Localization Commutes with Hom
这篇文章记录了我对 [2, Proposition 2.10] 的理解.
$$
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\Hom}{Hom}
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
$$
Some notations:
- $R$ is a commutative ring with $1$.
- $S$ is an $R$-algebra.
- $M, N$ are $R$-modules.
- $L$ is an $S$-module.
Since $S$ is an $R$-algebra, $L$ is naturally an $R$-moudle.
Lemma
$S \otimes_R L \cong L$ via $1 \otimes l \longleftrightarrow l$.
Proof.
对于普通的 $R$-module $M$, 张量 $S$ 后得到的 $S$-moudle
$S \otimes_R M$ 的 module 结构为 $s \cdot (s_1 \otimes m) = (ss_1) \otimes m$.
但因为 $L$ 具有 $S$-module 结构, 所以
$$
(ss_1) \otimes m = s\cdot (s_1 \otimes m) =
(s_1 \otimes m)\cdot s = s_1 \otimes (m\cdot s)
$$
也就是说 $s$ 能兼容地作用在两个分量上.
$\square$
Proposition
There is a unique $S$-module homomorphism
$$
\alpha_{M}: S \otimes_{R} \Hom_R(M, N) \to
\Hom_S(S\otimes_R M, S\otimes_R N)
$$
that takes an element $1\otimes \varphi$ (in LHS) to
$1 \otimes_R \varphi: S\otimes_R M \to S\otimes_R N$ (in RHS).
这里实际上就是通过两种视角来解读 $1 \otimes \varphi$. 一方面, 它可以解读为两个元素的张量积 (in LHS), 另一方面, 将 $1$ 理解成 $S$ 上的恒等映射, 则可以把 $1 \otimes \varphi$ 解读成两个映射的张量积.
Proof.
The map of sets
$$
\alpha': \Hom_R(M, N) \to \Hom_S(S\otimes M, S \otimes N)
$$
taking a homomorphism $\varphi$ to $1 \otimes \varphi$ is easily seen to be
a map of $R$-modules.
The result follows by applying Lemma (5ae6389)
with $L = \Hom_S(S\otimes M, S \otimes N)$.
$\square$
Proposition
If $S$ is flat over $R$ and $M$ is finitely presented, then $\alpha_M$ is an isomorphism.
Proof.
具体的细节去看书, 这里只简略说一下思路.
第一步, 我们先证明 $M = R$ 的情形. 这里用到了同构 $\mathrm{Hom}_R(R, N) = N$ and $S \otimes_R R = S$.
第二步, 我们证明 $M = \bigoplus_{i=1}^n R$ is a free module of finite rank 的情形. 这里利用了 $\Hom$ commutes with finite direct sum, 以及 $\otimes$ commutes with direct sum. 于是 $\alpha_M$ 实际上分解成直和 $\alpha_1 \oplus \cdots \oplus \alpha_n$, 然后每一个分量 $\alpha_i$ 都是 isomorphism.
第三步, 对于一般的 finitely presented 的 $M$. 它有一个 finite presentation $$ F \xrightarrow{\varphi} G \xrightarrow{\psi} M \xrightarrow{} 0 $$ in which $F$ and $G$ are free module of finite rank. We can first apply $\Hom(\bullet, N)$ then apply $S \otimes \bullet$ or, in an opposite order, apply $S \otimes \bullet$ first, then $\Hom(\bullet, S\otimes N)$. We therefore obtain a commutative diagram $$ \xymatrix { 0 \ar[d] \ar[r] & 0 \ar[d] \ar[r] & S\otimes \Hom(M,N) \ar[r] \ar[d]^{\alpha_M} & S\otimes \Hom(G, N) \ar[r] \ar[d]^{\alpha_G} & S\otimes \Hom(F, N) \ar[d]^{\alpha_F} \\\ 0 \ar[r] & 0 \ar[r] & \Hom(S\otimes M, S\otimes N) \ar[r] & \Hom(S\otimes G, S\otimes N) \ar[r] & \Hom(S\otimes F, S\otimes N) } $$ in which the horizontal maps are induced from $\varphi$ and $\psi$. The first row is exact because $S$ is flat over $R$. The second row is exact because $\otimes$ and $\Hom$ are right and left exact functor. The diagram commutes because of the definition of $\alpha_\bullet$. We hence deduce from the well-known five lemma that $\alpha_M$ is isomorphism. $\square$
第一步, 我们先证明 $M = R$ 的情形. 这里用到了同构 $\mathrm{Hom}_R(R, N) = N$ and $S \otimes_R R = S$.
第二步, 我们证明 $M = \bigoplus_{i=1}^n R$ is a free module of finite rank 的情形. 这里利用了 $\Hom$ commutes with finite direct sum, 以及 $\otimes$ commutes with direct sum. 于是 $\alpha_M$ 实际上分解成直和 $\alpha_1 \oplus \cdots \oplus \alpha_n$, 然后每一个分量 $\alpha_i$ 都是 isomorphism.
第三步, 对于一般的 finitely presented 的 $M$. 它有一个 finite presentation $$ F \xrightarrow{\varphi} G \xrightarrow{\psi} M \xrightarrow{} 0 $$ in which $F$ and $G$ are free module of finite rank. We can first apply $\Hom(\bullet, N)$ then apply $S \otimes \bullet$ or, in an opposite order, apply $S \otimes \bullet$ first, then $\Hom(\bullet, S\otimes N)$. We therefore obtain a commutative diagram $$ \xymatrix { 0 \ar[d] \ar[r] & 0 \ar[d] \ar[r] & S\otimes \Hom(M,N) \ar[r] \ar[d]^{\alpha_M} & S\otimes \Hom(G, N) \ar[r] \ar[d]^{\alpha_G} & S\otimes \Hom(F, N) \ar[d]^{\alpha_F} \\\ 0 \ar[r] & 0 \ar[r] & \Hom(S\otimes M, S\otimes N) \ar[r] & \Hom(S\otimes G, S\otimes N) \ar[r] & \Hom(S\otimes F, S\otimes N) } $$ in which the horizontal maps are induced from $\varphi$ and $\psi$. The first row is exact because $S$ is flat over $R$. The second row is exact because $\otimes$ and $\Hom$ are right and left exact functor. The diagram commutes because of the definition of $\alpha_\bullet$. We hence deduce from the well-known five lemma that $\alpha_M$ is isomorphism. $\square$
Corollary
(Localization commutes with Hom)
If $M$ is finitely presented, then $\alpha_M$ provides a natural isomorphism
$$
S^{-1}\Hom_R(M, N) \cong \Hom_{S^{-1}R}(S^{-1}M, S^{-1}N).
$$
Proof.
$S^{-1}R$ is flat over $R$ c.f.
[1, Corollary 3.6]
.
Also $S^{-1}M \cong S^{-1}R \otimes_R M$ c.f.
[2, Lemma 2.4]
.
$\square$
In the cases of primary interest to us, $R$ is Noetherian. 在这个条件下, finitely presented 和 finitely generated 是一回事.
References
- 1
- M.F.Atiyah and I.G.MacDonald. Introduction to Commutative Algebra. Addison-Wesley series in mathematics.
- 2
- David Eisenbud. Commutative Algebra with a View Toward Algebraic Geometry. GTM 150.