MATH 114: GALOIS THEORY SPRING 2008/09 PROBLEM SET 1 SOLUTIONS Throughout the problem set, unless stated otherwise, a ring will mean a commutative ring with unity 1 and will be denoted by R. We will assume that 0 6= 1 in R and will write R∗ for the set of units. Zn will denote the ring of integers modulo n. The ideal generated by a set S ⊆ R will be denoted by hSi. 1. Show that the following statements are equivalent: i. R is a field; ii. the only ideals in R are h0i and h1i; iii. for any ring R0, a homomorphism ' : R ! R0 is injective. Solution. (i) ) (ii): Let h0i= 6 I E R. Then I contains a nonzero element x; x is a unit since R is a field, hence I ⊇ hxi = R = h1i, hence I = h1i. (ii) ) (iii): Let ' : R ! R0 be a ring homomorphism. Then by our results in the lectures, ker ' is an ideal of R. Since '(1) = 1 6= 0, 1 2= ker ' and ker ' 6= h1i, hence ker ' = h0i and so ' is injective. (iii) ) (i): Let x 2 R and x2 = R∗. Then hxi 6= h1i. Let R0 := R=hxi and ' : R ! R0 be the map '(a) = a + R. It is easy to check that ' is a homomorphism, im ' = R0, and ker ' = hxi. By assumption, ' is injective, hence hxi = h0i, hence x = 0. n 2. We say that a 2 R is nilpotent if a = 0 for some n 2 N. (a) Show that the set of all nilpotent elements in R is an ideal. We will write N(R) for this set. It is called the nilradical of R. Solution. If a 2 N(R) and r 2 R, then since (ra)n = rnan, it is clear that ra 2 N(R). Let a; b 2 N(R); suppose am = 0, bn = 0. By the binomial expansion (which is valid in any commutative ring), m+n−1 X m + n − 1 (a + b)m+n−1 = akbm+n−k−1 k k=0 m−1 X m + n − 1 = bn akbm−1+k k k=0 m+n−1 X m + n − 1 + am ak−mbm+n−k−1 k k=m = 0 and so a + b 2 N(R). Hence N(R) is an ideal of R. (b) Find N(Z), N(Z12), and N(Z32). Solution. N(Z) = f0g, N(Z12) = f0; 6g, and N(Z32) = f0; 2; 4; 6; 8;:::; 30g. (c) Show that N(R=N(R)) = f0 + N(R)g (note that this set has only one element, namely, the coset 0 + N(R)). In other words, R=N(R) has no nilpotent elements other than its zero element. Date: February 17, 2009 (Version 1.0). 1 n Solution. Let a + N(R) 2 N(R=N(R)). So there exists n 2 N such that (a + N(R)) = n n n m 0 + N(R), so a + N(R) = 0 + N(R), so a 2 N(R), so (a ) = 0 for some m 2 N, so amn = 0, and so a 2 N(R). Hence a + N(R) = 0 + N(R). 3. Suppose R satisfies the following properties i. R has only one maximal ideal M; ii. R∗ = f1g. ∼ Show that R = Z2. Solution. Note that 1 2= M since M 6= R. Let a 2 1 + M. Then a2 = M (otherwise 1 = a − m 2 M) and so the ideal hai is not contained in M. Since every proper ideal must be contained in a maximal ideal, hai is not proper and so hai = R. Hence a is a unit. So we have shown that every element in 1 + M is a unit. Since f1g ⊆ 1 + M ⊆ R∗ = f1g, we conclude that 1 + M = f1g and so M = f0g. So R is a field. Since every nonzero element in a field is a unit, ∗ ∼ R = R [ f0g = f0; 1g = Z2. 4. If R has only one maximal ideal, then R is called a local ring. (a) Show that every field is a local ring. Solution. By our discussion in the lectures, every field has exacly one maximal ideal, namely, h0i. ∗ (b) Suppose M E R, M 6= R has the property that a2 = M implies a 2 R . Show that M is a maximal ideal and R is a local ring. Solution. Observe that every ideal I 6= R must contain only non-units (if it contains any units, it will be equal to R). Observe also that the given condition may be rephrased as a2 = R∗ implies a 2 M. Hence M contains every proper ideal of R. Hence M is the only maximal ideal of R. (c) Suppose M is a maximal ideal of R and has the property that a 2 1 + M implies a 2 R∗. Show that R is a local ring. Solution. Let a 2 R and a2 = M. Since M is maximal and the ideal ha; Mi ⊇ M, so ha; Mi = R. Hence there exist r 2 R and m 2 M such that ra + m = 1; hence ra = 1 − m: But 1 − m 2 1 + M and so 1 − m 2 R∗. Hence we may write [(1 − m)−1r]a = 1 which implies that a 2 R∗. So we have shown that any maximal ideal M of R has the property in (b). Thus R is a local ring. 5. (a) Find all the ideals of Q. Solution. By our result in the lectures or Problem 1. Q is a field and so its only ideals are f0g and Q. (b) Find all the subrings of Q (we require that a subring contains the unity). Solution. Let R be a subring of Q. Then necessarily Z ⊆ R since 1 2 R. Define n 1 o S := p 2 nf0g 2 R : Z p Note that S is a multiplicatively closed subset of Z that contains 1, i.e. s1; s2 2 S implies s1s2 2 S. Define na o := 2 a 2 ; p 2 S . ZS p Q Z For any a=p 2 ZS, since a 2 R and 1=p 2 R, a=p 2 R. Hence ZS ⊆ R: 2 On the other hand, for any a=p 2 R, we may assume that gcd(a; p) = 1 and ma + np = 1 for some m; n 2 Z. Then 1 ma + np a = = n + m · 2 R: p p p So p 2 S and a=p 2 ZS. Hence R ⊆ ZS: It follows that all subrings of Q are of the form R = ZS where S is a multiplicatively closed subset. If we really want to describe S explicitly, we may do so as follows. Let P ⊆ N denote the set of prime numbers. For any Σ ⊆ P, define a ZΣ := 2 Q a 2 Z; p1; : : : ; pk 2 Σ : d1 dk p1 ··· pk It is clear that ZΣ contains 0 and 1, and is closed under addition, multiplication, and additive inverses. For any Σ; Υ ⊆ P where Σ 6= Υ, there exists either p 2 ΣnΥ or p 2 ΥnΣ and so either 1=p 2 ZΣ and 1=p2 = ZΥ or vice versa. In other words Σ 6= Υ implies ZΣ 6= ZΥ. (c) Can Q=Z, regarded as an additive quotient group, be a commutative ring with unity? Solution. No. If this were true, then there exists a unity e + Z 2 Q=Z. But e + Z = 1 · e + Z = (1 + Z)(e + Z) = (0 + Z)(e + Z) = 0 + Z and so the unity (multiplicative identity) of Q=Z is the same as the zero (additive identity) of Q=Z. So for any a + Z 2 Q=Z, a + Z = (a + Z)(e + Z) = (a + Z)(0 + Z) = 0 + Z and we are led to the conclusion that Q=Z has only one element, namely, 0 + Z. This is a contradiction. 3.