Commutative Algebra
Lecture 4: Supports of Modules
• At the end of last lecture, we discussed localization of a ring R with respect to a
prime ideal p, and how there is an order-preserving bijection
o
n
o n
primes of R contained in p ∼
= primes of Rp
sending a prime q ⊆ p to the ideal qRp generated by its image in Rp . (This uses
the general notation for multiplying an ideal I by a module M : we denote by IM
the submodule of M generated by {im : i ∈ I, m ∈ M }. Then for any R-algebra
A, the ideal IA of A is the one generated by the image of I in A.)
In particular, Rp has a unique maximal ideal, pRp . A ring with a unique maximal
ideal is called a local ring.
• Here’s an extended example of a local ring that hopefully explains the terminology
“local” (and “localization” more generally).
Example 1. Let X be the unit interval [0, 1], although many topological spaces
will do, and let R = C(X, R) be the ring of continuous functions X → R. Let
x ∈ X be some point. Then
m := {f ∈ R : f (x) = 0}
is a maximal ideal, because it is the kernel of the surjective ring homomorphism
R → R : f 7→ f (x). I claim that the local ring Rm is the ring of germs of
continuous functions X → R at x, i.e. equivalence classes of pairs (U, f ) where U
is an open neighborhood of x and f : U → R is a continuous function, and where
(U, f ) = (V, g) if there is some neighborhood W ⊆ U ∩ V of x on which f |W = g|W .
Germs can be added, subtracted, and multiplied, and hence form a ring themselves,
the ring of “local” information about continuous functions near x.
Every continuous function f : X → R gives a germ (X, f ), and this constitutes
a ring homomorphism R → {germs}. Since any f ∈ R \ m is nonzero on an
open neighborhood of x, and on that open neighborhood it has a multiplicative
inverse, the germ of such an f is a unit and we obtain a map from the localization
Rm → {germs}.
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This map is injective: if f ∈ R and g ∈ R \ m, and the germ of f /g near x is
zero, then f vanishes on some neighborhood U of x. Let b be a “bump function”:
a continuous function that is zero outside of U but nonzero at x. Then b ∈ R \ m
and b · f = 0, so f /g = 0 in the localization Rm .
The map is also surjective: given any germ (U, f ), we can use our bump function
b to define a new function
(
f (y)b(y) if y ∈ U
g : y 7→
0
if y ∈
/U
which is continuous on all of X. Then the germ of g/b at x is equal to (U, f ), by
restricting both f and g to the neighborhood of x where b 6= 0. So (U, f ) is in the
image of Rm .
• Given a continuous function f : X → R, we can define its support as the set of
points x ∈ X where the germ of f at x is nonzero. This set is always closed, since
“having zero germ at x” is an open condition on x: if it’s true at x it’s also true on
a neighborhood of x.
There is an analogous definition for the support of an element of any module over
a ring:
Definition 2. Let R be a ring and M an R-module. Let m ∈ M . The support of
m, denoted Supp(m), is the set of prime ideals p ⊆ R such that m 67→ 0 in Mp .
(For those of you who haven’t already taken some algebraic geometry, the set of
prime ideals of a ring R is called its spectrum and denoted Spec(R). This set is
given a topology where the closed sets are those of the form V (I) := {primes p ⊆
R that contain I} for some ideal I. We think of elements of R as functions on this
topological space; the subset V (I) is the set of points where all the functions in I
vanish.)
• Today’s main theorems are about these supports of module elements. The first
result is that an element is zero if and only if its support is empty, just as a
function is identically zero if and only if its support is empty:
Lemma 3. Let R be a ring and M an R-module with an element m. Let f1 , . . . , fn
be elements of R generating the unit ideal. Then the following are equivalent:
1. m = 0.
2. m 7→ 0 in Mfi for each i ∈ {1, . . . , n}.
3. m 7→ 0 in Mp for each prime ideal p ⊆ R (i.e. Supp(m) = ∅.)
4. m 7→ 0 in Mm for each maximal ideal m ⊆ R.
Proof. Each statement cyclically implies the next:
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1⇒2. If m = 0, then m 7→ 0 in every localization Mf .
2⇒3. Given a prime ideal p, not every fi can be contained in p, so choose fi ∈
/ p
with m 7→ 0 in Mfi . The key is that Mp ∼
= (Mfi )p , since
(Mfi )p ∼
= Rp ⊗R (Rfi ⊗R M )
∼
= (Rp ⊗R Rf ) ⊗R M
i
∼
= (Rp )fi ⊗R M
but (Rp )fi = Rp since fi is already a unit in Rp , so
(Mfi )p ∼
= Rp ⊗ R M ∼
= Mp .
