Galois Theory Spring 2008/09 Problem Set 1 Solutions

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 ϕ : 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 ϕ : R → R0 be a ring homomorphism. Then by our results in the lectures, ker ϕ is an ideal of R. Since ϕ(1) = 1 6= 0, 1 ∈/ ker ϕ and ker ϕ 6= h1i, hence ker ϕ = h0i and so ϕ is injective. (iii) ⇒ (i): Let x ∈ R and x∈ / R∗. Then hxi= 6 h1i. Let R0 := R/hxi and ϕ : R → R0 be the map ϕ(a) = a + R. It is easy to check that ϕ is a homomorphism, im ϕ = R0, and ker ϕ = hxi. By assumption, ϕ is injective, hence hxi = h0i, hence x = 0.

n 2. We say that a ∈ R is nilpotent if a = 0 for some n ∈ 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 ∈ N(R) and r ∈ R, then since (ra)n = rnan, it is clear that ra ∈ N(R). Let a, b ∈ 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 ∈ N(R). Hence N(R) is an ideal of R. (b) Find N(Z), N(Z12), and N(Z32). Solution. N(Z) = {0}, N(Z12) = {0, 6}, and N(Z32) = {0, 2, 4, 6, 8,..., 30}. (c) Show that N(R/N(R)) = {0 + N(R)} (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) ∈ N(R/N(R)). So there exists n ∈ N such that (a + N(R)) = n n n m 0 + N(R), so a + N(R) = 0 + N(R), so a ∈ N(R), so (a ) = 0 for some m ∈ N, so amn = 0, and so a ∈ 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∗ = {1}. ∼ Show that R = Z2. Solution. Note that 1 ∈/ M since M 6= R. Let a ∈ 1 + M. Then a∈ / M (otherwise 1 = a − m ∈ 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 {1} ⊆ 1 + M ⊆ R∗ = {1}, we conclude that 1 + M = {1} and so M = {0}. So R is a field. Since every nonzero element in a field is a unit, ∗ ∼ R = R ∪ {0} = {0, 1} = 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 a∈ / M implies a ∈ 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 a∈ / R∗ implies a ∈ 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 ∈ 1 + M implies a ∈ R∗. Show that R is a local ring. Solution. Let a ∈ R and a∈ / M. Since M is maximal and the ideal ha, Mi ⊇ M, so ha, Mi = R. Hence there exist r ∈ R and m ∈ M such that ra + m = 1; hence ra = 1 − m. But 1 − m ∈ 1 + M and so 1 − m ∈ R∗. Hence we may write [(1 − m)−1r]a = 1 which implies that a ∈ 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 {0} 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 ∈ R. Deﬁne n 1 o S := p ∈ \{0} ∈ R . Z p

Note that S is a multiplicatively closed subset of Z that contains 1, i.e. s1, s2 ∈ S implies s1s2 ∈ S. Deﬁne na o := ∈ a ∈ , p ∈ S . ZS p Q Z

For any a/p ∈ ZS, since a ∈ R and 1/p ∈ R, a/p ∈ R. Hence

ZS ⊆ R.

2 On the other hand, for any a/p ∈ R, we may assume that gcd(a, p) = 1 and ma + np = 1 for some m, n ∈ Z. Then 1 ma + np a = = n + m · ∈ R. p p p So p ∈ S and a/p ∈ 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 := ∈ Q a ∈ Z, p1, . . . , pk ∈ 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 ∈ Σ\Υ or p ∈ Υ\Σ and so either 1/p ∈ ZΣ and 1/p∈ / 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 ∈ 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 ∈ 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.