The Stacks project

Lemma 10.36.2. Let $\varphi : R \to S$ be a ring map. Let $y \in S$. If there exists a finite $R$-submodule $M$ of $S$ such that $1 \in M$ and $yM \subset M$, then $y$ is integral over $R$.

Proof. Consider the map $\varphi : M \to M$, $x \mapsto y \cdot x$. By Lemma 10.16.2 there exists a monic polynomial $P \in R[T]$ with $P(\varphi ) = 0$. In the ring $S$ we get $P(y) = P(y) \cdot 1 = P(\varphi )(1) = 0$. $\square$

Comments (2)

Comment #6602 by Jonas Ehrhard on

Is that not just an application of Lemma 05BT to the morphism ? From there we get a polynomial with , and then .

There are also:

  • 2 comment(s) on Section 10.36: Finite and integral ring extensions

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Your email address will not be published.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 052I. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 10.36.2 as pdf