Then since m 7→ 0 in Mfi , it maps to zero in Mp as well.
(Or in other words, since fi acts invertibly on Mp , the canonical homomorphism M → Mp factors via the homomorphism M → Mfi sending m to zero,
so m 7→ 0 in Mp . Or in still other words, knowing that m/1 = 0 in Mfi , we
can find a power fik of fi for which fik m = 0 in M ; then fik ∈
/ p since p is
prime, so m 7→ 0 in Mp .)
3⇒4. Every maximal ideal is prime, so statement 4 is just a weaker version of
statement 3.
4⇒1. For each maximal m, the fact that m 7→ 0 in Mm tells us that there is fm ∈
/m
such that fm · m = 0. Then the ideal generated by all the fm is contained
P in no
maximal ideal, hence is the unit ideal and 1 can be written as 1 = ni=1 ri fmi
for some
Pn r1 , . . . , rn ∈ R and specific maximal ideals m1 , . . . , mn ⊆ R. Then
m = i=1 ri fmi m = 0.
• The support of an element is closely related to its annihilator :
Definition 4. Let R be a ring, M an R-module, and m ∈ M an element. The
annihilator of m in R is the ideal
AnnR (m) := {r ∈ R : r · m = 0}.
Sometimes we will leave off the R subscript if there can be no confusion.
Lemma 5. Let R be a ring and p ⊆ R a prime ideal. Let M be an R-module and
m ∈ M . Then p ∈ Supp(m) if and only if p ⊇ Ann(m).
Proof. Suppose that m 67→ 0 in Mp , that is, m/1 6= 0/1 in Mp . This is the same
as saying that for all f in the multiplicative subset R \ p, we have f · m 6= 0. Or
taking the contrapositive, this is the same as saying that for all f ∈ R such that
f · m = 0, we must have f ∈ p, i.e. Ann(m) ⊆ p.
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A geometric consequence of this lemma is that the support of any single element
of an R-module is a closed subset of Spec(R), namely V (Ann(m)).
Note also that the knowledge that the annihilator of an element m ∈ M is always
an ideal, which is easy to verify, can clean up the proof of 4 ⇒ 1 in Lemma 3:
if m 7→ 0 in Mm , then Ann(m) 6⊆ m; if this holds for all maximal ideals m, then
Ann(m) must be the unit ideal (1) and so 1 · m = 0.
• We have the following module-wide version of Lemma 3:
Lemma 6. Let R be a ring and M an R-module. Let f1 , . . . , fn ∈ R together the
unit ideal. Then the following are equivalent:
1. M = 0.
2. Mfi = 0 for each i ∈ {1, . . . , n}.
3. Mp = 0 for each prime ideal p ⊆ R.
4. Mm = 0 for each maximal ideal m ⊆ R.
Proof. The implications 1 ⇒ 2 ⇒ 3 ⇒ 4 run as before. For 4 ⇒ 1, let m ∈ M .
Them m 7→ 0 in each localization Mm at a maximal ideal (because each of these
localizations is itself zero), so m = 0. Since this holds for each element of M , we
have M = 0.
• We can also make a module-wide definitions of support and annihilator:
Definition 7. Let R be a ring and M an R-module. The support of M , denoted
Supp(M ), is the set of prime ideals p ⊆ R such that Mp 6= 0. The annihilator of
M is the ideal AnnR (M ) := {r ∈ R : rM = 0}.
• For example, M = 0 if and only if Ann(M ) = (1) and, and by Lemma 6, if and
only if Supp(M ) = ∅. Is it always the case that Supp(M ) = V (Ann(M )), as in
Lemma 5? No:
Example 8. Consider the support and annihilator of the Z-module
M
M=
Z/pZ.
p∈N prime
For which prime ideals p ⊆ Z is Mp nonzero? Localizing at p is the same as
tensoring with Zp , which preserves direct sums, so we have
M
Mp =
(Z/pZ)p .
p∈N prime
If p = (p) for some prime number p, then since everything in Z \ (p) already acts
invertibly on Z/pZ, we have (Z/pZ)(p) = Z/pZ 6= 0. So these prime ideals are all
in the support of Z.
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But the prime ideal (0) ⊆ Z is not in the support: Z(0) = Q, and (Z/pZ) ⊗ Q = 0
for all prime numbers p, so M(0) = 0.
