# Arbitrary Collection of Modules

Finally, we consider the case where we have potentially infinitely many modules.

Proposition 1.

For a collection of A-modules $(M_i)_{i\in I}$, we have

$S^{-1}(\oplus_i M_i) \cong \oplus_i (S^{-1} M_i).$

Proof

First claim: we will show that the LHS satisfies the universal property for direct sums. Now for any $S^{-1}A$-module N, we have:

\begin{aligned} \mathrm{Hom}_{S^{-1}A}(\oplus_i (S^{-1}M_i), N)&\cong \prod_{i\in I}\mathrm{Hom}_{S^{-1}A} (S^{-1}M_i, N) \\ &\cong \prod_{i\in I} \mathrm{Hom}_A(M_i, N) \\ &\cong \mathrm{Hom}_A(\oplus_i M_i, N) \\ &\cong \mathrm{Hom}_{S^{-1}A}(S^{-1}(\oplus_i M_i), N).\end{aligned}

• The first correspondence follows from the universal property of direct sums.
• The second follows from the universal property of localization $S^{-1}M_i$.
• The third follows from the universal property of direct sums.
• The last follows from the universal property of localization $S^{-1}(\oplus_i M_i)$.

This is a natural isomorphism, with both sides functorial in N. Now apply the following to get an isomorphism of A-modules

$S^{-1}(\oplus_i M_i) \cong \oplus_i (S^{-1} M_i).$

Exercise A

Prove the following corollary of Yoneda lemma: if $A, B \in \mathcal C$ are objects such that there is a natural isomorphism

$\mathrm{hom}_{\mathcal C}(A, -) \cong \mathrm{hom}_{\mathcal C}(B, -)$

then $A \cong B$ in $\mathcal C$.

In general, localization does not commute with direct products:

$S^{-1} (\prod_i M_i) \ne \prod_i S^{-1} M_i.$

To be specific, the projection maps $\prod_i M_i \to M_j$ for indices j induce $S^{-1}(\prod_i M_i) \to S^{-1}M_j$, and by the universal property of products

$S^{-1}(\prod_i M_i) \longrightarrow \prod_i (S^{-1} M_i)$

which is not an isomorphism in general. For example, we can set $A = \mathbb Z$ and $S = \mathbb Z - \{0\}$. If $M_1 = M_2 = \ldots = \mathbb Z$, then $(1, \frac 1 2, \frac 1 3, \ldots)$ in the RHS is not in the image of the map.

Exercise B

1. Find A-submodules $N_i \subseteq M$ such that $S^{-1}(\cap_i N_i) \ne \cap_i S^{-1} N_i$ as submodules of $S^{-1}M$.

2. Prove that for a collection of A-submodules $(N_i)_{i\in I}$ of M, we have

$S^{-1}(\sum_i N_i) = \sum_i S^{-1} N_i$,

as submodules of $S^{-1}M$. [Hint: use a straightforward proof.]

# Local Properties

Another important property we wish to emphasize is locality. As a motivation suppose V is an irreducible k-variety (k algebraically closed) with $A = k[V]$, which is a domain.

• Let $\phi : V\to k$ be a function such that for each $P\in V$, $\phi$ is regular at P.

Thus V is a union of open subsets $U_i$ such that each restriction $\phi |_{U_i}$ can be locally written as $P\mapsto \frac {f(P)}{g(P)}$ where $f,g \in k[V]$ and $g(P) \ne 0$ for all $P\in U_i$. We claim that this implies $\phi\in k[V]$, or to be specific, there exists $f\in k[V]$ such that $f(P) = \phi(P)$ for all $P\in V$.

Indeed, the preceding condition says $\phi \in k[V]_{\mathfrak m_P}$ for all $P\in V$, and since all maximal ideals of $k[V]$ are of the form $\mathfrak m_P$, we only need to show:

Proposition 2.

For any integral domain A, $A = \cap_{\mathfrak m \text{ maximal }} A_{\mathfrak m}$, where intersection occurs in the field of fractions $\mathrm{Frac}(A)$ of A.

Proof

(⊆) is obvious. For (⊇), suppose $f\in K$ lies in all $A_{\mathfrak m}$. Let $\mathfrak a = \{a \in A : af \in A\}$. Note that $\mathfrak a$ is an ideal of A, since if $af, a'f \in A$ then $(a+a')f = af+a'f \in A$, and if $af\in A$ then for any $b\in A$ we have $baf \in A$ as well.

