$\xymatrix{ S \ar[r] & S' \\ R \ar[r] \ar[u] & R' \ar[u] }$

be a commutative diagram of local homomorphisms of local rings. Assume that $S'$ is a localization of the tensor product $S \otimes _ R R'$. Let $M$ be an $S$-module and set $M' = S' \otimes _ S M$.

1. If $M$ is flat over $R$ then $M'$ is flat over $R'$.

2. If $M'$ is flat over $R'$ and $R \to R'$ is flat then $M$ is flat over $R$.

In particular we have

1. If $S$ is flat over $R$ then $S'$ is flat over $R'$.

2. If $R' \to S'$ and $R \to R'$ are flat then $S$ is flat over $R$.

Proof. Proof of (1). If $M$ is flat over $R$, then $M \otimes _ R R'$ is flat over $R'$ by Lemma 10.38.7. If $W \subset S \otimes _ R R'$ is the multiplicative subset such that $W^{-1}(S \otimes _ R R') = S'$ then $M' = W^{-1}(M \otimes _ R R')$. Hence $M'$ is flat over $R'$ as the localization of a flat module, see Lemma 10.38.19 part (5). This proves (1) and in particular, we see that (3) holds.

Proof of (2). Suppose that $M'$ is flat over $R'$ and $R \to R'$ is flat. By (3) applied to the diagram reflected in the northwest diagonal we see that $S \to S'$ is flat. Thus $S \to S'$ is faithfully flat by Lemma 10.38.17. We are going to use the criterion of Lemma 10.38.5 (3) to show that $M$ is flat. Let $I \subset R$ be an ideal. If $I \otimes _ R M \to M$ has a kernel, so does $(I \otimes _ R M) \otimes _ S S' \to M \otimes _ S S' = M'$. Note that $I \otimes _ R R' = IR'$ as $R \to R'$ is flat, and that

$(I \otimes _ R M) \otimes _ S S' = (I \otimes _ R R') \otimes _{R'} (M \otimes _ S S') = IR' \otimes _{R'} M'.$

From flatness of $M'$ over $R'$ we conclude that this maps injectively into $M'$. This concludes the proof of (2), and hence (4) is true as well. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).