The annihilator of M , however, is zero: Given any nonzero n ∈ Z, choose a prime
number p not dividing n, which thus has n · Z/pZ = Z/pZ 6= 0. So the only element
of Z that kills all of M is zero: Ann(M ) = 0. Then in particular:
Ann(M ) ⊆ (0) but (0) ∈
/ Supp(M ),
so Supp(M ) 6= V (Ann(M )).
In fact, Supp(M ) is not even a closed subset of Spec(R), since the only closed
subsets are of the form V ((0)) = Spec(R) and, for each n 6= 0, the subset V ((n)) =
{(p) : p is a prime dividing n}.
• We shall see that the equivalence between p ∈ Supp(M ) and p ⊇ Ann(M ) can only
fail if M is not finitely generated, like the direct sum above with infinitely many
summands. The first step is a pair of dual identities given a list of generators for
our module:
Lemma 9. Let R be a ring and M an R-module, and suppose {mi : i ∈ I} is a
generating set for M . Then
T
1. Ann(M ) = i∈I Ann(mi ).
S
2. Supp(M ) = i∈I Supp(mi ).
Proof. For the first claim,
T we have an easy containment Ann(M ) ⊆ Ann(mi ) for
each i, so Ann(M ) ⊆ i Ann(mi ). Conversely, if r ∈ Ann(mi ) for all i ∈ I, then
r · mi = 0 for all i ∈ I. Given any m ∈ M , we can write m as a finite linear
combination of the mi :
n
X
rj · mij ,
m=
j=1
and then r · m =
P
j rj (r
· mij ) = 0. So rM = 0 and r ∈ Ann(M ).
S
For the second claim, we again have the easy inclusion Supp(M ) ⊇ i Supp(mi ).
For the reverse containment, suppose p ∈
/ Supp(mi ) for all i ∈ I. Then each
mi 7→ 0 in Mp ; since the mi generate M (and hence their images generate Mp as
an Rp -module), we must have Mp = 0 and p ∈
/ Supp(M ).
As an easy corollary, if p ∈ Supp(M ) then p ∈ Supp(m) for some m ∈ M , so p ⊇
Ann(m) ⊇ Ann(M ). So we always have the inclusion Supp(M ) ⊆ V (Ann(M )).
• So we find that if {mi : i ∈ I} generates M , then p ∈ Supp(M )Tif and only if
p ⊇ Ann(mi ) for some i ∈ I. And p ⊇ Ann(M ) if and only if p ⊇ i∈I Ann(mi ).
Are these equivalent? We know they can’t be in general, but if I is finite they are:
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Lemma 10. Let I and J be ideals of a ring R, and let p ⊆ R be a prime ideal.
The following are equivalent:
1. p ⊇ I ∩ J.
2. p ⊇ IJ.
3. p ⊇ I or p ⊇ J.
T
Thus by induction, a prime ideal p containing a finite intersection ni=1 Ii is equivalent to p containing oneTof the ideals Ii . Note that this is not true for infinite
intersections in general: p∈N prime (p) = (0), but (0) contains no prime p. Also,
infinite products of ideals don’t make sense.
Proof. (1 ⇒ 2): Follows from I ∩ J ⊇ IJ.
(2 ⇒ 3): If J 6⊆ p, then choose j ∈ J \ p. Then ij ∈ p for all i ∈ I, so i ∈ p for all
i ∈ I and I ⊆ p.
(3 ⇒ 1): If p contains I or J, it certainly contains the smaller ideal I ∩ J, which is
contained in both I and J.
(Another proof of 1 ⇔ 2: since (I ∩ J)2 ⊆ IJ ⊆ I ∩ J, they have the same radical
and are contained in the same prime ideals.)
• So we obtain the following corollary analogous to Lemma 5:
Corollary 11. Let R be a ring and p ⊆ R a prime ideal. Let M be a finitely
generated R-module. Then p ∈ Supp(M ) if and only if p ⊇ Ann(M ).
The geometric consequence is that if M is a finitely generated A-module, the support of M is a closed subset of Spec(A), namely the closed subset V (Ann(M )).
Proof. We said already that if {m1 , . . . , mn } generates M as an A-module, then
p ∈ Supp(M ) if and only if pTcontains Ann(mi ) for some i ∈ {1, . . . , n}. This is
true if and only if p contains ni=1 Ann(mi ) = Ann(M ).
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