Homework 5 Solution. Math 113 Summer 2016

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) √ (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 ∈ 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. √ 3. Prove that Q[ 2] is a field. (Hint: think about division in C!) √ √ Solution: Recall that Q[ 2] = {a + b 2 | a, b ∈ Q} ⊂ R. We already know√ it’s a ring, so just have to check that nonzero elements are units. If 0 6= x ∈ Q[ 2] and x ∈ Q (ie b = 0), then x has a√ multiplicative inverse, as you learned in elementary school. If b 6= 0, so that x = a + b 2, then define √ a − b 2 y = . a2 − 2b2 √ √ 2 2 This is a well-defined element of Q[ 2] (since a − 2b 6= 0 - else 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 ∈ R. Solution: Suppose that r 2 = r, for every r ∈ R. Then, (1 + 1)2 = 1 + 1, so that 1+1+1+1=1+1 =⇒ 1 + 1 = 0 ∈ R. Hence, −1 = 1 ∈ R and, for any r ∈ R, we have r + r = r(1 + 1) = 0, so that −r = r. Now, let a, b ∈ 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) {p(x, y) | p(x, x) = 0} ⊂ C[x, y] (b) {p(x, y) | p(x, y) = p(y, x)} ⊂ C[x, y] Solution: Denote the given set in each case by I .

(a) Ideal - if p, q ∈ I then (p − q)(x, x) = p(x, x) − q(x, x) = 0 − 0 = 0. Hence, p − q ∈ I . If r ∈ C[x], p ∈ I then (rp)(x, x) = r(x, x)p(x, x) = r(x, x)0 = 0, so that rp ∈ 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 ∈ 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 ∈ 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 ∈ I . Hence, as R is a field, x is a unit. Thus, 1 = x−1x ∈ 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 )i∈I is a family of subrings of C (ie, some arbitrary collection of T subrings of C). Prove that i∈I 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 ]) √ (d) Let Y = { 2} ⊂ 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 = {π} ⊂ 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 ∈ 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 ∈ R then n = 1 + ... + 1 ∈ R, for any n ∈ 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 ∈ R. Then, r, s ∈ Ri , for every i ∈ I . Hence, r + s ∈ Ri , for every i ∈ I , so that r + s ∈ R. Similarly, we can show that −r ∈ R and 0 ∈ R. Hence, R is a subgroup of (C, +). Moreover, rs ∈ Ri , for every i ∈ I , so that rs ∈ R, and 1 ∈ Ri , for every i ∈ I , so that 1 ∈ R. T (c) Let I = {S ⊂ C | Y ⊂ S, S subring}. Then, R = S∈I S is a subring of C, by 0 (b), and, as Y ⊂ S, for every S ∈ I , Y ⊂ R. Moreover, if S ⊂ C is a subring containing Y then S0 ∈ I , and R ⊂ S0. (d) Define the homomorphism of rings √ f : Z[x] → C;; x 7→ 2. √ Then, im f = Z[Y ], because Z[Y ] = {a + b 2 | a, b ∈ Z} -√ this is true because√ the given set is a subring (as can be easily verified) containing 2. Hence, {a + b 2} is contained in the√ set I described in (c). Moreover, since√ Z is a subring of every subring of C, and √2 ∈ Z[Y ], then we must have {a + b 2} ⊂ Z[Y ]. This√ implies that Z[Y ] = {a + b 2} since Z[Y ] is the smallest subring of C containing 2. P i If p = ai x ∈ ker f then √ √ √ √ X i 0 = p( 2) = ai ( 2) =⇒ (a0 +2a2 +4a4 +...)+ 2(a1 +2a3 +...) = a+b 2. √ √ √ Since √2 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 ∈ 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 = {a0 + a1π + ... + anπ | ai ∈ Z}, and ker f = 0 using the given fact - if p ∈ 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 ∈ R such that uv = 1 ∈ R, since 1 ∈ (u). Thus, u is a unit.