Number theory lecture

8/12/2019 Number theory lecture

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MATH 361: NUMBER THEORY TENTH LECTURE

The subject of this lecture is nite elds.

1. Root Fields

Let k be any eld, and let f (X ) k [X ] be irreducible and have positive degree.We want to construct a supereld K of k in which f has a root. To do so, considerthe quotient ring

R = k [X ]/ f ,

where f is the principal ideal f (X )k [X ] of k[X ]. That is, R is the usual ringof polynomials over k subject to the additional rule f (X ) = 0. Specically, thering-elements are cosets and the operations are

(g + f ) + ( h + f ) = ( g + h) + f ,(g + f )(h + f ) = gh + f .

The fact that f is irreducible gives R the structure of a eld, not only a ring. Theonly matter in question is multiplicative inverses. To see that they exist, supposethat

g + f = f in R.

The condition is that g / f , i.e., f g. so (f, g ) = 1 (because f is irreducible),and so there exist F, G k [X ] such that

F f + Gg = 1 .

That is, f | Gg 1, i.e., Gg 1 f , so that

Gg + f = 1 + f in R.

That is,

(G + f )(g + f ) = 1 + f in R,

showing that G + f inverts g + f in R.Now use the eld R to create a set K of symbols that is a superset of k and is

in bijective correspondence with R. That is, there is a bijection : R K , (a + f ) = a for all a k .

Endow K with addition and multiplication operations that turn the set bijectioninto a eld isomorphism. The operations on K thus extend the operations on k .Name a particular element of K ,

r = (X + f ).1


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2 MATH 361: NUMBER THEORY TENTH LECTURE

Then

f (r ) = f ((X + f )) by denition of r

= (f (X + f )) since algebra passes through = (f (X ) + f ) by the nature of algebra in R= ( f ) by the nature of algebra in R= 0 by construction of .

Thus K is a supereld of k containing an element r such that f (r ) = 0.

For example, since the polynomial f (X ) = X 3 2 is irreducible over Q , thecorresponding quotient ring

R = Q [X ]/ X 3 2 = {a + bX + cX 2 + X 3 2 : a,b, c Q }

is a eld. And from R we construct a eld (denoted Q (r ) or Q [r ]) such that r 3 = 2.Yes, we know that there exist cube roots of 2 in the supereld C of Q , but the

construction given here is purely algebraic and makes no assumptions about thenature of the starting eld k to which we want to adjoin a root of a polynomial.

2. Splitting Fields

Again let k be a eld and consider a nonunit polynomial f (X ) k [X ]. We canconstruct an extension eld

k 1 = k (r 1),where r 1 satises some irreducible factor of f . Let

f 2(X ) = f (X )/ (X r 1) k 1[X ].

We can construct an extension eld

k 2 = k 1(r 2) = k (r 1 , r 2),

where r2 satises some irreducible factor of f 2 . Continue in this fashion untilreaching a eld where the original polynomial f factors down to linear terms. Theresulting eld is the splitting eld of f over k , denoted

splk (f ).

Continuing the example of the previous section, compute thatX 3 2X r

= X 2 + rX + r2 in Q (r )[X ].

Let s = rt where t3 = 1 but t = 1. Then, working in Q (r, t ) we have

s2 + rs + r2 = r 2 t2 + r2 t + r2 = r 2(t2 + t + 1) = r 2 0 = 0 ,

Thus s = rt satises the polynomial X 2

+ rX + r2

, and now compute thatX 2 + rX + r2

X rt = X rt 2 in Q (r, t )[X ].

That is,X 3 2 = ( X r )(X rt )(X rt 2) Q (r, t )[X ],

showing thatsplQ (X 3 2) = Q (r, t ).


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MATH 361: NUMBER THEORY TENTH LECTURE 3

3. Examples of Finite Fields

We already know the nite elds

F p = Z /p Z , p prime.

For any positive integer n we can construct the extension eld

K = spl F p (X pn X ).

Furthermore, the roots of X pn

X in K form a subeld of K . To see this, checkthat if a p

n

= a and b pn

= b then

(ab) pn

= a pn

b pn

= ab,

(a + b) pn

= a pn

+ b pn

= a + b,

(a 1) pn

= ( a pn

) 1 = a 1 if a = 0 ,

( a) pn

= ( 1) pn

a pn

= a (even if p = 2) .

(The modulo p result that ( a + b) p = a p + b p is sometimes called the freshmans dream .) Thus the splitting eld K is exactly the roots of X p

n

X . The roots aredistinct because the derivative

(X pn

X ) = 1

is nonzero, precluding multiple roots. Thus K contains pn elements. We give it aname,

Fq = spl F p (X pn X ) where q = pn .

4. Exhaustiveness of the Examples

In fact the elds

Fq , q = pn , n 1

are the only nite elds, up to isomorphism. To see this, let K be a nite eld.The natural homomorphism

Z K , n n 1K

has for its kernel an ideal I = nZ of Z such that

Z /I K ,

and so Z /I is a nite integral domain. This forces I = pZ for some prime p, and so

F p K .

