Math 210A – Fall 2015
Homework 9
Alexander J. Wertheim
Section 01, MWF, 11:00-11:50 AM
December 3rd, 2015
Contents
1 Problem 1
2
2 Problem 2
2
3 Problem 3
2
4 Problem 4
3
5 Problem 5
3
6 Problem 6
4
7 Problem 7
4
8 Problem 8
4
9 Problem 9
5
10 Problem 10
5
1
1
Problem 1
Every subring of Z is an additive subgroup which contains 1, and hence must be all of Z.
2
Problem 2
Let R be a subring of Z × Z. Then R is an additive subgroup of Z × Z which contains
(1, 1), and is closed under multiplication. Hence, every subring R contains the diagonal
subset D = {(n, n) | n ∈ Z}, which is itself a subring Z × Z, as it satisfies each of the
above properties. Consider the subset K = {|k − l| : (k, l) ∈ R} ⊂ Z, and let I be the ideal
generated by K in Z. We claim that R = D + I × I = {α + β | α ∈ D, β ∈ I × I}.
Since D ⊂ R, it is clear that (|k − l|, 0), (0, |k − l|) ∈ R for each (k, l) ∈ R, whence
R ⊇ K × {0}, {0} × K. Since R is closed under addition, R contains the additive closures of
both K × {0} and {0} × K. Since multiplication by elements of Z and repeated addition are
the same in Z, it follows that the additive closure of (say) K × {0} in R is exactly I × {0}.
Thus, R ⊇ I × {0}, {0} × I, so R ⊇ D + I × I.
Now suppose (k, l) ∈ R. Without loss of generality, suppose k > l. Then (k, l) = (l, l) +
(k − l, 0) ∈ D + I × I, so R ⊆ D + I × I. Since Z is a principal ideal domain, every ideal
I of Z is nZ for some n ∈ Z. Thus, every subring of Z × Z is given by D + nZ × nZ for
some n ∈ Z. Note Rn = D + nZ × nZ is in fact a subring of Z × Z for every n ∈ Z, as Rn
is clearly an additive subgroup of Z × Z, contains (1, 1), and is closed under multiplication.
Hence, {D + nZ × nZ | n ∈ Z} is exactly the set subrings of Z × Z.
3
Problem 3
We begin with the following more general fact.
Qn
Lemma 3.0.1. Let
R
=
k=1 Rk be a finite product of rings. Then every ideal I of R is a
Qn
product of ideals k=1 Ik such that Ik is an ideal of Rk for each k = 1, . . . , n.
Proof. Let {ek }nk=1 be the “standard basis” of idempotents for R, and let πk : R → Rk be
the canonical (set-theoretic) projection map for each k = 1, 2, . . . , n. Let I be an ideal of R.
Note that πk is trivially a surjective ring homomorphism, and by homework 8, problem 8(b),
πk (I) =: Ik is an ideal of Rk for each k. It is clear that I ⊆ I1 × I2 × · · · × In , since
Pn each Ik is
th
just the k projected component
P of I. Now note that ek I ⊆ I for each I, so k=1 ek I ⊆ I.
But clearly I1 × I2 × · · · × In ⊆ nk=1 ek I, since ek I = {0} × · · · × {0} × Ik × {0} × · · · × {0},
so we have I = I1 × I2 × · · · × In .
Hence, every ideal of Z × · · · × Z (n times) is a product of ideals I1 × · · · × In , where Ik
is an ideal of Z for each k = 1, . . . , n. Since Z is a principal ideal domain, the ideals of Z are
precisely kZ for some k ∈ Z. Thus, every ideal I ⊂ Z × · · · × Z is k1 Z × · · · × kn Z for some
k1 , . . . , kn ∈ Z.
2
4
Problem 4
Let R = Z, and take I = 2Z, J = 4Z. Then J ⊆ I, so I ∩ J = 4Z; but it is clear that
IJ = (2Z)(4Z) = 8Z, so IJ 6= I ∩ J. In general, since IJ ⊆ RJ = J and IJ ⊆ IR = I, we
have IJ ⊂ I ∩ J, so all counterexamples take the form of this strict inclusion. The previous
example generalizes as follows: if J = nZ and I = mZ for n, m ∈ N, then IJ = nmZ, and
I ∩ J = lcm(n, m)Z, so IJ = I ∩ J if and only if gcd(n, m) = 1.
