of minimal degree that has u as a root,we see that either g( x) or h( x) has the same degree as p( x), and so we have shown that p( x) is irreducible.

These facts about the polynomials that have a givenelement as a root can be proved in another way byusing results on rings of polynomials. The approachusing ring theory lends itself to more powerfulapplications, and so we will adopt that point of view.

Recall that the nonzero ideals of F [ x] are all principal, of the form( f ( x))={q( x) f ( x)|q( x)ÎF [ x]} ,

where f ( x) is any polynomial of minimal degree inthe ideal. A nonzero ideal is prime (and hencemaximal) if and only if its generator f ( x) isirreducible. Furthermore, if p( x) is irreducible, thenthe factor ring F [ x]/ ( p( x)) is a field.

Definition 1.2 Let F be an extension field of K

and let uÎF . If there exists a nonzero polynomial f ( x)ÎK [ x] such that f (u)=0, then u is said to be

algebraic over K . If there does not exist such a

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polynomial, then u is said to be transcendental over K .

The familiar constants e and π aretranscendental. These are not easy results, and theanalytic proofs lie beyond the scope of this course.That e is transcendental was proved by CharlesHermite (1822-1882) in 1873, and the correspondingresult for π was proved by Ferdinand Lindemann(1852-1939) in 1882.

Proposition 1.3 Let F be an extension field of K , and let uÎF be algebraic over K . Then thereexists a unique monic irreducible polynomial p( x)ÎK [ x] such that p(u)=0. It is characterized as themonic polynomial of minimal degree that has u as a

root. Furthermore, if f ( x) is any polynomial in K [ x]with f (u) = 0, then p( x) │ f ( x).Proof. Let I be the set of all polynomials

f ( x)ÎK [ x] such that f (u)=0. It is easy to see that I isclosed under sums and differences, and if f ( x)Î I,

then g( x) f ( x)Î I for all g( x)ÎK [ x]. Thus I is anideal of K [ x], and so I =( p( x)) for any nonzero polynomial p( x)Î I that has minimal degree. If f ( x),g( x)ÎK [ x] with f ( x)g( x)Î I, then we have f (u)g(u)=0, which implies that either f (u)=0 or g(u)=0, and so we see that I is a prime ideal. Thisimplies that the unique monic generator p( x) of I

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must be irreducible. Finally, since I =( p( x)) , we have p( x)| f ( x) for any f ( x)Î I .

Definition 1.4 Let F be an extension field of K ,and let u be an algebraic element of F . The monic

polynomial p( x) of minimal degree in K [ x] such that p(u)=0 is called the minimal polynomial of u over K .The degree of the minimal polynomial of u over K iscalled the degree of u over K .

Example 14.5 ( 2 has degree 2 over Q)Considering the set of real numbers R as an

extension field of the set of rational numbers Q, the

number 2 has minimal polynomial x2 — 2 over Q,

and so it has degree 2 over Q.

Example 1.6 ( 2 + 3 has degree 4 over Q)In this example we will compute the minimal

polynomial of 32 + over Q. If we let 32 += x ,then we must find a nonzero polynomial with rationalcoefficients that has x as a root. We begin by

rewriting our equation as 32 =- x Squaring bothsides gives 32222 =+- x x . Since we still need toeliminate the square root to obtain coefficients over Q, we can again rewrite the equation to obtain

x x 2212 =- . Then squaring both sides and rewriting

the equation gives x4 — 10 x2 + 1 = 0.

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To show that x4 —10 x2+1 is the minimal

polynomial of 32 + over Q, we must show that itis irreducible. It is easy to check that there are no

rational roots, so it could only be the product of twoquadratic polynomials, which can be assumed to haveinteger coefficients. Unfortunately, Eisenstein'sirreducibility criterion cannot be applied, and so wemust try to verify directly that the polynomial isirreducible. A factorization over Q of the form

x4 —10 x2+1=( x2+ax+b)( x2+cx+d)leads to the equations a+c=0, b+ac+d =-10,ad+bc=0, and bd = 1. If c=0, then b+d=-10 and bd =1, so b

2+ bd =-10b, or b

2+10b+1=0. The last

equation has no rational roots, so we may assume

that c≠0. Letting c=-a, we see that b=d, and sob=±1. It follows that a

2=8 or a2=12, a contradiction,

since aÎQ. We conclude that x4-10 x2+1 is irreducible

over Q, and so it must be the minimal polynomial of

32 + over Q.

