Homework 5 Solution. Math 113 Summer 2016. 1. Give examples of (commutative) subrings (with unity) R of C satisfying the following containments and non-containments. If it's not possible, explain why not. Note the symbol ⊂ means \proper subset"1 (a) Q ⊂ R, R ⊂ R (b) Q ⊂ R, R 6⊆ R. (c) Z 6⊆ R. Solution: (there are many possible correct answers) p (a) Could take R = Q[ 2] (b) Could take R = Q[i] (or whatever you chose for (a) willl work as well) (c) This is not possible - every subring of C contains Z. Since a subring must contain 1, and be closed under addition, it contains 1 + 1 + ··· + 1, so all positive integers are contained in the subring. Since it's closed under additive inversion, −1 must be in there as well, and hence (−1) + (−1) + ··· + (−1), so all negative integers are in there as well. 2. Let R = Z=nZ, n > 1, and k 2 R be a nontrivial element (meaning r 6= 0). Prove (a) k is a unit if and only if gcd(k, n) = 1. (b) k is a zerodivisor if and only if gcd(k, n) > 1. p 3. Prove that Q[ 2] is a field. (Hint: think about division in C!) p p Solution: Recall that Q[ 2] = fa + b 2 j a, b 2 Qg ⊂ R. We already knowp it's a ring, so just have to check that nonzero elements are units. If 0 6= x 2 Q[ 2] and x 2 Q (ie b = 0), then x has ap multiplicative inverse, as you learned in elementary school. If b 6= 0, so that x = a + b 2, then define p a − b 2 y = . a2 − 2b2 p p 2 2 This is a well-defined element of Q[ 2] (since a − 2b 6= 0 - else 2 2 Q), and you can check that xy = yx = 1. 4. Let R be a ring (not assumed commutative) in which every element is idempotent. Prove that R is in fact commutative, and satisfies r + r = 0 for every r 2 R. Solution: Suppose that r 2 = r, for every r 2 R. Then, (1 + 1)2 = 1 + 1, so that 1+1+1+1=1+1 =) 1 + 1 = 0 2 R. Hence, −1 = 1 2 R and, for any r 2 R, we have r + r = r(1 + 1) = 0, so that −r = r. Now, let a, b 2 R. Then, (a+b)2 = a+b =) a2 +ab +ba+b2 = a+b =) ab +ba = 0 =) ab = −ba = ba. Thus, R is commutative. 5. Decide which of the following sets are ideals in the given ring: 1For this problem, at least. (a) fp(x, y) j p(x, x) = 0g ⊂ C[x, y] (b) fp(x, y) j p(x, y) = p(y, x)g ⊂ C[x, y] Solution: Denote the given set in each case by I . (a) Ideal - if p, q 2 I then (p − q)(x, x) = p(x, x) − q(x, x) = 0 − 0 = 0. Hence, p − q 2 I . If r 2 C[x], p 2 I then (rp)(x, x) = r(x, x)p(x, x) = r(x, x)0 = 0, so that rp 2 I . (b) Not an ideal: xy is in the set, but if we scale it up by x, we get x2y, which is not in the set. 6. Let R be a commutative ring with unity. (a) Prove that Hom(Z, R) contains one element. (b) Give an example of a commutative ring with unity R such that Hom(R, Z) = ?. (c) Give en example of a commutative ring with unity R such that Hom(R, Z) is infinite. Solution: (a) The function n times i : Z :! R ; n 7!1R + ... + 1R is a ring homomorphism. If f 2 Hom(Z, R) is any other ring homomorphism then f (n) = f (1 + ... + 1) = f (1) + ... + f (1) = 1R + ... + 1R = i(n) =) i = f . Hence, there is exactly one ring homomorphism Z ! R. (b) R = Z=2Z - we would require that 0 = f (0) = f (1 + 1) = 1 + 1 = 2 2 Z, which is absurd. (c) Take R = Z[x]. Then, Hom(R, Z) is in bijection with Z, as we saw in class. 