## 10.100 Base change and flatness

Some lemmas which deal with what happens with flatness when doing a base change.

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

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 10.39.7. 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 10.39.18 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 10.39.17. We are going to use the criterion of Lemma 10.39.5 (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. $\square$

Here is yet another application of the local criterion of flatness.

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. $\square$

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