In a field F, the intersection of any collection of subfields of F is again a subfield. In particular, if F isan extension field of K and uÎF, then theintersection of all subfields of F that contain both K

and u is a subfield of F. This intersection iscontained in any subfield that contains both K and u.

This guarantees the existence of the field defined below.

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Definition 1.7 Let F be an extension field of K ,and let u1, u2 , ..., unÎF . The smallest subfield of F

that contains K and u1, u2 , ..., un will be denoted byK (u

1,u

2,...,u

n). It is called the extension field of K

generated by u1,u2,...,un. If F=K (u) for a singleelement uÎF , then F is said to be a simple extensionof K . If F is an extension field of K, and u1 , u2 , ...,

unÎF, then it is possible to construct K (u1 ,u2 ,...,un) by adjoining one ui at a time. That is, we first

construct K (u1) , and then consider the smallestsubfield of F that contains K (u1) and u2. This would

be written as K (u1)(u2), but it is clear from thedefinition of K (u1,u2) that the two fields are equal.This procedure can be continued to construct

K (u1 ,u2 ,...,un). It is thus sufficient to describe K (u) for a single element, which we do in the next

proposition.Proposition 1.8 Let F be an extension field of

K , and let uÎF .

(a)If u

is algebraic over K

, thenK

(u

)=K

[ x

]/‹ p

( x

)›,where p( x) is the minimal polynomial of u over K .(b)If u is transcendental over K , then K (u) = K ( x),where K ( x) is the quotient field of the integraldomain K [ x].

Proof. Define j u: K [ x]→F by j u ( f ( x))=f (u) , for

all polynomials f ( x)ÎK [ x]. This defines a ringhomomorphism, and ker(j u) is the set of all

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polynomials f ( x) with f (u)=0. The image of j u is asubring of F consisting of all elements of the forma0+ a1u + ... + anu

n , and it must be contained in

every subring of F that contains K and u. In particular, the image of j u must be contained inK (u).

(a) If u is algebraic over K, then ker j u=( p( x)) , for the minimal polynomial p( x) of u over K. In this casethe fundamental homomorphism theorem for rings

implies that the image of j u is isomorphic toK [ x] / ( p( x)), which is a field since p( x) is irreducible.But then the image of j u must in fact be equal toK (u) , since the image is a subfield containing K and u.

(b) If u is transcendental over K, then ker j u={0},and so the image of j u is isomorphic to K [ x ] . Since F

is a field there exists an isomorphism θ from thefield of quotients of K [ x] into F. Since everyelement of the image of θ is a quotient of elements

that belong to K (u) , it follows that this image must be contained in K(u). Then since it is a field thatcontains u, it must be equal to K (u)

To help understand the field K (u) , it may beuseful to approach its construction from a moreelementary point of view. Given an extension field F

of K and an element uÎF, any subfield that contains

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K and u must be closed under sums and products, soit must contain all elements of the forma0+a1u+...+anu

n , where anÎK for 0≤i≤n.

Furthermore, since it must be closed under division,it must contain all elements of the form

(ao+a1u+...+anun)/( b0+b1u+...+bmu

m)such that the denominator is nonzero. This set can beshown to be a subfield of F, and so it must be equalto K (u).

If u is algebraic of degree n over K, let theminimal polynomial of u over K be p(x)=co+c1 x+...+cn x

n. Since co+c1u+...+cnu

n=0, we

can solve for un and obtain a formula that allows usto reduce any expression of the form

ao+a1u+...+amu

m

to one involving only u,u

2

,…,un-1

and elements of K. Given any expression of the formao+a1u+...+an-1u

n-1, we let f ( x) be the corresponding polynomial

ao+a1 x+...+an-1 xn-1.

