Properties of Localization

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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

Localization and Hom, Tensor

We begin by proving a lemma. Define an \(A\)-module homomorphism \(S^{-1}A\times_A M \rightarrow S^{-1}M\) by \((r/u, x)\mapsto rx/u\); this is an \(A\)-bilinear map, and therefore induces an \(A\)-linear map \(S^{-1}A\otimes_A M \rightarrow S^{-1}M\). (§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, we first define a function from \(M\times S\) to \(S^{-1}A\otimes_AM\) by

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

This function determines a well-defined \(A\)-linear map from \(S^{-1}M\) to \(S^{-1}A\otimes_AM\). To verify this, we need only check that the function respects the equivalence relation on \(M\times S\). Thus, suppose two elements of \(M\times S\) satisfy \((x,s)\sim (x',s')\). Then there exists some \(t\in S\) such that \(tsx'=ts'x\), and from this we obtain

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

Since \(ts',ts\in A\), moving \(ts'\) and \(ts\) to the left of \(\otimes\) on each side yields

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

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

In particular, this allows us to establish the functoriality of module localization. For any \(u: M \rightarrow M'\), we define \(S^{-1}M \rightarrow S^{-1}M'\) by identifying both sides with localization via the map

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

In general, tensor products are right exact, but in this case the functor is exact.

Proposition 2 \(S^{-1}A\) is a flat \(A\)-module. (§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. By Lemma 1, it suffices to show that the induced linear map \(S^{-1}M \rightarrow S^{-1}M'\) is injective. Take any \(x/s\in S^{-1}M\) and suppose its image \(u(x)/s\) in \(S^{-1}M'\) is zero. 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 we must have \(tx=0\) in \(M\). Hence 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, if \(u:M \rightarrow M'\) is injective (resp. surjective, bijective), then the induced map \(S^{-1}M \rightarrow S^{-1}M'\) is also injective (resp. surjective, bijective). Proposition 4 can be regarded as a sort of (strong) converse to this. To prove it, we first establish the following lemma.

Lemma 3 Let \(M\) be an \(A\)-module and consider the localization \(\epsilon_\mathfrak{m}:M \rightarrow M_\mathfrak{m}\) at a maximal ideal \(\mathfrak{m}\) of \(A\). Then an element \(x\) of \(M\) is zero if and only if for every maximal ideal \(\mathfrak{m}\) of \(A\), 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 \(\mathfrak{m}\), the localized map \(u_\mathfrak{m}: M_\mathfrak{m} \rightarrow N_\mathfrak{m}\) is a monomorphism (resp. epimorphism, isomorphism).

The proof follows by applying Lemma 3 to the kernel and cokernel.

The following proposition will be used frequently hereafter, so it is worth remembering the statement even if one does not dwell on the proof.

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 map

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

fits into the following commutative diagram

so the proposition holds in this case. Here the vertical maps come from the isomorphisms

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

Next, since \(\Hom\) and \(\otimes\) commute with finite direct sums, the proposition also holds for any flat \(A\)-module \(E\) and any finitely generated free \(A\)-module \(M\). Finally, for the case where \(M\) is finitely presented, take a free presentation

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

and apply the four lemma to the following commutative diagram

In particular, suppose a short exact sequence

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

is given. Then this exact sequence is split 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 splitting exact sequence. Moreover, examining the proof of §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 split. Thus we obtain the following.

Corollary 6 Suppose a 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

The following result is not strictly related to localization, but we mention it here because the proof uses a multiplicative subset.

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

Let \(a_1,a_2\) be two elements of \(A\); we 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 each intersect \(S\), so there exist suitable \(b_1,b_2\in A\) and \(x_1,x_2\in \mathfrak{a}\) such that \(a_ib_i+x_i\in S\). Since \(S\) is closed under multiplication, the 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 lie in \(\mathfrak{a}\), contradicting the assumption that \(\mathfrak{a}\) and \(S\) are disjoint.

In a similar vein we obtain the following.

Corollary 8 For an ideal \(\mathfrak{a}\) of a ring \(A\), define the radical \(\sqrt{\mathfrak{a}}\) 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. For the converse, if \(a\not\in \sqrt{\mathfrak{a}}\) then set \(S=\{a^k\mid k\geq 1\}\) 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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