Lemma 10.101.7. Let $R \to S$ be a ring map. Let $M$ be an $R$-module. Assume

1. $R$ is Artinian

2. $R \to S$ is injective, and

3. $M \otimes _ R S$ 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 $R$. $\square$

There are also:

• 2 comment(s) on Section 10.101: Flatness criteria over Artinian rings

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 051L. Beware of the difference between the letter 'O' and the digit '0'.