If f ( x) is nonzero, then it must be relatively primeto p( x) since p( x) is irreducible and

deg( f ( x))<deg( p( x)).Thus there exist polynomials g( x) and q( x) such

that f ( x)g( x)+p( x)q( x)=1. Substituting u gives f (u)g(u)=1 since p(u)=0, and so g(u)=1/ f (u). Thus

the denominators can be eliminated in our description of K (u). We conclude that when u is

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algebraic over K of degree n, each element of K (u)has the form ao+a1u+.. .+an-1u

n-1 for elementsao , a1 ,…,an-1ÎK.

Example 1.9 (Computations in Q(3

2 ))We can let u be any root of the polynomial x3 — 2 ,

since the computations will be the same as for

u= 3 2 . The extension field Q(u) of Q isisomorphic to the factor ring Q[ x]/ ( x3 — 2), and so

computations can be done in either field. For example, let us compute (1+u2)-1. In Q(u) we can set

up the equation(1+u

2)(a+bu+cu2)=1. Using the identities u

3=2 and u

4=2u to multiply out the left-hand side, we obtainthe equations a+2b=1 , b+2c=0 , and a+c=0. These

lead to the solution a = 1/5, b = 2/5, and c = —1/5.On the other hand, to find the multiplicative inverseof the element [1+ x2] in Q[ x]/( x3-2), we can use theEuclidean algorithm to solve for

gcd ( z2 + l ,z3-2).

We obtain x3

-2 = x( x2

+1)-( x+2) and then x

2+1=( x-2)( x+2)+5.

Solving for the linear combinations that give thegreatest common divisor yields the equation

1=(-x2/5+2 x/5+1/5)( x2+1)+( x/5-2/5)( x3-2)

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We can reduce the equation to a congruence modulo x

3 — 2 and use the isomorphism to obtain the same

answer we got previously:(1+u

2)-1=-u

2/5+2u/5+1/5Proposition 1.10 Let K be a field and let

p( x)ÎK [ x] be any irreducible polynomial. Then thereexists an extension field F of K and an element uÎF

such that the minimal polynomial of u over K is p( x).

Proof. This is simply a restatement of Kronecker's theorem. Recall that the extension field F is constructed as K [ x]/( p( x)), and K is viewed asisomorphic to the subfield consisting of all cosetsdetermined by the constant polynomials. The

element u is the coset determined by x, and itfollows that p(u)=0. Since p( x) is irreducible,Proposition 4.1.3 shows that p( x) is the minimal

polynomial for u over K.

4.2 Finite and Algebraic Extensions

If F is an extension field of K, then themultiplication of F defines a scalar multiplication,considering the elements of K as scalars and theelements of F as vectors. This gives F the structureof a vector space over K, and allows us to make useof the concept of the dimension of a vector space.

Proposition 2.1 Let F be an extension field of K and let uÎF be an element algebraic over K . If the

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minimal polynomial of u over K has degree n, thenK (u) is an n-dimensional vector space over K .

Proof. Let p( x)=c0+c1 x + ...+cn xn be the

minimal polynomial of u over K. We will show thatthe set B = {1 , u, u2 ,... ,un-1} is a basis for K (u)over K. By Proposition 4.1.6, K (u)=K [ x]/( p( x)), and since each coset of K [ x]/ ( p( x)) contains a uniquerepresentative of degree less than n, it follows fromthis isomorphism that each element of K (u) can be

represented uniquely in the form aol+a1u+...+an-

1un-1

. Thus B spans K (u), and the uniqueness of representations implies that B is also a linearlyindependent set of vectors.

Definition 2.2 Let F be an extension field of K.

If the dimension of F as a vector space over K isfinite, then F is said to be a finite extension of K.

The dimension of F as a vector space over K iscalled the degree of F over K , and is denoted by[F :K ].In the next proposition, by using the notion of the

degree of an extension, we are able to give a usefulcharacterization of algebraic elements. The

proposition implies, in particular, that every elementof a finite extension must be algebraic.

Proposition 2.3 Let F be an extension field of

K and let uÎF . The following conditions areequivalent:

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(1) u is algebraic over K ;(2) K (u) is a finite extension of K ;(3) u belongs to a finite extension of K.

Proof. It is clear that (1) implies (2) and (2)implies (3). To prove (3) implies (1), suppose thatuÎ E, for a field E with K Í E and [ E :K ]=n. The set1, u, u

2 ,..., u

n contains n+1 elements, and thesecannot be linearly independent in an n-dimensionalvector space. Thus there exists a relation

ao+a1u+...+anun=0 with scalars from K that are notall zero. This shows that u is a root of a nonzero

polynomial in K [ x].Counting arguments often provide very useful

tools. In case we have extension fields K Í E ÍF, we

can consider the degree of E over K and the degreeof F over E. The next theorem shows that there is avery simple relationship between these two degreesand the degree of F over K. Theorem 4.2.4 will playa very important role in our study of extension fields.

Theorem 2.4 Let E be a finite extension of K

and let F be a finite extension of E . Then F is afinite extension of K, and

[F :K ] = [F : E ][ E : K ] .

Proof. Let [F : E ]=n and let [ E :K ]=m. Letu1 ,u2 ,…,un be a basis for F over E and let v1, v2 ,,…,vm

be a basis for E over K. We claim that the set B of

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nm products uiv j (where 1≤i ≤ n and 1 ≤ j ≤ m) is a basis for F over K.

We must first show that B spans F over K. If u is

any element of F, then u = å=

n

i 1aiui for elements ai in

E. For each element ai we have ai = å=

m

j

jijvc1

where

cijÎK. Substituting gives u =

å=

n

i 1

å=

m

j 1 cijv j ui so Bspans F over K.

To show that B is a linearly independent set, suppose

that å ji, cijv j ui = 0 for some linear combination of the elements of B, with coefficients in K. This

expression can be written as å å == =

n

i

m

ji jij uvc

1 10)( . Since

the elements u1 , u2 ,…, un form a basis for F over E,

each of the coefficients å=

n

j jijvc

1 (which belong to E)

must be zero. Then since the elements v1 ,v2 ,… ,vm

form a basis for E over K, for each i we must havecij=0 for all j.

Example 2.5 .4]:)32([ =+ QQ

In Example 6.1.2 we showed that 32 + has

degree 4 over Q by showing that it has the minimal polynomial x

4 — 10 x2 +1. The previous theorem can

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be used to give an alternate proof, using the fact

that )3,2()32( QQ =+ . To show that the twofield extensions are equal, we first observe that

)3,2()32( QQ Ì+ since we have)3,2(32 QÎ+ and the field )32( +Q is

defined by the property that it contains 32 + and is contained in any extension field that contains Q

and 32 + . On the other hand,

1)23()32( =-×+ , and so )23(23 +Î- Q

since it is the multiplicative inverse of 23 + .

Because

2

)23()32(3and

2

)23()32(2

-++=

--+=

it follows that )23()2,3( +Ì QQ . The minimal

polynomial of 2 over Q is x2 — 2, which is

irreducible, and so 2 has degree 2 over Q. To

compute)]2(:)3,2([ QQ

, we need to show that x

2 — 3 is irreducible over )2(Q , in which case it will

be the minimal polynomial of 3 over )2(Q . The

roots 3± of the polynomial do not belong to

)2(Q , because they cannot be written in the form

Qa,bba Î+ for 2 . Thus

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4]:)2([)]2(:)2,3([]:)23([ =×=+ QQQQQQ

This argument using degrees shows that any monic

polynomial of degree 4 that has 32 + as a rootmust be its minimal polynomial.

Corollary 2.6 Let F be a finite extension of K .Then the degree of any element of F is a divisor of [F :K ].

If F is an extension field of K, then themultiplication of F defines a scalar multiplication,considering the elements of K as scalars and theelements of F as vectors. This gives F the structureof a vector space over K, and allows us to make use

of the concept of the dimension of a vector space.Proposition 3.1 Let F be an extension field of K

and let uÎF be an element algebraic over K . If theminimal polynomial of u over K has degree n, thenK (u) is an n-dimensional vector space over K .

Proof. Let p( x)=c0+c1 x + ...+cn xn

be theminimal polynomial of u over K. We will show thatthe set B = {1 , u, u

2 ,... ,u

n-1} is a basis for K (u) over K. ByProposition 4.1.6, K (u)=K [ x]/( p( x)), and since eachcoset of K [ x]/ ( p( x)) contains a unique representativeof degree less than n, it follows from this iso-

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morphism that each element of K (u) can berepresented uniquely in the form

aol+a1u+...+an-1un-1

.

Thus B spans K (u), and the uniqueness of representations implies that B is also a linearlyindependent set of vectors.□

Definition 3.2 Let F be an extension field of K.

If the dimension of F as a vector space over K isfinite, then F is said to be a finite extension of K.

The dimension of F as a vector space over K iscalled the degree of F over K , and is denoted by[F :K ].In the next proposition, by using the notion of thedegree of an extension, we are able to give a useful

characterization of algebraic elements. The proposition implies, in particular, that every elementof a finite extension must be algebraic.

Proposition 3.3 Let F be an extension field of K and let uÎF . The following conditions areequivalent:

(4) u is algebraic over K ;(5) K (u) is a finite extension of K ;(6) u belongs to a finite extension of K.

Proof. It is clear that (1) implies (2) and (2)implies (3). To prove (3) implies (1), suppose that

uÎ E, for a field E with K Í E and [ E :K ]=n. The set1, u, u

2 ,..., u

n contains n +1 elements, and these

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cannot be linearly independent in an n-dimensionalvector space. Thus there exists a relationao+a1u+...+anu

n=0 with scalars from K that are not

all zero. This shows that u is a root of a nonzero polynomial in K [ x].Counting arguments often provide very useful tools.In case we have extension fields K Í E ÍF, we canconsider the degree of E over K and the degree of F

over E. The next theorem shows that there is a very

simple relationship between these two degrees and the degree of F over K. Theorem 4.2.4 will play avery important role in our study of extension fields.

Theorem 3.4 Let E be a finite extension of K

and let F be a finite extension of E . Then F is a

finite extension of K , and [F :K ] = [F : E ][ E : K ] .

Proof. Let [F : E ]=n and let [ E :K ]=m. Letu1 ,u2 ,…,un be a basis for F over E and let v1, v2 ,,…,vm

be a basis for E over K. We claim that the set B of nm products uiv j (where 1≤i ≤ n and 1 ≤ j ≤ m) is a

basis for F over K.

We must first show that B spans F over K. If u is

any element of F, then u = å=

n

i 1aiui for elements ai in

E. For each element ai we have ai = å=

m

j

jijvc1

where

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cijÎK. Substituting gives u = å=

n

i 1å=

m

j 1 cijv j ui so B

spans F over K.

To show that B is a linearly independent set, supposethat å ji, cijv j ui = 0 for some linear combination of the elements of B, with coefficients in K. This

expression can be written as å å == =

n

i

m

ji jij uvc

1 10)( . Since

the elements u1 , u2 ,,…, un form a basis for F over E,

each of the coefficients å=

n

j jijvc

1 (which belong to E)

must be zero. Then since the elements v1 ,v2 ,… ,vm

form a basis for E over K, for each i we must have

cij=0 for all j.Example 3.5 .4]:)32([ =+ QQ

In Example 6.1.2 we showed that 32 + hasdegree 4 over Q by showing that it has the minimal

polynomial x4 — 10 x2 +1. The previous theorem can

be used to give an alternate proof, using the factthat )3,2()32( QQ =+ . To show that the twofield extensions are equal, we first observe that

