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 $P(y) = P(y) \cdot 1 = P(\varphi )(1) = 0$.
$\square$

## 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Your email address will not be published.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.

## Comments (2)

Comment #6602 by Jonas Ehrhard on

Comment #6848 by Johan on

There are also: