Solutions Ark3
From the book: Number 1, 2, 3, 5 and 6 on page 43 and 44.
Number 1: Let S be a multiplicatively closed subset of a ring A, and let M be a
finitely generated A-module. Then S −1 M = 0 if and only if there exists an element
s ∈ S such that sM = 0.
Solution: If sM = 0 for an element s ∈ S it follow immediatly from the equivalence
relation used to define S −1 M that S −1 M = 0.
Assume S −1 M and let m1 , . . . , mr be a generating set of elements in M . Each one
of them maps to zero in S −1 M meaning that for each i, with 1 ≤ i ≤ r, there exists an
si ∈ S such that si mi = 0. Take s = s1 s2 · · · · · sr . Then s ∈ S since S is closed under
multiplication, and smi = 0 for all i. But the mi ’s forming a generating set for M , this
implies that sM = 0.
�
Number 2: Let a be an ideal of a ring A, and let S = 1 + a. Show that S −1 a is
contained in the Jacobson radical of S −1 A.
Use this to give a proof of Nakayama’s lemma, version (2.5).
Solution: The prime ideals in S −1 A are all of the form S −1 p for p a prime ideal in
A disjoint from S, and maximal ideals correspond to maximal ideals.
The pime ideals p satisfying p ∩ (1 + a) �= ∅ are those for which we may find a y ∈ p
with y = 1 + a for an a ∈ A, i.e., those which are comaximal with a.
Hence the maximal ideals in S −1 A are all of the form S −1 m where m is a maximal
ideal in A not comaximal with a. But a not being comaximal with a maximal ideal m
means that a⊆ m. Hence S −1 a is contained in all maximal ideals in S −1 A, i.e., , it is
contained in the Jacobson radical.
We shall show that Nakayama version (2.6) implies Nakayama (2.5): Assume
aM = M for a�
finitely generated A-module M . Let
S −1 aS −1 M =
� S−1= 1 + a. Clearly
−1
−1
S−1 M (if m =
ai mi with ai ∈ a, them s m =
s ai mi where s ai ∈ S −1 a.) We
just checked that S −1 a lies in the Jacobson radical of S −1 A, so Nakayama version
(2.6) gives us that S −1 M = 0. By exercise (1) it follows that there is an element x in
S = 1 + a — i.e., that satisfies x ≡ 1 mod a — with xM = 0.
�
Number 3: Let A be a ring and let S and T be two multiplicatively closed subsets of
A, and let U be the image of T in S −1 A. Show that the ring (ST )−1 A and U −1 (S −1 A)
are isomorphic.
Solution: This is an exercise in the yoga of universal properties and in the jui-juisti
of diagram chasing1 , using the following diagram, where the arrow at the top going to
1
It is also an exercise in the aikido toho iai of writing diagrams in LATEX
Solutions Ark3
MAT4200 — autumn 2011
the right and the two arrows pointing down are canonical localisation maps:
� S −1 A
�
���
�
�
�� ��
��������
�
�
�
�
��
�
� −1
(ST )−1 A �
U (S −1 A)
A ���
The four other maps are all induced by the universal property of an appropriate localisation. The two horisontal maps in the lower row — one each way — are the ones that
interest us the most. They give the required isomorphism (do the details yourself!).
Alternatively — if your inclination is less universal — you may do this directly on
elements; i.e., check that sending ( u1 )−1 (s−1 a) to (us)−1 a and vice versa are two well
defined maps being mutually inverses.
�
Number 5: Let A be a ring. Suppose that for each prime ideal p , the local ring Ap
has no nilpotent element �= 0. The A has no nilpotent element �= 0. If each Ap is an
integral domain, is it true that A is an integral domain?
Solution: Assume that all the localisations Ap are reduced (newspeak for not having
non-zero nilpotents), and assume that there is an x ∈ A with x �= 0 and xn = 0. We
may as well assume that xn−1 �= 0.
Let m be a maximal ideal containing the ideal Ann(xn−1 ) (which is proper because
n−1
x
�= 0.) Then xn−1 �= 0 in Am; indeed if it were zero, there would be an s �∈ m with
sxn−1 = 0 in A which is impossible since Ann(xn−1 )⊆ m. But clearly xn = 0 also holds
in Am, so Am has non-zero nilpotents!
Even if all localisations Ap are integral domains, it is not true that A is. The simplest
example is the product of two fields: A = k1 × k2 which is not an integral domain since
(x, 0)(0, y) = 0 for any x ∈ k1 and y ∈ k2 .
This ring has exactly two ideals which both are maximal, namely m1 = { (x, 0) |
x ∈ k1 } and m2 = { (0, y) | y ∈ k2 } — indeed every element (x, y) with both x and y
different from zero, is invertible.
Let us argue that Am2 ≈ k1 . Then by symmetry Am2 ≈ k1 , and we have our example
since they both are without zero-divisors.
All elements (0, y) map to 0 in Am2 since s = (1, 0) �∈ m2 and s(0, 1) = 0. This
shows that the inclusion map k1 → A (which is a ring homomorphism and sends x to
(x, 0)) composed with the localisation map A → Am2 is surjective — indeed, (x, y) =
(x, 0) + (0, y) and (x, 0) have a common image in Am2 . Hence it is an isomorphism.
�
Number 6: Let A be a ring and let Σ be the set of all multiplicatively closed subsets
s of A such that 0 �∈ S. Then S has maximal elements, and S is maximal in S if and
—2—
Solutions Ark3
MAT4200 — autumn 2011
only if S = A \ p for a minimal prime ideal p of A.
Solution:
We intend to use Zorn’s lemma, so let { Si }i∈I be a chain in Σ. Clearly
�
�
i∈I Si is multiplicatively close, since if s ∈ Si and s ∈ Si� they both belong to Sj for
any j ∈ I with j > i and j > i� . Hence Zorn gives us maximal elements in Σ.
Now let us check that p = A \ S is an ideal. Let a �∈ S and b ∈ A. We assume
that ab ∈ S and take a look at the set S � = { an c | n ∈ N and c ∈ S }. It is clearly
a multiplicatively closed subset since S is, and as a ∈ S � but a �∈ S, there is a strict
inclusion S ⊂ S � . Now we claim that 0 ∈ S � ; indeed if an c = 0, we get (ab)n c = 0, but
ab ∈ S, hence (ab)n c ∈ S which gives a contradiction as 0 �∈ S by hypothesis. As S is
maximal, it follows that ab �∈ S, and A \ S is an ideal. It is easy to se that it must be
a prime ideal, and by the maximality of S, it must be a minimal prime ideal.
�
Oppgave 1. Let n ∈ Z be an integer and let S be the multiplicativvely closed set
S = { m ∈ Z | (m, n) = 1 }. We denote the ring S −1 Z by Z(n) . If p is prime, then Z(p)
is the ring Z localised at the prime ideal (p). This is a local ring with maximal ideal
(p).
a) Let n = 6. And show that Z(6) has two maximal ideals, namely m1 = (3)Z6 and
m2 = (2)Z(6) .
b) Show that Z(6) /m1 � F3 and Z(6) /m2 � F2 .
c) What is the Jacobson radical to Z6 ? If J denotes the Jacobson radical, describe
Z6 /J.
d) In general, for any n ∈ Z, show that Z(n) is a semilocal ring, i.e., is a ring with
only finitely many maximal ideals. Describe those ideals and their residue fields. (The
residue field to a maximal ideal m in a ring A is A/m.)
Solution:
a) Let a⊆ Z6 be an ideal. The elements in a are of the form nm−1 where neither 2 nor 3
is a factor of m. If the same applies to n, i.e., (n, 6) = 1, then nm−1 is invertibel in Z6
and a = Z6 . This shows that the only ideals in Z6 are the principal ideals (2a 3b )Z6 — a
and b non-negative integers — and among them (2)Z6 and (3)Z6 are the only maximal
ones.
b) We have F3 = Z/(3)Z ≈ Z6 /(3)Z6
The inclusion Z⊆ Z6 induces a map Z/(3)Z → Z6 /(3)Z6 wich is injective since
(3)Z6 ∩ Z = (3)Z.
If m is an integer with (m, 3) = 1 we may write mx+3a = 1 where x and a are in Z.
Hence x ≡ m−1 mod (3)Z6 , and hence any element satisfies nm−1 ≡ nx mod (3)Z6
where nx ∈ Z, and the map above is also surjective.