)3,2()32( QQ Ì+ since we have)3,2(32 QÎ+ and the field )32( +Q is

defined by the property that it contains 32 + and

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118

is contained in any extension field that contains Q

and 32 + . On the other hand,1)23()32( =-×+ , and so )23(23 +Î- Q

since it is the multiplicative inverse of 23 + .Because

,2

)23()32(3and

2

)23()32(2

-++=

--+=

it follows that )23()2,3( +Ì QQ . The minimal

polynomial of 2 over Q is x2

— 2, which isirreducible, and so 2 has degree 2 over Q. To

compute )]2(:)3,2([ QQ , we need to show that

x2 — 3 is irreducible over )2(Q , in which case it will

be the minimal polynomial of 3 over )2(Q

. Theroots 3± of the polynomial do not belong to

)2(Q , because they cannot be written in the form

Qa,bba Î+ for 2 . Thus

4]:)2([)]2(:)2,3([]:)23([ =×=+ QQQQQQ

This argument using degrees shows that any monic

polynomial of degree 4 that has 32 + as a rootmust be its minimal polynomial.

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Corollary 3.6 Let F be a finite extension of K .Then the degree of any element of F is a divisor of [F :K ].

Proof. If uÎF, then [F :K ]=[F :K (u)][K (u):K ].Example 3.7

The field )2(3Q does not contain 2 , since

the degree of 2 over Q is not a divisor of .3]:)2([ 3 =QQ

Corollary 3.8 Let F be an extension field of K , with algebraic elements u1,u2,...,unÎF . Thenthe degree of K (u1,u2,...,un) over K is at most the

product of the degrees of ui over K , for 1 < i < n.Proof. We give a proof by induction on n. By

Proposition 6.2.1, the result is true for n = 1. If theresult is assumed to be true for the case n — 1, thenlet E = K (u1 ,u2 ,…,un-1). Since

K (u1 ,u2 ,…,un-1,un)=E (un),

the desired conclusion will follow from the equality[ E (un):K ] =[ E (un): E ] [ E :K ]

if we can show that [ E (un): E ] is less than the degreeof un over K. Now the minimal polynomial of un

over K is in particular a polynomial over E, and thusthe minimal polynomial of un over E must be adivisor of it. Applying Proposition 6.2.1 completes

the proof. □

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Corollary 3.9 Let F be an extension field of K .The set of all elements of F that are algebraic over K forms a subfield of F .

Proof. If u, v are algebraic elements of F, thenK (u,v) is a finite extension of K by Theorem 6.2.4.Since u + v, u — v, and uv all belong to K (u,v), theseelements are algebraic by Proposition 6.2.3. Thesame argument applies to u/v, if v¹0. □

Definition 3.10 An extension field F of K is

said to be algebraic over K if each element of F isalgebraic over K .

Proposition 3.11 Every finite extension is analgebraic extension.

Proof. This follow immediately from Proposition

6.2.3The following example shows that an algebraicextension need not be a finite extension.

Example 3.12 (Algebraic numbers)Let C be the set of complex numbers uÎC such thatu is algebraic over Q. Then Q is a subfield of C byCorollary 6.2.7, called the field of algebraic numbers.We showed in Corollary 4.3.7 that for any prime p

the polynomial 1+ x +... +x p-2

+x p-1 is irreducible over

Q. The roots of this polynomial exist in C (they arethe primitive pth roots of unity) and thus C contains

algebraic numbers of arbitrarily large degree over Q,which shows that it cannot have finite degree over Q.

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Thus C is an algebraic extension of Q, but not afinite extension.

Proposition 3.13 Let F be an algebraicextension of E and let E be an algebraic extensionof K . Then F is an algebraic extension of K .

Proof. If F is algebraic over E, then any elementuÎF must satisfy some nonzero polynomial f ( x)=a0+a1 x+...+an x

n over E. Since each element ai

is algebraic over K, for 1 < i < n, it follows that

K (a1 ,a2 ,.. .,an ,u) is a finite extension of K, and so uis algebraic over K by Proposition 6.2.3.

EXERCISES 1. Find the degree and a basis for each of thegiven field extensions:

(a) )3(Q over Q

(b) )7,3(Q over Q

(c) )73( +Q over Q

(d) )22( +Q over Q

(e) )(v Q over Q, where 2

31 i+-=v

2. Find the degree and a basis for each of the

given field extensions:(a) )7,3(Q over )3(Q

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(b) )7,5(3Q over )7(Q

(c) )7,3(Q over )73( +Q ]

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