Math 330, Abstract Algebra I Solutions to Homework 8 Problems

Chapter 13. Problem 14. Show that the nilpotent elements of a commutative ring forma subring.

Solution: Let S be the set of nilpotent elements of a commutative ring. Since 0 ∈ S, S isnonempty. So, suppose a, b ∈ S, then there exist positive integers h and k with ah = bk = 0.Notice that if bk = 0, (−b)k = 0 since for even k the two are equal and for odd k, they areadditive inverses (and the additive inverse of 0 is 0). Now, consider (a− b)h+k :

(a− b)h+k =h+k∑i=0

(h + k

i

)ah+k−i(−b)i

=k∑

i=0

(h + k

i

)ah+k−i(−b)i +

h+k∑i=k+1

(h + k

i

)ah+k−i(−b)i

Since ah = 0 and (−b)k = 0, and each term in the first summation is the product of ah withsome other ring element, and each term in the second summation is the product of (−b)k

with some other ring element, each term in each sum is equal to 0, as is this sum. Hence,(a− b) ∈ S.

Now, let a and b be as above. Since R is a commutative ring, we can write (ab)h = ahbh =0 · bh = 0 = bh · 0 = bhah = (ba)h, hence ba = ab ∈ S. So, by the subring test, S is a subringof R.

Chapter 13. Problem 42. Let R be a commutative ring with unity 1 and prime charac-teristic. If a ∈ R is nilpotent, prove that there is a positive integer k such that (1 + a)k = 1.

Solution: Let R be as specified above, and let p be the characteristic of R. Let a be anilpotent element of R with an = 0. Since p is prime, p > 1, and so we can find somepositive integer i with pi > n. Then, by Problem 41 part b) we have (1 + a)pi

= 1pi+ api

=1 + anapi−n = 1 + 0 = 1.

Chapter 14. Problem 34. Let R be a ring and let I be an ideal of R. Prove that thefactor ring R/I is commutative if and only if rs− sr ∈ I for all r and s in R.

Solution: (⇒) Suppose R/I is a commutative ring. Let r, s ∈ R. Then rs+ I = (r + I)(s+I) = (s + I)(r + I) = sr + I, and adding the additive inverse of sr to both sides, we havers− sr + I = sr − sr + I = 0 + I = I. Hence, rs− sr ∈ I.

(⇐) Let r + I and s + I be arbitrary elements of R/I. Now, suppose rs− sr ∈ I. Thenrs− sr + I = I, and rs+ I = sr + I. But rs+ I = (r + I)(s+ I) and sr + I = (s+ I)(r + I),so we have (r + I)(s+ I) = (s+ I)(r + I) for these arbitrary elements of the ring R/I, henceR/I is commutative.

Chapter 14. Problem 42. Let R be a commutative ring and let A be any ideal of R.Show that the nil radical of A, N(A) = {r ∈ R|∃n ∈ Z s.t. rn ∈ A}, is an ideal of R.

Solution: Notice that any element of A will be an element of N(A), hence N(A) is nonempty.Now, suppose a, b ∈ N(A). Then there exist positive integers h and k with ah, bk ∈ A. Notice

Math 330, Abstract Algebra I Solutions to Homework 8 Problems

that if bk ∈ A, so is (−b)k since for even k, (−b)k = bk and for odd k, (−b)k = −bk, and Ais closed under additive inverses. Now, consider (a− b)h+k:

(a− b)h+k =h+k∑i=0

(h + k

i

)ah+k−i(−b)i

=k∑

i=0

(h + k

i

)ah+k−i(−b)i +

h+k∑i=k+1

(h + k

i

)ah+k−i(−b)i

Notice that every term in the first summation is the product of ah with some other ringelement and every term in the second summation is the product of (−b)k with some otherring element. Since A is an ideal of R and we know ah ∈ A and (−b)k ∈ A, we know thatevery term in each summation is contained in A, hence the sum itself is contained in A.That is, (a− b)h+k ∈ A, hence a− b ∈ N(A).

Now, for a ∈ N(A) and some arbitrary element r ∈ R, we know that there is somepositive integer h with ah ∈ A. Since R is a commutative ring, we can write (ar)h = ahrh =rhah = (ra)h, which we know is an element of A, since ah ∈ A and A is an ideal. So, by theideal test N(A) is an ideal of R.

Chapter 14. Problem 56. Let R be a commutative ring with unity and let I be a properideal with the property that every element of R that is not in I is a unit of R. Prove that Iis the unique maximal ideal of R.

Solution: Let R and I be given as above. Suppose A is an ideal of R with I ⊂ A ⊆ R (sothat I is properly contained in A). Then there is some element a ∈ A, a 6∈ I. Since a 6∈ I, amust be a unit. So there is some a−1 ∈ R, and the product of this ring element with a mustbe contained in the ideal A: a · a−1 = 1 ∈ A. Since 1 is an element of A, A = R, hence I ismaximal.

Now, suppose B 6= I is a maximal ideal of R. As B is maximal and B 6= I, B is not asubset of I (for then B ( I ( R implies that B is not maximal). Thus there must be anelement b ∈ B so that b 6∈ I, hence b is a unit. By the above argument, we see that B mustin fact be all of R, as B contains 1. Hence, B is not actually a maximal ideal, and I is, infact, the unique maximal ideal of R.