Properties of Localization

We now examine further properties of localization. The first goal of this post is to prove that there is a close relationship between the localization of modules and the localization of rings discussed in the previous post. In this post, we fix a ring \(A\), a multiplicative subset of \(A\), \(S\), and an \(A\)-module \(M\).

Localization and Hom, Tensor

We begin by proving a lemma. If we define a function \(S^{-1}A\times M \rightarrow S^{-1}M\) by \((r/u, x)\mapsto rx/u\), this is an \(A\)-bilinear map, and therefore induces an \(A\)-linear map \(S^{-1}A\otimes_A M \rightarrow S^{-1}M\). ([Algebraic Structures] §Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Theorem 5)

Lemma 1 The \(A\)-linear map defined above is an isomorphism.

Proof

It suffices to construct an inverse. To this end, let us first define a function from \(M\times S\) to \(S^{-1}A\otimes_AM\) by

\[(x,s)\mapsto \frac{1}{s}\otimes x\]

Then this function defines a well-defined map from \(S^{-1}M\) to \(S^{-1}A\otimes_AM\) that is \(A\)-linear. To see this, it suffices to show that this function is well-behaved with respect to the equivalence relation defined on \(M\times S\). Thus, suppose \((x,s)\sim (x',s')\) holds for two given elements of \(M\times S\). Then there exists some \(t\in S\) such that \(tsx'=ts'x\) holds, and from this

\[\frac{1}{tss'}\otimes ts'x=\frac{1}{tss'}\otimes tsx'\]

holds. But since \(ts',ts\in A\), moving \(ts'\) and \(ts\) on the left- and right-hand sides, respectively, to the left of \(\otimes\) yields

\[\frac{1}{s}\otimes x=\frac{1}{s'}\otimes x'\]

It is clear that this function is an \(A\)-linear map and is the inverse of \(S^{-1}A\otimes_A M \rightarrow S^{-1}M\) defined above.

In particular, using this we can also show the functoriality of the localization of modules. This is because for any \(u: M \rightarrow M'\), we can define \(S^{-1}M \rightarrow S^{-1}M'\) by identifying both sides of the function

\[S^{-1}A\otimes_A u: S^{-1}A\otimes_AM \rightarrow S^{-1}A\otimes_AM'\]

with localizations. In general, the tensor product is right exact, but in this case it becomes an exact functor.

Proposition 2 \(S^{-1}A\) is a flat \(A\)-module. ([Multilinear Algebra] §Projective, Injective, and Flat Modules, ⁋Definition 7)

Proof

Suppose an injective \(A\)-linear map \(u:M \rightarrow M'\) is given; we must show that \(S^{-1}A\otimes_A u\) is injective. However, by Lemma 1, it suffices to show that the linear map \(S^{-1}M \rightarrow S^{-1}M'\) is injective. For some \(x/s\in S^{-1}M\), suppose that the element sent to \(S^{-1}M'\), namely \(u(x)/s\), is in \(S^{-1}M'\) equal to \(0\). Then from \(u(x)/s=0/1\), there exists some \(t\in S\) such that

\[tu(x)=u(tx)=0\]

holds, and since \(u\) is injective, in \(M\) we must have \(tx=0\). Then in \(S^{-1}M\),

\[\frac{x}{s}=\frac{tx}{ts}=\frac{0}{ts}=0\]

so we obtain the desired result.

Properties Determined by Localization

By Proposition 2 above and the right exactness of the tensor product, we know that if \(u:M \rightarrow M'\) is injective (resp. surjective, bijective), then the map \(S^{-1}M \rightarrow S^{-1}M'\) induced by this is also such. Proposition 4 can be regarded as a kind of (strong) converse to this. To this end, we first prove the following lemma.

Lemma 3 Consider an \(A\)-module \(M\) and, at a maximal ideal of \(A\), \(\mathfrak{m}\), the localization \(\epsilon_\mathfrak{m}:M \rightarrow M_\mathfrak{m}\). Then an element of \(M\), \(x\), is \(0\) if and only if for every maximal ideal of \(A\), \(\mathfrak{m}\), the map \(\epsilon_\mathfrak{m}\) defined above satisfies \(\epsilon_\mathfrak{m}(x)=0\).

Proof

One direction is obvious, so it suffices to prove the converse. Fix a maximal ideal \(\mathfrak{m}\) and suppose \(\epsilon_\mathfrak{m}(x)=0\). This is equivalent to \(\ann(x)\) not being contained in \(\mathfrak{m}\). Then by the given condition, \(\ann(x)\) is an ideal not contained in any maximal ideal of \(A\), and the only such ideal is \(A\) itself. Thus \(\ann(x)=A\), which completes the proof.

Therefore, the following holds.

Proposition 4 An \(A\)-linear map \(u:M \rightarrow N\) is a monomorphism (resp. epimorphism, isomorphism) if and only if for every maximal ideal \(\mathfrak{m}\), \(u_\mathfrak{m}: M_\mathfrak{m} \rightarrow N_\mathfrak{m}\) is such.

To prove this, it suffices to apply Lemma 3 to the kernel and cokernel.

The following proposition will be used frequently hereafter, so even if one does not worry about the proof, it is good to remember the result.

