ALG2 HW1

Evan NashMATH 818Homework 1 Bin Chen

7.4.30 Let I be an ideal of the commutative ring R and define

rad I = {r ∈ R | rn ∈ I for some n ∈ Z+}

called the radical of I. Prove that rad I is an ideal containing I.

Proof. We first note that if i ∈ I is arbitrary, then i1 ∈ I, so indeed I ⊆ rad I.Next, we shall show that rad I is a subring of R. Because I ⊆ rad I, rad I isnon-empty, so we must demonstrate that it is closed under addition, additiveinverses and multiplication.

Let a, b ∈ rad I be arbitrary such that am ∈ I and bn ∈ I. We first seek toshow that (a + b)k ∈ I for some k ∈ Z+. I claim that k = m + n satisfies this.To see this, we consider the expansion:

(a + b)m+n =m+n∑i=0

(m + n

i

)aibm+n−i

=

(n∑i=0

(m + n

m + i

)aibn−i

)am +

(m∑j=0

(m + n

n + j

)am−jbj

)bn

Because am, bn ∈ I and I is an ideal, the terms(∑n

i=0

(m+nm+i

)aibn−i

)am and(∑m

j=0

(m+nn+j

)am−jbj

)bn are both in I. As I is closed under addition, we there-

fore have that (a + b)m+n ∈ I and so rad I is closed under addition.

We must next show that if a ∈ rad I then −a ∈ rad I. If am ∈ I, then(−a)2m = (−1)2ma2m = a2m. Then a2m = (am)2 ∈ I as I has closure. Thus,rad I is closed under additive inverses.

To finish the proof, we shall show that r (rad I) ⊆ rad I for arbitrary r ∈ R.This will jointly demonstrate that rad I is closed under multiplication and thatrad I is an ideal because R is commutative. We let a ∈ rad I be arbitrary witham ∈ I and consider ra. Observe the following algebra:

(ra)m = rmam ∈ rmI ⊆ I

This demonstrates that ra ∈ rad I and so rad I is an ideal of R.

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7.4.32 Let I be an ideal of the commutative ring R and define

Jac I to be the intersection of all maximal ideals of R that contain I

where the convention is that Jac R = R.

a) Prove that Jac I is an ideal of R containing I.

Proof. We let {Mi} be the set of maximal ideals of R that contain I so thatJac I =

⋂iMi. We must demonstrate that Jac I is nonempty, closed un-

der subtraction and multiplication and that r(Jac I) ⊆ Jac I for all r ∈ R.Certainly Jac I is nonempty as it must contain I, so we let a, b ∈ Jac I bearbitrary. We let M be an arbitrary maximal ideal of R that contains I. ThenJac I ⊆M , so a, b ∈M . Because M is an ideal, it is closed under subtraction,so a− b ∈M . As M was arbitrary, this means that a− b ∈Mi for all i. Thus,a− b ∈

⋂iMi = Jac I and so Jac I is closed under subtraction.

We let r ∈ R be arbitrary and seek to show that r(Jac I) ⊆ Jac I. This willdemonstrate that Jac I is closed under multiplication and that Jac R is anideal as R is commutative. We let a ∈ Jac I and r ∈ R be arbitrary. Asa ∈ Jac I, we know that a ∈ M , so ra ∈ M as M is an ideal. Because M wasarbitrary, it must be the case that ra ∈ Mi for all i, so ra ∈ Jac I and thusr(Jac I) ⊆ Jac I and the claim is proven.

b) Prove that rad I ⊆ Jac I.

Proof. Let Jac I =⋂iMi for maximal ideals Mi ⊂ R. Then suppose r ∈ rad I

is arbitrary so that rm ∈ I for some m ∈ Z+. Consider some arbitrary M inthe set of maximal ideals that contain I. If we can show that r ∈ M , thenrad I ⊆ Mi for all Mi as M was arbitrary, so rad I ⊆ Jac I and we will bedone. We shall assume that r /∈M and show that this delivers a contradiction.

Because rm ∈ I and I ⊆ Jac I ⊆ M , we know rm ∈ M . By Corollary 14, wealso have that M is a prime ideal of R as R is commutative. Then becauserm ∈ M , we know that either r ∈ M or rm−1 ∈ M . But because r /∈ M ,we must have rm−1 ∈ M . But then we observe once again that either r ∈ Mor rm−2 ∈ M and we may follow a similar line of reasoning. After m − 2iterations of this argument, we are left with the case that r2 ∈ M , whichmeans it is unavoidable that r is in M . This contraction shows that r ∈M , sorad I ⊆ Jac I and the claim is proven.

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c) Let n > 1 be an integer. Describe Jac nZ in terms of the prime factorizationof n.

We note that the maximal ideals of Z are precisely those subrings of the formpZ for p a prime. This is because any subring of Z is of the form nZ and theonly subrings that will contain pZ must be of the form nZ where n divides p.The only possibilities are pZ and Z itself, so pZ is indeed a maximal ideal.

Thus, if n = pα11 pα2

2 · · · pαkk for distinct primes pi, we set m = p1p2 · pk. I claim

that mZ is Jac nZ. We note that every maximal ideal of Z that contains nZwill be of the form piZ. Thus, the intersection of all of these maximal idealswill be p1p2 · · · pkZ = mZ as claimed.

7.4.37 A commutative ring R is called a local ring if it has a unique maximal ideal.Prove that if R is a local ring with maximal ideal M then every element ofR −M is a unit. Prove conversely that if R is a commutative ring with 1 inwhich the set of nonunits forms an ideal M , then R is a local ring with uniquemaximal ideal M .

Proof. Let R be a local ring and suppose M is the unique maximal ideal ofR. Suppose that there exists some r /∈ M such that r is a nonunit. Then (r),the ideal generated by r, is not contained in M . But because r is not a unit,there is no element a ∈ R with ar = 1, so 1 /∈ (r). Thus (r) 6= R, so (r) is aproper ideal of R. By Proposition 11, (r) must therefore be contained in somemaximal ideal of R as R has 1. But (r) * M , so this implies that there issome maximal ideal of R other than M . This defies our assumption that M isunique, so every element of R not in M must be a unit, proving the claim.

Conversely, suppose that R is a commutative ring with 1 in which the setof nonunits forms an ideal M . We note by Proposition 9 that every idealcontaining a unit equals R, so every proper ideal of R must have only nonunits.Thus, every proper ideal of R is contained in M , so M must be the uniquemaximal ideal of R.

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7.5.3 Let F be a field. Prove that F contains a unique smallest subfield F0 and thatF0 is isomorphic to either Q or Z/pZ for some prime p.

Proof. We first show that F contains a unique smallest subfield. Let {Si} bethe set of subfields of F and let F0 =

⋂i Si. I claim that F0 is a subfield of

F . As every subfield must contain 1, 1 ∈ F0 so we let a, b ∈ F0 be arbitraryand seek to show that a − b ∈ F0 and ab−1 ∈ F0. This will demonstrate thatF0 is a subring with 1 that is closed under multiplicative inverses, making it asubfield. If S ∈ {Si} is arbitrary, then a, b ∈ S, so a− b ∈ S and ab−1 ∈ S as Sis a subfield. Because S was arbitrary, we have that a− b ∈ Si and ab−1 ∈ Sifor all i and so a− b ∈ F0 and ab−1 ∈ F0, confirming that F0 is a subfield of F .As F0 is contained in every subfield of F , it must also be the unique smallestsubfield of F .

To show the second part of the claim, we consider the characteristic of F . Iclaim that this value is either 0 or p for some prime p. It is certainly possiblethat the characteristic of F is 0 as the field Q fulfills this property. We thereforeconsider the possibility that the characteristic of F is n for some compositenumber n so that n1 = 0. Then n = ab and so ab1 = 0, which implies thata1 · b1 = 0. But because a, b < n, we know that a1 6= 0 and b1 6= 0, whichimplies that F contains zerodivisors, which is impossible as F is a field. Thus,if the characteristic of F is not zero, it must be prime.

