## Chapter 5: Commutative Rings: Solved problems

1. Let R be the ring with 8 elements consisting of all 3 × 3 matrices with entries in Z2 which have the following form:

You may assume that the standard laws for addition and multiplication of matrices are valid.

(a) Show that R is a commutative ring (you only need to check closure and commutativity of multiplication).

Solution: It is clear that the set is closed under addition, and the following computation checks closure under multiplication.

Because of the symmetry
a <--> x,
b <--> y,
c <--> z,
the above computation also checks commutativity.

(b) Find all units of R, and all nilpotent elements of R.

Solution: Four of the matrices in R have 1's on the diagonal, and these are invertible since their determinant is nonzero. Squaring each of the other four matrices gives the zero matrix, and so they are nilpotent.

(c) Find all idempotent elements of R.

Solution: By part (b), an element in R is either a unit or nilpotent. The only unit that is idempotent is the identity matrix (in a group, the only idempotent element is the identity) and the only nilpotent element that is also idempotent is the zero matrix.

2. Let R be the ring Z2 [x] / < x2 + 1 >. Show that although R has 4 elements, it is not isomorphic to either of the rings Z4 or Z2 Z2.

Solution: In R we have a+a = 0, for all a in R, so R is not isomorphic to Z4. On the other hand, in R we have [x+1] [0] but [x+1]2 = [x2+1] = [0]. Thus R cannot be isomorphic to Z2 Z2, since in that ring (a,b)2 = (0,0) implies a2 = 0 and b2 = 0, and this implies a = 0 and b = 0 since Z2 is a field.

3. Find all ring homomorphisms from Z120 into Z42.

Solution: Let : Z120 -> Z42 be a ring homomorphism. The additive order of (1) must be a divisor of gcd (120,42) = 6, so it must belong to the subgroup 7Z42 = { 0,7,14,21,28,35 }. Furthermore, (1) must be idempotent, and it can be checked that in 7Z42, only 0,7,21,28 are idempotent.

If (1) = 7, then the image is 7Z42 and the kernel is 6Z120. If (1) = 21, then the image is 21Z42 and the kernel is 2Z120. If (1) = 28, then the image is 14Z42 and the kernel is 3Z120.

4. Are Z9 and Z3 Z3 isomorphic as rings?

Solution: The answer is no. The argument can be given using either addition or multiplication. Addition in the two rings is different, since the additive group of Z9 is cyclic, while that of Z3 Z3 is not. Multiplication is also different, since in Z9 there is a nonzero solution to the equation x2 = 0, while in Z3 Z3 there is not. (In Z9 let x = 3, while in Z3 Z3 the equation (a,b)2 = (0,0) implies a2 = 0 and b2 = 0, and then a = 0 and b = 0.)

5. In the group Z180× of units of the ring Z180, what is the largest possible order of an element?

Solution: Since 180 = 22 32 5, it follows from Theorem 3.5.4 that the ring Z180 is isomorphic to the ring Z4 Z9 Z5. Then Example 5.2.10 shows that
Z180× Z4× × Z9× × Z5× Z2 × Z6 × Z4 .
In the latter additive group, the order of an element is the least common multiple of the orders of its components. It follows that the largest possible order of an element is lcm [2,6,4] = 12.

6. For the element a = (0,2) of the ring R = Z12 Z8, find Ann (a) = { r in R | ra=0 }. Show that Ann (a) is an ideal of R.

Solution: We need to solve (x,y)(0,2) = (0,0) for (x,y) in Z12 Z8. We only need 2y 0 (mod 8), so the first component x can be any element of Z12, while y = 0,4. Thus Ann ((0,2)) = Z12 4 Z8. This set is certainly closed under addition, and it is also closed under multiplication by any element of R since 4 Z8 is an ideal of Z8.

7. Let R be the ring Z2 [x] / < x4 + 1 >, and let I be the set of all congruence classes in R of the form [ f(x) (x2 + 1)].

(a) Show that I is an ideal of R.

(b) Show that R/I Z2 [x] / < x2 + 1 >.

Hint: If you use the fundamental homomorphism theorem, you can do the first two parts together.

Solution: Define : Z2 [x] / < x4 + 1 > -> Z2 [x] / < x2 + 1 > by (f(x) + < x4 + 1 >) = (f(x) + < x2 + 1 >). This mapping is well-defined since x2 + 1 is a factor of x4 +1 over Z2. It is not difficult to show that is an onto ring homomorphism, with kernel equal to I.

(c) Is I a prime ideal of R?

Solution: No: (x+1)(x+1) 0 (mod x2 + 1).

