Lemma 10.100.2. Consider a commutative diagram of local rings and local homomorphisms

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

Let $M$ be a finite $S$-module. Assume that

1. the horizontal arrows are flat ring maps

2. $M$ is flat over $R$,

3. $\mathfrak m_ R R' = \mathfrak m_{R'}$,

4. $R'$ and $S'$ are Noetherian.

Then $M' = M \otimes _ S S'$ is flat over $R'$.

Proof. Since $\mathfrak m_ R \subset R$ and $R \to R'$ is flat, we get $\mathfrak m_ R \otimes _ R R' = \mathfrak m_ R R' = \mathfrak m_{R'}$ by assumption (3). Observe that $M'$ is a finite $S'$-module which is flat over $R$ by Lemma 10.39.9. Thus $\mathfrak m_ R \otimes _ R M' \to M'$ is injective. Then we get

$\mathfrak m_ R \otimes _ R M' = \mathfrak m_ R \otimes _ R R' \otimes _{R'} M' = \mathfrak m_{R'} \otimes _{R'} M'$

Thus $\mathfrak m_{R'} \otimes _{R'} M' \to M'$ is injective. This shows that $\text{Tor}_1^{R'}(\kappa _{R'}, M') = 0$ (Remark 10.75.9). Thus $M'$ is flat over $R'$ by Lemma 10.99.7. $\square$

Comment #6059 by BB on

Should the tensor defining $M'$ should take place over $S$ instead of $R$? If not, the finiteness of $M'$ over $S'$ asserted in the second sentence seems confusing.

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