Finitely Generated Flat Modules over a Local Noetherian Ring are Free

Theorem 1 (Finitely Generated Modules over a Noetherian Ring are Finitely Presented).

Let $R$ be a Noetherian ring and $M$ be a finitely generated $R$-module. Then $M$ is finitely presented.

Proof. Since $M$ is finitely generated, there exists a surjective $R$-module homomorphism $\varphi :R^n\to M$ for some integer $n\ge 0$. Let $K=\ker (\varphi )$. The module $R^n$ is Noetherian because $R$ is. As a submodule of a Noetherian module, $K$ is finitely generated. Thus, there exists a surjective homomorphism $\psi :R^m\to K$ for some integer $m\ge 0$. Composing the inclusion $i:K\to R^n$ with $\psi$ gives a map $R^m\to R^n$ whose image is $K$. This yields the finite presentation of $M$:

$$R^m\to R^n\to M\to 0$$

$\square$

Theorem 2 (Finitely Generated Projective is Finitely Presented).

Let $R$ be a ring. If an $R$-module $P$ is finitely generated and projective, then $P$ is finitely presented.

Proof. Since $P$ is finitely generated, there exists a surjective homomorphism $\varphi :R^n\to P$. Let $K=\ker (\varphi )$. The short exact sequence $0\to K\to R^n\stackrel{\varphi }{\to }P\to 0$ splits because $P$ is projective. Therefore, $R^n\cong P\oplus K$. As a direct summand of a finitely generated module, $K$ is finitely generated. Thus there exists a surjection $\psi :R^m\to K$. This gives a finite presentation $R^m\stackrel{i\circ \psi }{\to }{R}^n\to P\to 0$, where $i:K\hookrightarrow R^n$ is the inclusion. $\square$

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Theorem 3 (Localization of a Flat Module is Flat).

Let $R$ be a commutative ring, $M$ an $R$-module, and $S$ a multiplicative subset of $R$. If $M$ is a flat $R$-module, then the localization $S^{-1}M$ is a flat $S^{-1}R$-module.

Proof. Let $0\to N^{\prime }\to N\to N^{\prime \prime }\to 0$ be an exact sequence of $S^{-1}R$-modules. Viewing this as a sequence of $R$-modules, tensoring with the flat $R$-module $M$ yields the exact sequence $0\to N^{\prime }{\otimes }_{R}M\to N{\otimes }_{R}M\to N^{\prime \prime }{\otimes }_{R}M\to 0$. Applying the exact functor $S^{-1}(-)$ and using the canonical isomorphism $S^{-1}(X{\otimes }_{R}Y)\cong S^{-1}X{\otimes }_{S^{-1}R}{S}^{-1}Y$ gives the exact sequence:

$$0\to S^{-1}(N^{\prime }){\otimes }_{S^{-1}R}{S}^{-1}M\to S^{-1}(N){\otimes }_{S^{-1}R}{S}^{-1}M\to S^{-1}(N^{\prime \prime }){\otimes }_{S^{-1}R}{S}^{-1}M\to 0$$

Since each $N_i$ is an $S^{-1}R$-module, $S^{-1}{N}_i\cong N_i$. The sequence is $0\to N^{\prime }{\otimes }_{S^{-1}R}{S}^{-1}M\to N{\otimes }_{S^{-1}R}{S}^{-1}M\to N^{\prime \prime }{\otimes }_{S^{-1}R}{S}^{-1}M\to 0$, which is the tensor product of the original sequence with $S^{-1}M$ over $S^{-1}R$. Its exactness implies $S^{-1}M$ is a flat $S^{-1}R$-module. $\square$

Theorem 4 (Localization of a Finitely Presented Module is Finitely Presented).

Let $R$ be a commutative ring, $M$ a finitely presented $R$-module, and $S$ a multiplicative subset of $R$. The localization $S^{-1}M$ is a finitely presented $S^{-1}R$-module.