We then observe that 1 is contained in every subfield of F and so because asubfield must be closed under addition and additive inverses, every subfieldof F must contain this set of all possible sums of 1 and its additive inverse−1. We will notate this set of elements of the form n1 with n ∈ Z as (1). Inthe case that the characteristic of F is p, the map ϕ : (1) → Z/pZ definedby ϕ(a1) = a (mod p) can be easily shown to be an isomorphism. If thecharacteristic is zero, the map ϕ : (1) → Z defined by ϕ(a1) = a is also anisomorphism. Thus, if the characteristic of F is p, then (1) is a subfield as Z/pZis a field for all prime p. This field will be contained in every other subfieldand so it is the smallest subfield of F . If the characteristic is zero, then (1) isisomorphic to Z, and Corollary 16 says that the smallest field that contains Z isits field of fractions, Q, which will similarly be contained in all other subfieldsof F . Thus, the smallest subfield of F is isomorphic to either Q or Z/pZ forsome prime p.

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7.5.6 Prove that the real numbers, R, contain a subring A with 1 ∈ A and A maximal(under inclusion) with respect to the property that 1

2/∈ A.

Proof. We shall prove this through use of Zorn’s Lemma. Let S be the set ofall subrings of R with 1 that do not contain 1

2. We know that S is non-empty

as Z is a subring of R with 1 that does not contain 12. Then let C = {Ri} be

an arbitrary chain in S ordered by inclusion. I claim that⋃iRi is a subring of

R with 1 not containing 12.

To show this, we let a, b ∈⋃iRi arbitrary so that a ∈ Ra and b ∈ Rb for some

Ra, Rb ∈ C. Because C is ordered by inclusion, we may assume without loss ofgenerality that Rb ⊆ Ra so that a, b ∈ Ra. Then because Ra is a subring of R,we know that a− b ∈ Ra and ab ∈ Ra. This further implies that a− b ∈

⋃iRi

and ab ∈⋃iRi, so

⋃iRi is a subring of R. Because 1 ∈ Ri and 1

2/∈ Ri for all i,

we also know that 1 ∈⋃iRi and 1

2/∈⋃iRi. Then

⋃iRi is a maximal subring

of the chain C with 1 and not containing 12, and so Zorn’s Lemma tells us that

R must contain a maximal subring with 1 that does not contain 12.

8.1.3 Let R be a Euclidean Domain. Let m be the minimum integer in the set ofnorms of nonzero elements of R. Prove that every nonzero element of R ofnorm m is a unit. Deduce that a nonzero element of norm zero (if such anelement exists) is a unit.

Proof. Let a ∈ R be an element of norm m. Because R is a Euclidean Domain,there exist some q, r ∈ R such that 1 = qa+ r and r = 0 or N(r) < N(a) = m.If r is nonzero, then N(r) < m is impossible as m is the minimum integer inthe set of norms of nonzero elements of R. Thus, r must equal zero and so1 = qa for some q ∈ R, which means that a is a unit.

If a ∈ R is a nonzero element with norm zero, then its norm must be theminimum integer in the set of norms of nonzero elements as norms are bydefinition non-negative. The proof above therefore shows that a must be aunit, as claimed.

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8.1.10 Prove that the quotient ring Z[i]/I is finite for any nonzero ideal I of Z[i].

Proof. We first note that Z[i] is a Euclidean Domain through use of the normN(a + bi) = a2 + b2. Then Proposition 1 tells us that every ideal I of Z[i] isprincipal so that I = (d) for some nonzero d ∈ I of minimum norm. Thenconsider some g ∈ Z[i] with N(g) > N(d). If g = dq for some q ∈ Z[i], theng ∈ I, so we shall assume that g = dq + r for some nonzero r ∈ Z[i] such thatN(r) < N(d). Then g − r = dq, so g − r ∈ I. This implies that g − r + I = I,so g + I = r + I. Thus, every element of Z[i] is either in I or is in the samecoset of I as some element of Z[i] with norm less than that of d. But thereare only a finite number of elements in Z[i] with norm less than d, so there areonly a finite number of cosets of I in Z[i]. Thus, Z[i]/I is finite and the claimis proven.

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ALG2 HW1