Processing math: 100%

The Stacks project

Lemma 10.43.7. Let k be a field and let S be a k-algebra. Assume that S is reduced and that S_{\mathfrak p} is geometrically reduced for every minimal prime \mathfrak p of S. Then S is geometrically reduced.

Proof. Since S is reduced the map S \to \prod _{\mathfrak p\text{ minimal}} S_{\mathfrak p} is injective, see Lemma 10.25.2. If K/k is a field extension, then the maps

S \otimes _ k K \to (\prod S_\mathfrak p) \otimes _ k K \to \prod S_\mathfrak p \otimes _ k K

are injective: the first as k \to K is flat and the second by inspection because K is a free k-module. As S_\mathfrak p is geometrically reduced the ring on the right is reduced. Thus we see that S \otimes _ k K is reduced as a subring of a reduced ring. \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Your email address will not be published.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.