10.100 Base change and flatness
Some lemmas which deal with what happens with flatness when doing a base change.
Lemma 10.100.1. Let
\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.
If M is flat over R then M' is flat over R'.
If M' is flat over R' and R \to R' is flat then M is flat over R.
In particular we have
If S is flat over R then S' is flat over R'.
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
the horizontal arrows are flat ring maps
M is flat over R,
\mathfrak m_ R R' = \mathfrak m_{R'},
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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