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Problem 5
Since every ring is also an abelian group, every finite ring of 2 or 3 elements is also an abelian
group of 2 or 3 elements, respectively. There is exactly one abelian group of order 2 up to
isomorphism, the cyclic group Z/2Z, so any finite ring R of with 2 elements arises from
a ring structure on Z/2Z. Since [0]2 ∈ Z/2Z must be the additive identity of R, and the
multiplicative identity of R must be distinct from the additive identity (else R is the zero
ring by homework 8, problem 1), the multiplicative identity of R must be [1]2 . This uniquely
determines a ring structure on R, as we must have
[0]2 · [0]2 = [0]2 ; [0]2 · [1]2 = [0]2
[1]2 · [0]2 = [0]2 ; [1]2 · [1]2 = [1]2
Hence, there is one ring with two elements up to isomorphism (namely, Z/2Z with the usual
quotient ring structure). Similarly, there is one abelian group of order 3 up to isomorphism,
the cyclic group Z/3Z, so any finite ring R of with 3 elements arises from a ring structure
on Z/3Z. Since [0]3 ∈ Z/3Z must be the additive identity of R, and the multiplicative
identity of R must be distinct from the additive identity as before, we must have 1R = [1]3
or 1R = [2]3 . Suppose 1R = [1]3 ; then we have the usual ring structure on Z/3Z given by the
multiplications:
[0]3 · [0]3 = [0]3 ; [0]3 · [1]3 = [0]3 ; [0]3 · [2]3 = [0]3
[1]3 · [0]3 = [0]3 ; [1]3 · [1]3 = [1]3 ; [1]3 · [2]3 = [2]3
[2]3 · [0]3 = [0]3 ; [2]3 · [1]3 = [2]3 ; [2]3 · [2]3 = [1]3
Note that all of the equalities except the last are given by the axiom 1R · r = r · 1R = r for
all r ∈ R, and the rule 0 · r = r · 0 = 0 proved in homework 8, problem 1. The last equality
is required by distributivity, as
[2]3 · ([2]3 + [2]3 ) = [2]3 · [1]3 = [2]3
and so
[2]3 · ([2]3 + [2]3 ) = [2]3 · [2]3 + [2]3 · [2]3 = [2]3
so any other choice of [2]3 · [2]3 leads to a contradiction. Suppose that 1R = [2]3 . Then we
must have [1]3 · [2]3 = [1]3 ; but we get a contradiction, as distributivity requires:
([1]3 + [2]3 ) · [1]3 = [1]3 · [1]3 + [1]3 = [0]3
3
([1]3 + [1]3 ) · [1]3 = [1]3 · [1]3 + [1]3 · [1]3 = [2]3
The first equality requires [1]3 · [1]3 = [2]3 , but the second requires [1]3 · [1]3 = [1]3 . Hence,
there is one ring with three elements up to isomorphism (namely, Z/3Z with the usual ring
structure)
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Problem 6
Let R be a commutative ring, r ∈ R. Suppose f : Z[X] → R is a ring homomorphism such
that f (X) = r. Since f (1) must equal 1R (which determines f on Z - see homework 8,
problem 2), for any p(X) = an xn + an−1 xn−1 + · · · + a0 ∈ Z[X]
f (p(X)) =f (an xn + an−1 xn−1 + · · · + a0 )
=f (an )f (x)n + f (an−1 )f (xn−1 ) + · · · + f (a0 )
=an (1R )rn + an−1 (1R )rn−1 + · · · + a0 (1R )
=an rn + an−1 rn−1 + · · · + a0
where ai ri represents (ri )ai in (R, +). Note that a homomorphism f exists, since the assignment f (X) = r extends to a ring homomorphism by the computation above. Hence, there
is a unique ring homomorphism f : Z[X] → R such that f (X) = r. The above computation
also shows that Im(f ) = {p(r) | p(X) ∈ Z[X]}. Let S be the smallest subring of R containing
r. Note that S ⊂ Im(f ), since Im(f ) is a subring of R which contains r. By contrast, since
S is a subring of R which contains r, it is additively and multiplicatively closed, and thus
contains all sums of powers of ±r; that is, it contains all integer polynomials in r, whence
S ⊇ Im(f ). Hence, Im(f ) is the smallest subring of R which contains r.
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Problem 7
Let R be an integral domain, and suppose R[X] is a principal ideal domain. Let r 6= 0 ∈ R,
and consider the ideal hr, Xi. Since R[X] is a PID, hr, Xi = hα(X)i for some α(X) ∈ R. In
particular, we have α(X)f (X) = r for some f (X) ∈ R[X]. Since R is an integral domain,
so is R[X], so a simple degree argument shows that deg(α(X)) = 0, i.e. α(X) = s for some
s ∈ R. Further, we also have α(X)g(X) = s · g(X) = X for some g(X) ∈ R[X]. A similar
degree argument shows that deg(g(X)) = 1, i.e. g(X) = r0 X + s0 , whence s · (r0 X + s0 ) =
sr0 X +ss0 = X. We thus obtain that sr0 = 1, i.e. s is a unit. Thus, hr, Xi = hsi = R[X], so in
particular there exists p(X), q(X) such that rp(X)+Xq(X) = 1 for some p(X), q(X) ∈ R[X].