If $\mathfrak a \ne (1)$, then $\mathfrak a\subseteq \mathfrak m$ for some maximal ideal $\mathfrak m \subset A$. But $f \in A_{\mathfrak m}$ so for some $s \in A-\mathfrak m$ we have $sf\in A$. Thus $s \in \mathfrak a - \mathfrak m$, a contradiction. ♦

When we look at $A=\mathbb Z$, this becomes quite obvious: when p is prime, $\mathbb Z_{(p)}$ is the ring of $\frac a b \in \mathbb Q$ where ab are integers and b is not a multiple of p. Hence if $\frac a b\in \mathbb Q$ (reduced) lies in all $\mathbb Z_{(p)}$ it just means b is not divisible by any prime, so $b = \pm 1$.

# More Local Properties

In all the following results, $\mathfrak m$ denotes a maximal ideal of A.

Proposition 3.

Let M be an A-module. Then $M=0$ if and only if $M_{\mathfrak m} = 0$ for all $\mathfrak m$.

Proof

(⇒) is obvious. For (⇐) fix $m\in M$ and let $\mathfrak a = \{a \in A : am = 0\}$. It is easy to show that this is an ideal of A. Now for any maximal ideal $\mathfrak m$, since $\frac m 1 \in M_{\mathfrak m}$ is zero there exists $s\in A-\mathfrak m$ such that $sm = 0$. Thus $s \in \mathfrak a - \mathfrak m$. Since $\mathfrak a$ is an ideal not contained in any maximal ideal, it is (1), i.e. $1\in \mathfrak a$ so $m=0$. ♦

Note

The ideal $\mathfrak a$ above is called the annihilator of m; we will have more to say about it later.

Proposition 4.

Let $f:M\to N$ be a homomorphism of A-modules. Then $f$ is zero if and only if $f_{\mathfrak m} : M_{\mathfrak m} \to N_{\mathfrak m}$ is zero for all $\mathfrak m$.

Proof

(⇐) : for each $\mathfrak m$ we have $(\mathrm{im} f)_{\mathfrak m} = \mathrm{im} f_{\mathfrak m} = 0$ since $f_{\mathfrak m} = 0$. Thus by proposition 3, $\mathrm{im} f = 0$ and we have $f=0$. ♦

Corollary 1.

Let $f, g:M\to N$ be homomorphisms of A-modules. Then $f=g$ if and only if $f_{\mathfrak m} = g_{\mathfrak m}$ for all $\mathfrak m$.

Proof.

Apply proposition 4 to $f-g$. ♦

Proposition 5.

Let $N\stackrel f\to M \stackrel g\to P$ be A-linear maps. The sequence is exact if and only if

$N_{\mathfrak m} \stackrel {f_{\mathfrak m}} \longrightarrow M_{\mathfrak m} \stackrel {g_{\mathfrak m}} \longrightarrow P_{\mathfrak m}$

is exact for each $\mathfrak m$.

Proof

(⇒) We saw in theorem 1 here that localization is an exact functor.

(⇐) First we show that $\mathrm{im} f \subseteq \mathrm{ker} g$, or equivalently $g\circ f = 0$. But $(g\circ f)_{\mathfrak m} = g_{\mathfrak m} \circ f_{\mathfrak m} = 0$ for each $\mathfrak m$ so $g\circ f = 0$.

For the reverse inclusion, consider the module $(\mathrm{ker} g) / (\mathrm{im} f)$. We have

$((\mathrm{ker} g) / (\mathrm{im} f))_{\mathfrak m} \cong (\mathrm{ker} g)_{\mathfrak m} / (\mathrm{im} f)_{\mathfrak m} \cong (\mathrm{ker} g_{\mathfrak m}) / (\mathrm{im} f_{\mathfrak m}).$

Since this is zero for all $\mathfrak m$, we have $\mathrm{ker} g = \mathrm{im} f$. ♦

Corollary 2.

An A-linear map $f: M\to N$ is injective (resp. surjective) if and only if $f_{\mathfrak m} : M_{\mathfrak m} \to N_{\mathfrak m}$ is injective (resp. surjective) for each $\mathfrak m$.

Proof

Apply proposition 5 to $0\to M \stackrel f\to N$ and $M\stackrel f\to N \to 0$ respectively. ♦

Note

As a general statement, we say the above properties are local. Philosophically, this means in order to check a certain property, it suffices to check at each maximal ideal so we can restrict ourselves to the case where the base ring is local. With Nakayama’s lemma, we will see that finitely generated modules over local rings are quite well-behaved.

Geometrically, this says we only need to look at each point of the variety to check the property. For example, if we interpret modules over the coordinate ring as vector bundles over the variety, then locality says: a bundle map $E \to F$ is injective if and only if $E_P \to F_P$ is injective at every point P.

Exercise C

Recall that the localization of a reduced ring is reduced, and that of an integral domain is an integral domain. Decide if each of the following is true and justify your answer.

• If $A_{\mathfrak m}$ is reduced for all $\mathfrak m$, then so is A.
• If $A_{\mathfrak m}$ is an integral domain for all $\mathfrak m$, then so is A.

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