Identify F p with its image in K . Then K is a nite-dimensional vector space over F p,so that |K | = pn for some n. Every nonzero element x K satises the conditionx p

n 1 = 1, and so every element x K satises the condition x pn

= x. In sum,K = Fq up to isomorphism where again q = pn .


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MATH 361: NUMBER THEORY TENTH LECTURE 5

The polynomial X 4 + X + 1 works: it has no roots, and it doesnt factor into twoquadratic terms because

(X 2

+ aX + 1)( X 2

+ bX + 1) = X 4

+ ( a + b)X 3

+ abX 2

+ ( a + b)X + 1 .Thus, again up to isomorphism,

F16 = F2[X ]/ X 4 + X + 1

= F2(r ) where r 4 + r + 1 = 0

= {a + br + cr2 + dr 3 : a,b,c, d F2}.

8. A Remarkable Polynomial Factorization

In general, to construct the eld Fq where q = pn , we need an irreducible monicpolynomial of degree n over F p. Are there such? How do we nd them?

Fix a prime p and a positive integer n . For any d 1, let MI p(d) denote the setof monic irreducible polynomials of degree d over F p. Then

X pn

X =d| n f MI p (d)

f (X ) in F p[X ].

To establish the identity, consider any monic irreducible polynomial f , and let ddenote its degree. We want to show that f (X ) | X p

n

X if and only if d | n.Working in a eld large enough to contain all roots of f and all roots of X p

n

X ,we have

f (X ) | X pn

X in F p[X ] each root of f (X ) is a root of X pn

X.

(The = direction is a little subtle, because conceivably f could have repeatroots. Indeed, there do exist so-called inseparable scenarios where the roots of anirreducible polynomial over a eld do repeat. But in our context, if even one root aof f is a root of X p

n

X then f (X ) | X pn

X . To see this, substitute a for X inthe relation X p

n

X = q (X )f (X ) + r(X ) (where r = 0 or deg( r ) < deg f ) to getr (a) = 0. Thus r = 0 since otherwise a satises some irreducible factor of r alongwith satisfying the irreducible f , but a can not satisfy two distinct irreducible poly-nomials because then it would satisfy their gcd, which is the constant polynomial 1.Since r = 0, f (X ) | X p

n

X as desired.) Recall that the set of roots of X pn

X is precisely F pn . Also, up to isomorphism, if we adjoin any root of f to F p then weget a eld F p(r ) = F p[X ]/ f = F pd (recall that d = deg( f )). Thus

f (X ) | X pn

X in F p[X ] F pd F pn .

And so, by the nature of nite eld containments, we have shown exactly what wewant: For any monic irreducible polynomial f over F p, letting d denote its degree,

f (X ) | X pn

X in F p[X ] d | n.

To count the monic irreducible polynomials over F p of a given degree, take thedegrees of both sides of the identity

X pn

X =d| n f MI p (d)

f (X ) in F p[X ]

to get pn =

d| n

d |MI p(d)|.


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6 MATH 361: NUMBER THEORY TENTH LECTURE

By Mobius inversion,

|MI p(n)| = 1

n d| n(n/d ) pd (where is the Mobius function) .

The sum on the right side is positive because it is a base- p expansion with topterm pn and the coefficients of the lower powers of p all in {0, 1}. So |MI p(n)| > 0for all n > 0. That is, there do exist monic irreducible polynomials of every degreeover every eld F p.

For example, taking p = 2 and n = 3,

X 8 X = X (X 7 1) = X (X 1)(X 6 + X 5 + X 4 + X 3 + X 2 + X + 1)

= X (X 1)(X 3 + X 2 + 1)( X 3 + X + 1) mod 2 ,a product of two linear factors and two cubic factors. And the counting formulafrom the previous section gives the results that it must,

|MI2(1) | = 11(1)2

1= 2 ,

|MI2(3) | = 13

((1)23 + (3)2 1) = 13

(8 2) = 2 .

Similarly, taking p = 3 and n = 2,

X 9 X = X (X 8 1) = X (X 4 1)(X 4 + 1)

= X (X 2 1)(X 2 + 1)( X 2 + X + 1)( X 2 X + 1) mod 3

= X (X 1)(X + 1)( X 2 + 1)( X 2 + X 2 + 1)( X 2 + X + 1) ,and

|MI3(1) | = 11

(1)31 = 3 ,

|MI3(2) | = 12((1)3

2

+ (3)31

) = 12 (9 3) = 3 .

9. Common Errors

The nite eld Fq where q = pn is neither of the algebraic structures Z/q Zand ( Z /p Z )n as a ring. As a vector space, Fq = F n p , but the ring (multiplicative)structure of F q is not that of Z/q Z or of (Z /p Z )n .

The nite eld F pm is not a subeld of F pn unless m | n, in which case it is.

10. Gauss Sums Again

Let p and q be odd primes.Working in Fq2 , the multiplicative generator g has order q 2 1 = 0 mod 8.

Consequently, g(q2 1) / 8 has order 8 and can serve as 8 in the Gauss sum calculation

of (2/q ).Working in Fqp 1 , the multiplicative generator g has order q p 1 1 = 0 mod p.Consequently, g(q

p 1 1) /p has order p and can serve as p in the Gauss sum proof of the main quadratic reciprocity law.

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Number theory lecture