Proof. Since $M$ is finitely presented, there exists an exact sequence $R^m\to R^n\to M\to 0$. The localization functor $S^{-1}(-)$ is exact, so applying it yields the exact sequence $(S^{-1}R{)}^m\to (S^{-1}R{)}^n\to S^{-1}M\to 0$. This is a finite presentation for $S^{-1}M$ over the ring $S^{-1}R$. $\square$

Theorem 5 (Hom Commutes with Localization for Finitely Presented Modules).

Let $R$ be a commutative ring, $M$ a finitely presented $R$-module, $N$ an $R$-module, and $\mathfrak{p}$ a prime ideal of $R$. There is a canonical isomorphism of $R_{\mathfrak{p}}$-modules:

$$({\mathrm{Hom}}_R(M,N){)}_{\mathfrak{p}}\cong {\mathrm{Hom}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}},N_{\mathfrak{p}})$$

Proof. A finite presentation $R^m\stackrel{f}{\to }{R}^n\to M\to 0$ gives rise to two exact sequences. First, applying the left-exact functor ${\mathrm{Hom}}_R(-,N)$ and then localizing at $\mathfrak{p}$ yields the exact sequence $0\to ({\mathrm{Hom}}_R(M,N){)}_{\mathfrak{p}}\to (N_{\mathfrak{p}}{)}^n\to (N_{\mathfrak{p}}{)}^m$. Second, localizing the presentation first and then applying ${\mathrm{Hom}}_{R_{\mathfrak{p}}}(-,N_{\mathfrak{p}})$ yields the exact sequence $0\to {\mathrm{Hom}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}},N_{\mathfrak{p}})\to (N_{\mathfrak{p}}{)}^n\to (N_{\mathfrak{p}}{)}^m$. The two modules $({\mathrm{Hom}}_R(M,N){)}_{\mathfrak{p}}$ and ${\mathrm{Hom}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}},N_{\mathfrak{p}})$ are the kernels of the same homomorphism $(N_{\mathfrak{p}}{)}^n\to (N_{\mathfrak{p}}{)}^m$ and are thus canonically isomorphic. $\square$

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Theorem 6 (Finitely Presented Flat Modules over a Local Ring are Free).

Let $(R,\mathfrak{m})$ be a local ring and let $M$ be a finitely presented flat $R$-module. Then $M$ is a free $R$-module of finite rank.

Proof. Let $k=R/\mathfrak{m}$ be the residue field. Let $\{{\bar{x}}_1,\dots ,{\bar{x}}_n\}$ be a basis for the $k$-vector space $M/\mathfrak{m}M$, and let $x_i\in M$ be lifts. By Nakayama’s Lemma, $\{x_1,\dots ,x_n\}$ generates $M$. This defines a surjective homomorphism $\varphi :R^n\to M$. Let $K=\ker (\varphi )$, yielding the short exact sequence $0\to K\to R^n\to M\to 0$. Since $M$ is finitely presented, $K$ is finitely generated. As $M$ is a flat $R$-module, ${\mathrm{Tor}}_1^R(M,k)=0$. The long exact sequence of $\mathrm{Tor}$ implies that the sequence obtained by tensoring with $k$ is also short exact:

$$0\to K{\otimes }_{R}k\to R^n{\otimes }_{R}k\stackrel{\bar{\varphi }}{\to }M{\otimes }_{R}k\to 0$$

This is the sequence $0\to K/\mathfrak{m}K\to k^n\stackrel{\bar{\varphi }}{\to }M/\mathfrak{m}M\to 0$. By construction, $\bar{\varphi }$ is an isomorphism, so $K/\mathfrak{m}K=0$. Since $K$ is finitely generated over a local ring, Nakayama’s Lemma implies $K=0$. Thus, $\varphi$ is an isomorphism and $M\cong R^n$. $\square$

Theorem 7 (Finitely Generated Projective Modules over a Local Ring are Free).

Let $(R,\mathfrak{m})$ be a local ring and let $M$ be a finitely generated projective $R$-module. Then $M$ is a free $R$-module of finite rank.

Proof. Since $M$ is a finitely generated projective $R$-module, it is finitely presented by Theorem 8. Every projective module is flat. Thus, $M$ is a finitely presented flat module over the local ring $R$. By Theorem 4, $M$ is a free $R$-module of finite rank. $\square$

Theorem 8 (Finitely Generated Flat Modules over a Local Noetherian Ring are Free).

Let $(R,\mathfrak{m})$ be a local Noetherian ring and let $M$ be a finitely generated flat $R$-module. Then $M$ is a free $R$-module of finite rank.

Proof. The ring $R$ is Noetherian and the module $M$ is finitely generated. By Theorem 1, $M$ is finitely presented. Thus, $M$ is a finitely presented flat module over the local ring $R$. By Theorem 4, $M$ must be a free $R$-module of finite rank. $\square$

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Theorem 9 (Finitely Presented Locally Free Modules are Projective).

Let $M$ be a finitely presented $R$-module. If $M_{\mathfrak{p}}$ is a free $R_{\mathfrak{p}}$-module for every prime ideal $\mathfrak{p}$ of $R$, then $M$ is a projective $R$-module.

Proof. An $R$-module is projective if and only if ${\mathrm{Hom}}_R(M,-)$ is an exact functor. This is equivalent to showing that for any surjection $\pi :F\to M$, the induced map ${\mathrm{Hom}}_R(M,\pi ):{\mathrm{Hom}}_R(M,F)\to {\mathrm{Hom}}_R(M,M)$ is surjective. A homomorphism is surjective if and only if it is surjective upon localization at every prime ideal $\mathfrak{p}$. Since $M$ is finitely presented, Theorem 5 provides a commutative diagram where the vertical maps are isomorphisms:

$$\begin{array}{ccc}({\mathrm{Hom}}_R(M,F){)}_{\mathfrak{p}}& \to & ({\mathrm{Hom}}_R(M,M){)}_{\mathfrak{p}}\\ \cong \downarrow & & \downarrow \cong \\ {\mathrm{Hom}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}},F_{\mathfrak{p}})& \to & {\mathrm{Hom}}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}},M_{\mathfrak{p}})\end{array}$$

The bottom map is induced by the surjection ${\pi }_{\mathfrak{p}}:F_{\mathfrak{p}}\to M_{\mathfrak{p}}$. By hypothesis, $M_{\mathfrak{p}}$ is a free $R_{\mathfrak{p}}$-module, hence projective. The projectivity of $M_{\mathfrak{p}}$ ensures the bottom map is surjective for every $\mathfrak{p}$. Consequently, the top map is surjective for every $\mathfrak{p}$, which implies the original map ${\mathrm{Hom}}_R(M,\pi )$ is surjective. Thus $M$ is projective. $\square$

Theorem 10 (Finitely Presented Flat Modules are Projective).

Let $R$ be a commutative ring. If $M$ is a finitely presented flat $R$-module, then $M$ is a projective $R$-module.

Proof. Let $M$ be a finitely presented flat $R$-module. For any prime ideal $\mathfrak{p}\subset R$, the ring $R_{\mathfrak{p}}$ is a local ring. By Theorem 2, the localization $M_{\mathfrak{p}}$ is a flat $R_{\mathfrak{p}}$-module. By Theorem 3, $M_{\mathfrak{p}}$ is a finitely presented $R_{\mathfrak{p}}$-module. By Theorem 4, any finitely presented flat module over a local ring is free. Thus, $M_{\mathfrak{p}}$ is a free $R_{\mathfrak{p}}$-module for every prime ideal $\mathfrak{p}$. The module $M$ is finitely presented and locally free. By Theorem 6, $M$ is projective. $\square$