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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.
2. We say that a ∈ R is nilpotent if an = 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 = rn an , 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 k m+n−k−1
a b
k
k=0
m−1
X m + n − 1
n
=b
ak bm−1+k
k
k=0
m+n−1
X m + n − 1
m
+a
ak−m bm+n−k−1
k
(a + b)m+n−1 =
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).
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Solution. Let a + N (R) ∈ N (R/N (R)). So there exists n ∈ N such that (a + N (R))n =
0 + N (R), so an + N (R) = 0 + N (R), so an ∈ N (R), so (an )m = 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, M i ⊇ M , so
ha, M i = 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)−1 r]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
1
o
n
S := p ∈ Z\{0} ∈ R .
p
Note that S is a multiplicatively closed subset of Z that contains 1, i.e. s1 , s2 ∈ S implies
s1 s2 ∈ S. Define
na
o
ZS :=
∈ Q a ∈ Z, p ∈ S .
p
For any a/p ∈ ZS , since a ∈ R and 1/p ∈ R, a/p ∈ R. Hence
ZS ⊆ R.
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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
∈ Q a ∈ Z, p1 , . . . , pk ∈ Σ .
ZΣ :=
pd11 · · · pdkk
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.
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