The Stacks project

Lemma 10.39.16. Let $R \to S$ be a flat ring map. The following are equivalent:

  1. $R \to S$ is faithfully flat,

  2. the induced map on $\mathop{\mathrm{Spec}}$ is surjective, and

  3. any closed point $x \in \mathop{\mathrm{Spec}}(R)$ is in the image of the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$.

Proof. This follows quickly from Lemma 10.39.15, because we saw in Remark 10.18.5 that $\mathfrak p$ is in the image if and only if the ring $S \otimes _ R \kappa (\mathfrak p)$ is nonzero. $\square$


Comments (0)

There are also:

  • 3 comment(s) on Section 10.39: Flat modules and flat ring maps

Post a comment

Your email address will not be published.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.




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 00HQ. Beware of the difference between the letter 'O' and the digit '0'.