The Stacks project

Lemma 10.39.18. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset.

  1. The localization $S^{-1}R$ is a flat $R$-algebra.

  2. If $M$ is an $S^{-1}R$-module, then $M$ is a flat $R$-module if and only if $M$ is a flat $S^{-1}R$-module.

  3. Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak p}$ is a flat $R_{\mathfrak p}$-module for all primes $\mathfrak p$ of $R$.

  4. Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak m}$ is a flat $R_{\mathfrak m}$-module for all maximal ideals $\mathfrak m$ of $R$.

  5. Suppose $R \to A$ is a ring map, $M$ is an $A$-module, and $g_1, \ldots , g_ m \in A$ are elements generating the unit ideal of $A$. Then $M$ is flat over $R$ if and only if each localization $M_{g_ i}$ is flat over $R$.

  6. Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$) for all primes $\mathfrak q$ of $A$.

  7. Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p = R \cap \mathfrak m$) for all maximal ideals $\mathfrak m$ of $A$.

Proof. Let us prove the last statement of the lemma. In the proof we will use repeatedly that localization is exact and commutes with tensor product, see Sections 10.9 and 10.12.

Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Assume that $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module for all maximal ideals $\mathfrak m$ of $A$ (with $\mathfrak p = R \cap \mathfrak m$). Let $I \subset R$ be an ideal. We have to show the map $I \otimes _ R M \to M$ is injective. We can think of this as a map of $A$-modules. By assumption the localization $(I \otimes _ R M)_{\mathfrak m} \to M_{\mathfrak m}$ is injective because $(I \otimes _ R M)_{\mathfrak m} = I_{\mathfrak p} \otimes _{R_{\mathfrak p}} M_{\mathfrak m}$. Hence the kernel of $I \otimes _ R M \to M$ is zero by Lemma 10.23.1. Hence $M$ is flat over $R$.

Conversely, assume $M$ is flat over $R$. Pick a prime $\mathfrak q$ of $A$ lying over the prime $\mathfrak p$ of $R$. Suppose that $I \subset R_{\mathfrak p}$ is an ideal. We have to show that $I \otimes _{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is injective. We can write $I = J_{\mathfrak p}$ for some ideal $J \subset R$. Then the map $I \otimes _{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is just the localization (at $\mathfrak q$) of the map $J \otimes _ R M \to M$ which is injective. Since localization is exact we see that $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module.

This proves (7) and (6). The other statements follow in a straightforward way from the last statement (proofs omitted). $\square$

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