—3—
Solutions Ark3
MAT4200 — autumn 2011
c) The Jacobson radical R (we all ♥ \mathfrak) of Z6 is (6)Z6 , and by the Chinese
remainder theorem, R = Z6 /(6)Z6 ≈ F3 × F2 .
d) This is a straight forward generalisation of what we have done so far. We know that
there is a one-to-one correspondence between maximal ideals in Zn = S −1 Z and ideals
p in Z maximal among those with p ∩ S = ∅, i.e., prime ideals (p) where p is a prime
not dividing n. So there is one maximal ideal S −1 (p)Z = (p)Zn in Zn for each prime
divisor p of n.
There is a natural map Fp = Z/(p)Z → Zn /(p)Zn (sending the residue class of
x mod (p)Z to the one mod (p)Zn ) which is injective because (p)Zn ∩ Z = (p)Z. It
is surjective: If (m, p) = 1 we may find integers x and a with mx + ap = 1. Hence
km−1 ≡ kx mod (p)Zn and consequently km−1 is in the image .
�
Oppgave 2. Let A be a subring of the ring B and let p ⊂ A be a prime ideal. Let S
be the multiplicatively closed set S = { x ∈ A | x �∈ p } = A \ p.
a) Show that the primes in S −1 B correspond to the primes q in B such that q ∩ A⊆ p
— more presicely, they are of the form S −1 q with q ∩ A⊆ p.
√
Let now A = Z and B = Z[i 5].
√
√
−1
b) If p = (3)Z,
√ show that S Z[i 5] has exactly the two maximal ideals (3, 1 + i 5)
and (3, 1 − i 5). What are the residue fields?
√
c) If p = (2) show that S −1 Z[i 5] is a local ring. What is the residue field?
Solution: a) This is the general principle — (3.11) on page 41 — describing how
prime ideals behave when localised, but adapted to the setting of this exercise.
b) We first check that the two ideals in question are maximal,√and by symmetry,
it
√
suffises to do that for one of them. We do that by computing Z[i 5]/(3, 1 + i 5)√which
as a bonus gives us that the residue field is F3 . By letting X correspond to i 5 we
have an isomorphism
√
√
Z[i 5]/(3, 1 + i 5) ≈ Z[X]/(X 2 + 5, 3, 1 + X).
Now (X 2 + 5, 3, 1 + X) = (3, 1 + X) since X 2 + 5 = (X − 1)(X + 1) + 3, and thus
Z[X]/(X 2 + 5, 3, 1 + X) ≈ Z[X]/(3, 1 + X) ≈ F3 .
√
√
Let p⊆ Z[i 5] be a prime ideal surviving in S −1 Z[i 5], i.e., such√that S −1 p stays
proper. Then a ∩ Z = (3), so 3 ∈ p. Pick another element w = a + ib 5 ∈ p. As 3 ∈ p,
we may replace a and b by any other integer in the same residue class mod 3 and still
have an element in p. In other we may assume that a and b both are in { ±1, 0 }.
Hence if√we there is
√ an element with ab �≡ 0 mod 3, the prime ideal p contains
either 1 + i 5 or 1 − i 5 and we are through!
—4—
Solutions Ark3
MAT4200 — autumn 2011
We have to eliminate
√ the case that ab ≡√0 mod 3 for all w ∈ p. If for one w we have
a ≡ 0 mod 3, then i 5 ∈ p, hence also (i 5)2 = −5 and therefore 2 · 3 − 5 = 1. This
can not be the case since p is a proper ideal. Hence a �≡ 0 mod 3 for any element w.
Then b ≡ 0 mod 3 for all w.√It follows that a ≡ 0 mod 3 since if not, √
p would not
√ be
proper,
and
hence
p
=
(3)Z[i
5].
But
this
is
not
a
prime
ideal:
6
=
(1+i
5)(1−i
5) ∈
√
(3)Z[i 5].
√
√
c) One checks that m = (2, 1 + i 5) is a maximal ideal in Z[i 5] with residue field F2
as above, using the following equalities between ideals
(X 2 + 1, 2, X + 1) = ((X + 1)(X − 1) + 6, 2, X + 1) = (2, X + 1).
Now
√
√
√
√
m2 = (22 , 2(1 + i 5), (1 + i 5)2 ) = (4, 2(1 + i 5), −4 + 2i 5) = (2),
so if p is any prime ideal with 2 ∈ p, we have m2 ⊆ p. Hence m⊆ p because p is prime,
and knowing that m is maximal
we conclude that p = m.
√
This shows that S −1 Z[i 5] is a local ring with residue field F2 .
�
—5—