7. (a) Let R = C[x, y], I = (y − x). What ring is R=I isomorphic to? (b) Let R = Z[x], I = (3, x). What ring is R=I isomorphic to? Solution: (there are many possible correct answers) (a) C[x] (or C[y]) (b) Z=3Z 8. Prove that every field has exactly two ideals. What are they? Solution Assume R is a field. The only two ideals are (0), R - if I 6= (0) is an ideal then there is some nonzero x 2 I . Hence, as R is a field, x is a unit. Thus, 1 = x−1x 2 I and I = R. Note that (0) and R are actually distinct ideals, as otherwise R would be the zero ring, and this is prohibited by the definition of a field. 9. In this problem you will give a precise meaning to the process of `adjoining elements' described in the notes. (a) Prove that any subring R ⊂ C contains Z. (b) Suppose that (Ri )i2I is a family of subrings of C (ie, some arbitrary collection of T subrings of C). Prove that i2I Ri is a subring of C. 2 2 (c) Let Y ⊂ C be a subset. Prove that there exists a subring R ⊂ C containing Y with the following property: if S ⊂ C is a subring such that Y ⊂ S, then R ⊂ S. (The subring R just determined is often denoted Z[Y ]) p (d) Let Y = f 2g ⊂ C. Prove that there is an isomorphism of rings 2 ∼ Z[x]=(x − 2) = Z[Y ]. (Hint: Give an explicit description of Z[Y ]) (e) Let Y = fπg ⊂ C. Prove that there is an isomorphism of rings ∼ Z[x] = Z[Y ]. You will need to use the following fact: there is no polynomial f 2 Z[x] such that f (π) = 0. (This means that π is a transcendental number; it is a surprisingly hard result to prove.) Solution: (a) Since 1 2 R then n = 1 + ... + 1 2 R, for any n 2 Z>0. Hence, −n is also in R, since it is an additive subgroup of C. (b) Denote the given intersection R. Let r, s 2 R. Then, r, s 2 Ri , for every i 2 I . Hence, r + s 2 Ri , for every i 2 I , so that r + s 2 R. Similarly, we can show that −r 2 R and 0 2 R. Hence, R is a subgroup of (C, +). Moreover, rs 2 Ri , for every i 2 I , so that rs 2 R, and 1 2 Ri , for every i 2 I , so that 1 2 R. T (c) Let I = fS ⊂ C j Y ⊂ S, S subringg. Then, R = S2I S is a subring of C, by 0 (b), and, as Y ⊂ S, for every S 2 I , Y ⊂ R. Moreover, if S ⊂ C is a subring containing Y then S0 2 I , and R ⊂ S0. (d) Define the homomorphism of rings p f : Z[x] ! C;; x 7! 2. p Then, im f = Z[Y ], because Z[Y ] = fa + b 2 j a, b 2 Zg -p this is true becausep the given set is a subring (as can be easily verified) containing 2. Hence, fa + b 2g is contained in thep set I described in (c). Moreover, sincep Z is a subring of every subring of C, and p2 2 Z[Y ], then we must have fa + b 2g ⊂ Z[Y ]. Thisp implies that Z[Y ] = fa + b 2g since Z[Y ] is the smallest subring of C containing 2. P i If p = ai x 2 ker f then p p p p X i 0 = p( 2) = ai ( 2) =) (a0 +2a2 +4a4 +...)+ 2(a1 +2a3 +...) = a+b 2. p p p Since p2 is irrational, we must have a = b = 0. As p(− 2) = a − b 2, we see 2 that ± 2 are roots of p. Hence, p = (x − 2)q, for some q 2 Z[x]. This implies that ker f = (x2 − 2), and the result follows by the Isomorphism Theorem. (e) Define f : Z[x] ! C ; x 7! π. n Then, in a similar way as (d), we have that im f = fa0 + a1π + ... + anπ j ai 2 Zg, and ker f = 0 using the given fact - if p 2 ker f then p(π) = 0. 2This was erroneously omitted from the original version. 3 10. Prove that if an element u in a commutative ring R with unity is not contained in any proper ideal, then u must be a unit. Solution: If u is not contained in any proper ideal then the ideal (u) = R. Hence, there is some v 2 R such that uv = 1 2 R, since 1 2 (u). Thus, u is a unit. 4.