8. Find all maximal ideals, and all prime ideals, of Z36 = Z / 36Z.

Solution: If P is a prime ideal of Z36, then Z36 / P is a finite integral domain, so it is a field, and hence P is maximal. Thus we only need to find the maximal ideals of Z36. The lattice of ideals of Z36 is exactly the same as the lattice of subgroups, so the maximal ideals of Z36 correspond to the prime divisors of 36. The maximal ideals of Z36 are thus 2Z36 and 3Z36.

An alternate approach uses Proposition 5.3.7, which shows that there is a one-to-one correspondence between the ideals of Z / 36Z and the ideals of Z that contain 36Z. In Z, every ideal is principal, so the relevant ideals correspond to the divisors of 36. Again, the maximal ideals that contain 36Z are 2Z and 3Z, and these correspond to 2Z36 and 3Z36.

9. Give an example to show that the set of all zero divisors of a ring need not be an ideal of the ring.

Solution: The elements (1,0) and (0,1) of Z × Z are zero divisors, but if the set of zero divisors were closed under addition it would include (1,1), an obvious contradiction.

10. Let I be the subset of Z[x] consisting of all polynomials with even coefficients. Prove that I is a prime ideal; prove that I is not maximal.

Solution: Define : Z[x] -> Z2 [x] by reducing coefficients modulo 2. This is an onto ring homomorphism with kernel I. Then R/I is isomorphic to Z2 [x], which is not a field, so I is not maximal.

11. Let R be any commutative ring with identity 1.

(a) Show that if e is an idempotent element of R, then 1-e is also idempotent.

Solution: We have (1-e)2 = (1-e)(1-e) = 1 -e -e + e2 = 1 -e -e + e = 1-e.

(b) Show that if e is idempotent, then R Re R(1-e).

Solution: Note that e(1-e) = e - e2 = e-e = 0. Define : R -> Re R(1-e) by (r) = (re,r(1-e)), for all r in R. Then is one-to-one since if (r) = (s), then re=se and r(1-e)=s(1-e), and adding the two equations gives r = s. Furthermore, is onto, since for any element (ae,b(1-e)) we have (ae,b(1-e)) = (r) for r = ae + b(1-e). Finally, it is easy to check that preserves addition, and for any r,s in R we have (rs) = (rse,rs(1-e)) and (r) (s) = (re,r(1-e))(se,s(1-e)) = (rse2,rs(1-e)2) = (rse,rs(1-e)).

12. Let R be the ring Z2 [x] / < x3 + 1 >.

Note: The multiplication table is given below. It is not necessary to compute the multiplication table in order to solve the problem.

(a) Find all ideals of R.

Solution: By Proposition 5.3.7, the ideals of R correspond to the ideals of Z2 [x] that contain < x3 + 1 >. We have the factorization x3 + 1 = x3 - 1 = (x-1)(x2+x+1), so the only proper, nonzero ideals are the principal ideals generated by [x+1] and [x2+x+1].

(b) Find the units of R.

Solution: We have [x]3 = [1], so [x] and [x2] are units. On the other hand, [x+1][x2+x+1] = [x3+1] =[0], so [x+1] and [x2+x+1] cannot be units. This also excludes [x2+x] = [x][x+1] and [x2+1] =[x2][1+x]. Thus the only units are 1, [x], and [x2].

(c) Find the idempotent elements of R.

Solution: Using the general fact that (a+b)2 = a2+2ab+b2 = a2+b2 (since Z2 [x] has characteristic 2) and the identities [x3] = [1] and [x4] = [x], it is easy to see that the idempotent elements of R are [0], [1], [x2+x+1], and [x2+x].

13. Let S be the ring Z2 [x] / < x3 + x >.

Note: Here is the multiplication table. It is not necessary to compute the multiplication table in order to solve the problem.

(a) Find all ideals of S.

Solution: Over Z2 we have the factorization x3 + x = x (x2+1) = x (x+1)2, so by Proposition 5.3.7, the proper nonzero ideals of S are the principal ideals generated by [x], [x+1], [x2+1] = [x+1]2, and [x2 + x] = [x][x+1].

< [x2+x] > = { [0], [x2+x] }
< [x2+1] > = { [0], [x2+1] }

< [x] > = { [0], [x], [x2], [x2+x] }
< [x+1] > = { [0], [x+1], [x2+1], [x2+x] }

(b) Find the units of R.

Solution: Since no unit can belong to a proper ideal, it follows from part (a) that we only need to check [x2+x+1]. This is a unit since [x2+x+1]2 = [1].

(c) Find the idempotent elements of R.

Solution: Since [x3] = [1], we have [x2]2 = [x2], and then [x2 + 1]2 = [x2+1]. These, together with [0] and [1], are the only idempotents.