Proposition 5 Fix a ring \(A\) and an \(A\)-algebra \(E\). Then for any \(A\)-modules \(M,N\), the following \(E\)-module homomorphism

\[\alpha: E\otimes_A\Hom_A(M,N) \rightarrow\Hom_E(E\otimes_A M, E\otimes_AN);\qquad (1\otimes f)\mapsto \id_E\otimes_A f\]

is well-defined. In particular, if \(E\) is a flat \(A\)-module and \(M\) is finitely presented, then \(\alpha\) is an isomorphism.

Proof

That \(\alpha\) is well-defined is obvious. Now suppose \(E\) is a flat \(A\)-module and \(M=A\). Then the given

\[\alpha: E\otimes_A\Hom_A(A, N) \rightarrow\Hom_E(E\otimes_AA, E\otimes_AN)\]

fits into the following commutative diagram

so the given proposition holds. Here the vertical maps come from the respective isomorphisms

\[\Hom_A(A,N)\cong N,\qquad \Hom_E(E\otimes_AA,E\otimes_AN)\cong\Hom_E(E,E\otimes_AN)\cong E\otimes_AN\]

Next, since \(\Hom\) and \(\otimes\) commute with finite direct sums, this proposition also holds for any flat \(A\)-module \(E\) and any finitely generated free \(A\)-module \(M\). Finally, in the case where \(M\) is finitely presented, after taking the following free presentation consisting of finitely generated free \(A\)-modules \(F,G\)

\[F \rightarrow G \rightarrow M \rightarrow 0\]

it suffices to apply the four lemma to the following commutative diagram

In particular, suppose the following short exact sequence

\[0 \rightarrow M \rightarrow L \rightarrow N \rightarrow 0\]

is given. Then this exact sequence is a split exact sequence if and only if for every \(A\)-module \(K\),

\[0 \rightarrow \Hom_\rMod{A}(K,M) \rightarrow \Hom_\rMod{A}(K,L)\rightarrow \Hom_\rMod{A}(K,N) \rightarrow 0\]

is a split exact sequence; moreover, examining the proof of [Multilinear Algebra] §Hom and the Tensor Product, ⁋Proposition 1, we see that in fact if the above sequence is exact when \(K=N\), that is, if

\[\Hom_\rMod{A}(N,L) \rightarrow \Hom_\rMod{A}(N,N) \rightarrow 0\]

is surjective, then the original exact sequence \(0 \rightarrow M \rightarrow L \rightarrow N \rightarrow 0\) is a split exact sequence. Thus we obtain the following.

Corollary 6 Suppose an arbitrary short exact sequence

\[0 \rightarrow M \rightarrow L \rightarrow N \rightarrow 0\]

is given. If \(N\) is finitely presented and for every maximal ideal \(\mathfrak{m}\),

\[0 \rightarrow M_\mathfrak{m} \rightarrow L_\mathfrak{m} \rightarrow N_\mathfrak{m} \rightarrow 0\]

is a splitting exact sequence, then the original exact sequence splits.

Radical of an Ideal

Strictly speaking, the following result is not related to localization, but because we use a multiplicative subset in stating it, we mention it here and move on.

Proposition 7 For a ring \(A\) and a multiplicative subset \(S\), suppose \(\mathfrak{a}\) is maximal among ideals disjoint from \(S\). Then \(\mathfrak{a}\) is a prime ideal.

Proof

First, since \(1\in S\), from \(\mathfrak{a}\cap S=\emptyset\) we have \(1\not\in \mathfrak{a}\), that is, \(\mathfrak{a}\neq A\). Now suppose two elements of \(A\), \(a_1,a_2\), are given, and let us show that if \(a_1,a_2\not\in \mathfrak{a}\), then \(a_1a_2\not\in \mathfrak{a}\). By the maximality of \(\mathfrak{a}\), the two ideals \(\mathfrak{a}+(a_1)\) and \(\mathfrak{a}+(a_2)\) must intersect \(S\), so for some \(b_1,b_2\in A\) and \(x_1,x_2\in \mathfrak{a}\) we must have \(a_ib_i+x_i\in S\). But since \(S\) is closed under multiplication, the following element

\[(a_1b_1+x_1)(a_2b_2+x_2)=a_1a_2b_1b_2+a_1b_1x_2+a_2b_2x_1+x_1x_2\]

must also belong to \(S\). If, contrary to the conclusion, \(a_1a_2\in \mathfrak{a}\), then all four terms on the right-hand side belong to \(\mathfrak{a}\), which contradicts the assumption that \(\mathfrak{a}\) and \(S\) do not intersect.

In a similar vein, we obtain the following.

Corollary 8 For a ring \(A\) and an ideal \(\mathfrak{a}\), define the radical of \(\mathfrak{a}\), \(\sqrt{\mathfrak{a}}\), by the formula

\[\sqrt{\mathfrak{a}}=\{a\mid a^k\in \mathfrak{a}\text{ for some $k\in \mathbb{N}$}\}\]

Then

\[\sqrt{\mathfrak{a}}=\bigcap_\text{\scriptsize$\mathfrak{p}$ prime containing $\mathfrak{a}$} \mathfrak{p}\]

holds.

Proof

One direction is obvious. Conversely, if \(a\not\in \sqrt{\mathfrak{a}}\), it suffices to set \(S=\{a^k\mid k\geq 0\}\) and apply §Localization, ⁋Proposition 8.

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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