Since deg(Xq(X)) > 1 if q(X) 6= 0, we must have q(X) = 0, whence rp(X) = 1. Similarly,
we must have deg(p(X)) = 0, i.e. p(X) = t for some t ∈ R, so r is a unit in R. Since r was
an arbitrary nonzero element of R, this means R is a field.
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Problem 8
Let R be a nonzero commmutative ring. Then R has a maximal ideal a (see problem 10),
so the quotient R[X]/hai ∼
= (R/a)[X] is a polynomial ring over a field. Put F = R/a. Then
4
for any f (x) ∈ R[X], we have R[X]/ha, f (X)i ∼
= F [X]/f (X), where f (X) ∈ F [X] is the reduction of f (mod a). Since F [X]/f (X) is a domain (field) if and only if f (X) is irreducible
over F [X], it suffices to show that any polynomial ring over a field contains infinitely many
(mutually nonassociate∗ ) irreducible polynomials, as every irreducible polynomial in F [X]
is expressible as the reduction of some polynomial f (X) ∈ R[X] mod a by the surjectivity
of the canonical quotient map π : R → F .
Recall that F [X] is a UFD (in fact, F [X] is a Euclidean domain). Suppose F [X] has
only finitely
Q many (nonassociate) irreducible polynomials p1 (X), . . . , pn (X) ∈ F [X]. Then
g(X) = ni=1 pi (X) + 1 is not divisible by any of the polynomials p1 , . . . , pn , and so must
be divisible by an irreducible polynomial k(X) 6= pi (X) for each i = 1, . . . , n. Since the
factorization of g(X) is unique up to units, it is clear that k(X) is nonassociate to any
pi (X), so this is a contradiction. Hence, F [X] has infinitely many nonassociate irreducible
polynomials, which completes the proof.
∗
: Note that nonassociate irreducible polynomials generate distinct ideals in F [X], as hf (X)i =
hg(X)i implies f (X) = p(X)g(X) for some p ∈ F [X], which contradicts the irreducibility of
f (X) if p is not a unit.)
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Problem 9
Let B ⊆ A be a subgroup of an abelian group A, and let
I = {f ∈ End(A) such that f (A) ⊂ B} ⊂ End(A)
Note that I is an additive subgroup of A, as for any two elements f, g ∈ I, we have (f +
g)(a) = f (a) + g(a) ∈ B for all a ∈ A, since f (A), g(A) ⊂ B and B is a subgroup of A.
Further, fix f ∈ I, and let h ∈ End(A). Then
(f ◦ h)(a) = f (h(a)) ∈ B
for all a ∈ A, since h ∈ End(A), so h(a) ∈ A, and thus f (h(a)) ∈ B. Thus, (f ◦ h)(A) ⊆ B,
whence f ◦ h ∈ I for all h ∈ End(A), so I is a right ideal of End(A).
10
Problem 10
First, we need a standard fact.
Lemma 10.0.2. Let R be a commutative ring, and let x ∈ R be a non-unit. Then x is
contained in some maximal ideal of R.
Proof. Let Ω be the set of proper ideals of R containing x, ordered by inclusion. Note that
Ω is nonempty, as the principal ideal hxi generated by x is proper and is S
not all of R as x is
not a unit. Consider any chain of ideals {ai }i∈I ⊆ Ω; we claim that a = i∈I ai is an upper
bound for this chain. For any α, β ∈ a, we have α ∈ ai , β ∈ aj for some i, j ∈ I. Without loss
of generality, suppose ai ⊂ aj . Then α, β ∈ aj , and since aj is an ideal, we have α + β ∈ aj ,
whence α + β ∈ a. Further, for any r ∈ R, rα ∈ ai , since ai is an ideal of R, so rα ∈ a.
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Hence, a is an ideal of R. Note a contains x, since x ∈ ai for each i ∈ I, and a is proper,
since 1 ∈
/ ai for each i ∈ I, so 1 ∈
/ a. Finally, it is clear that ai ⊆ a for each i ∈ I, so a is
indeed an upper bound for the chain {ai }i∈I , whence every chain in Ω has an upper bound.
By Zorn’s lemma, Ω has a maximal element, which completes the proof.
( =⇒ ) Suppose x ∈ Rad(R), but 1 − xy is not a unit. Every non-unit is contained in
some maximal ideal, so 1 − xy is in some maximal ideal m. Since x ∈ Rad(R), x ∈ m, i.e.
xy ∈ m, whence 1 ∈ m, a contradiction.
( ⇐= ) Suppose x ∈
/ Rad(R), i.e. there is some maximal ideal m such that x ∈
/ m. Then the
ideal generated by x and m is all of R, so in particular there is some m ∈ m, y ∈ R such that
m + xy = 1. Then 1 − xy ∈ m, so it not a unit.
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