Lemma 10.101.7. Let R \to S be a ring map. Let M be an R-module. Assume
R is Artinian
R \to S is injective, and
M \otimes _ R S is a flat S-module.
Then M is a flat R-module.
Lemma 10.101.7. Let R \to S be a ring map. Let M be an R-module. Assume
R is Artinian
R \to S is injective, and
M \otimes _ R S is a flat S-module.
Then M is a flat R-module.
Proof. First proof: Let I \subset R be the Jacobson radical of R. Then I is nilpotent and M/IM is flat over R/I as R/I is a product of fields, see Section 10.53. Hence M is flat by an application of Lemma 10.101.5.
Second proof: By Lemma 10.53.6 we may write R = \prod R_ i as a finite product of local Artinian rings. This induces similar product decompositions for both R and S. Hence we reduce to the case where R is local Artinian (details omitted).
Assume that R \to S, M are as in the lemma satisfying (1), (2), and (3) and in addition that R is local with maximal ideal \mathfrak m. Let A be a set and x_\alpha \in A be elements such that \overline{x}_\alpha forms a basis for M/\mathfrak mM over R/\mathfrak m. By Nakayama's Lemma 10.20.1 we see that the elements x_\alpha generate M as an R-module. Set N = S \otimes _ R M and I = \mathfrak mS. Then \{ 1 \otimes x_\alpha \} _{\alpha \in A} is a family of elements of N which form a basis for N/IN. Moreover, since N is flat over S we have \text{Tor}_1^ S(S/I, N) = 0. Thus we conclude from Lemma 10.101.3 that N is free on \{ 1 \otimes x_\alpha \} _{\alpha \in A}. The injectivity of R \to S then guarantees that there cannot be a nontrivial relation among the x_\alpha with coefficients in R. \square
Comments (0)
There are also: