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MAT4200 Fall 2012
Mandatory assignment
Suggested solutions to the problems
Problem 1
a): The sequence 0 → M 0 → M → M 00 → 0 is exact if and only if 0 →
→ Mp → Mp00 → 0 is exact for all p ⊂ A prime ideal. So clearly Mp 6= 0 if
and only Mp0 6= 0 or Mp00 6= 0.
Mp0
b): We have (A/a)p = Ap /ap , so (A/a)p 6= 0 if and only if ap 6= Ap . This
holds if and only if a ∩ (A \ p) = ∅, which again is equivalent to a ⊆ p, hence to
p ∈ Z(a).
c): Assume p + Ann M for some prime ideal p. Take x ∈ Ann M , x ∈
/ p. For
m
xm
0
∈
M
,
we
have
=
=
any m ∈ M , xm = 0. Therefore, given any m
p
s
s
xs
1 = 0,
hence Mp = 0.
If the A-module M is finitely generated, we know by Prop. 3.14 that
(Ann M )p = Ann(Mp ) for any prime ideal p. If p ⊇ Ann M , then pAp ⊇
(Ann M )p = Ann(Mp ), hence Ann(Mp ) 6= Ap , and therefore Mp 6= 0.
(Alternatively: Assume M is finitely generated, say by m1 , . . . , mn . Suppose
p ⊂ A is a prime ideal such that Mp = 0. Then there are elements s1 , . . . , sn ∈
A \ p such that si mi = 0, for i = 1, . . . , n. Set s = s1 · · · sn . Then clearly
sM = 0, so s ∈ Ann M , but s ∈
/ p.)
Problem 2
Assume m ∈ M , m 6= 0. Then, since 1 · m = m 6= 0, a := Ann(x) 6= A.
Multiplication with x gives an A-module homomorphism A → M with kernel
a, hence induces an injective homomorphism A/a → M . Since B is flat as an
A-module, we get an injective B-module homomorphism
B ⊗A A/a ∼
= B/aB → B ⊗A M = MB .
Therefore, in order to show that MB 6= 0, it suffices to show that B/aB 6= 0,
or, equivalently, that aB 6= B. Since a 6= A, a ⊆ m for some maximal ideal in
A. By assumption, mB ⊆ n for some maximal ideal n in B. Since aB ⊆ mB,
therefore aB 6= B.
Problem 3
a): The ring S −1 A = Ap is a local ring with maximal ideal S −1 p = pAp .
Since taking radicals commutes with localization, we get that r(S −1 (pn )) =
S −1 r(pn ) = S −1 (r(p)) = S −1 p is maximal, and hence S −1 (pn ) is S −1 p-primary.
The contraction of a primary ideal is primary, so p(n) := S(pn ) is p-primary
(since S(p) = p).
b); Clearly, if p(n) = pn , then also pn is p-primary. Conversely, we know
from Prop. 4.8 that if pn is p-primary, then S(pn ) = pn , hence p(n) = pn .
Problem 4
Since 3X divides both 9X and 3X 2 , we have (9X, 3X 2 ) ⊆ (3X) = (3)(X) ⊆
(3) ∩ (X). Also, (9X, 3X 2 ) ⊆ (9, X 2 ). Hence
(9X, 3X 2 ) ⊆ (3) ∩ (X) ∩ (9, X 2 ).
Let us show that
(9X, 3X 2 ) ⊇ (3) ∩ (X) ∩ (9, X 2 ).
Let f (X) ∈ (3) ∩ (X) ∩ (9, X 2 ). Then f (X) = 9g(X) + X 2 h(X), where 3|h(X)
and X|g(X). Then
f (X) = 9X
g(X)
h(X)
+ 3X 2
∈ (9X, 3X 2 ).
X
3
The ideals (3) and (X) are prime, since Z[X]/(3) ∼
= Z/(3)[X] and Z[X]/(X) = Z
re integral domains. The ideal (9, X 2 ) is primary, since its radical r(9, X 2 ) =
r(r(9) + r(X 2 )) = r(3, X) = (3, X) is a maximal ideal (because Z[X]/(3, X) ∼
=
Z/(3) is a field). Hence we have found a primary decomposition of (9X, 3X 2 ),
with distinct associated primes (3), (X), and (3, X). In order to assert that
the decomposition is minimal, it remains to show there is no redundancy. But
clearly (3) + (X) ∩ (9, X 2 ) (since e.g. X 2 ∈
/ (3)), (X) + (3) ∩ (9, X 2 ) (since e.g.
2
9∈
/ (X)), and (9, X ) + (X) ∩ (3) = (3X) (since e.g. 3X ∈
/ (9, X 2 )).
The primes (3) and (X) are minimal, whereas (3, X) contains each of the
other two and is embedded.
Problem 5
a): Let C denote the integral closure of A in K(A) = K(B). Since A ⊆ B,
C is contained in the integral closure of B, which by assumption is equal to B.
But B is integral over A, so B ⊆ C. Hence C = B.
b): Write A = k[x, y, z], where xy 2 = z 2 . Since the element yz ∈ K(A),
we have A ⊆ B := A[ yz ] ⊆ K(A). But B = k[x, y, z, yz ] ∼
= k[y, yz ] ∼
= k[ yz , y]
is isomorphic to a polynomial ring in two variables, hence is integrally closed
in its field of fractions k( yz , y). Moreover, since A ⊆ B ⊆ K(A), we have
K(A) ⊆ K(B) = k( yz , y) ⊆ K(A), and so the two fields of fractions are equal.
The element yz ∈ B is integral over A, since it satisfies the monic polynomial
relation ( yz )2 − x = 0. Hence B is integral over A, and we conclude by applying
a) that B is the integral closure of A.