14. Show that the rings R and S in Problem 12 and Problem 13 are isomorphic as abelian groups, but not as rings.

Solution: Both R and S are isomorphic to Z2 × Z2 × Z2, as abelian groups. They cannot be isomorphic as rings since R has 3 units, while S has only 2.

15. Let Z[i] be the subring of the field of complex numbers given by
Z[i] = { m+ni in C | m,n in Z } .
(a) Define : Z[i] -> Z2 by (m+ni) = [m+n]2. Prove that is a ring homomorphism. Find ker () and show that it is a principal ideal of Z[i].

Solution: We have the following computations, which show that is a ring homomorphism.

((a+bi)+(c+di)) = ((a+c)+(b+d)i) = [a+c+b+d]2
((a+bi))+ ((c+di)) = [a+b]2+[c+d]2 = [a+b+c+d]2

((a+bi)) ((c+di)) = [a+b]2 · [c+d]2 = [ac+ad+bc+bd]2.

We claim that ker () is generated by 1+i. It is clear that 1+i is in the kernel, and we note that (1-i)(1+i) = 2. Let m+ni be in ker() = { m+ni | m+n 0 (mod 2) }. Then m and n are either both even or both odd, and so it follows that m-n is always even. Therefore

m+ni = (m-n) + n + ni = (m-n) + n (1+i)
= ( (m-n)/(2) ) (1-i) (1+i) + n (1+i)
= [ (1)/(2) (m-n) (1-i) + n ] (1+i),

and so m+ni belongs to the principal ideal generated by 1+i.

(b) For any prime number p, define : Z[i] -> Zp [x] / < x2+1 > by (m+ni) = [ m+nx ] . Prove that is an onto ring homomorphism.

Solution: We have the following computations, which show that is a ring homomorphism. We need to use the fact that [x2] = [-1] in Zp [x] / < x2+1 >.

((a+bi)+(c+di)) = ((a+c)+(b+d)i) = [(a+c)+(b+d)x]
((a+bi))+ ((c+di)) = [a+bx]+[c+dx] = [(a+c)+(b+d)x]

((a+bi)) ((c+di)) = [a+bx] [c+dx] = [ac+(ad+bc)x+bdx2] .

Since the elements of Zp [x] / < x2+1 > all have the form [a+bx], for some congruence classes a and b in Zp, it is clear the is an onto function.

16. Let I and J be idealsin the commutative ring R, and define the function : R -> R/I R/J by (r) = (r+I,r+J), for all r in R.

(a) Show that is a ring homomorphism, with ker () = I J.

Solution: The fact that is a ring homomorphism follows immediately from the definitions of the operations in a direct sum and in a factor ring. Since the zero element of R/I R/J is (0+I,0+J), we have r in ker () if and only if r in I and r in J, so ker () = I J.

(b) Show that if I + J = R, then is onto, and thus R/(I J) R/I R/J.

Solution: If I+J = R, then we can write 1 = x + y, for some x in I and y in J. Given any element (a+I,b+J) in R/I R/J, consider r = bx + ay, noting that a = ax + ay and b = bx + by. We have a - r = a - bx - ay = ax - bx in I, and b - r = b - bx - ay = by - ay in J. Thus (r) = (a+I,b+J), and is onto. The isomorphism follows from the fundamental homomorphism theorem.

17. Considering Z[x] to be a subring of Q[x], show that these two integral domains have the same quotient field.

Solution: An element of the quotient field of Q[x] has the form (f(x))/(g(x)), for polynomials f(x) and g(x) with rational coefficients. If m is the lcm of the denominators of the coefficients of f(x) and n is the lcm of the denominators of the coefficients of g(x), then we have f(x)/g(x) = n/m h(x)(k(x) for h(x), k(x) in Z[x], and this shows that f(x)/g(x) belongs to the quotient field of Z [x].

18. Let p be an odd prime number that is not congruent to 1 modulo 4. Prove that the ring Zp[x] / < x2 + 1 > is a field.

Hint: Show that a root of x2 = -1 leads to an element of order 4 in the multiplicative group Zp×.

Solution: We must show that x2 + 1 is irreducible over Zp, or, equivalently, that x2 + 1 has no root in Zp. Suppose that a is a root of x2 + 1 in Zp. Then a2 -1 (mod p), and so a4 1 (mod p). The element a cannot be a root of x2 - 1, so it does not have order 2, and thus it must have order 4. By Lagrange's Theorem, this means that 4 is a divisor of the order of Zp×, which is p-1. Therefore p = 4q+1 for some q in Z, contradicting the assumption.