Processing math: 100%

The Stacks project

Lemma 10.128.10 (Critère de platitude par fibres: locally nilpotent case). Let

\xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] }

be a commutative diagram in the category of rings. Let I \subset R be a locally nilpotent ideal and M an S'-module. Assume

  1. R \to S is of finite type,

  2. R \to S' is of finite presentation,

  3. M is a finitely presented S'-module,

  4. M/IM is flat as a S/IS-module, and

  5. M is flat as an R-module.

Then M is a flat S-module and S_\mathfrak q is flat and essentially of finite presentation over R for every \mathfrak q \subset S such that M \otimes _ S \kappa (\mathfrak q) is nonzero.

Proof. If M \otimes _ S \kappa (\mathfrak q) is nonzero, then S' \otimes _ S \kappa (\mathfrak q) is nonzero and hence there exists a prime \mathfrak q' \subset S' lying over \mathfrak q (Lemma 10.18.6). Let \mathfrak p \subset R be the image of \mathfrak q in \mathop{\mathrm{Spec}}(R). Then I \subset \mathfrak p as I is locally nilpotent hence M/\mathfrak p M is flat over S/\mathfrak pS. Hence we may apply Lemma 10.128.9 to R_\mathfrak p \to S_\mathfrak q \to S'_{\mathfrak q'} and M_{\mathfrak q'}. We conclude that M_{\mathfrak q'} is flat over S and S_\mathfrak q is flat and essentially of finite presentation over R. Since \mathfrak q' was an arbitrary prime of S' we also see that M is flat over S (Lemma 10.39.18). \square


Comments (0)


Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.