Lemma 10.38.3. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. The set of elements of $S$ which are integral over $I$ form a $R$-submodule of $S$. Furthermore, if $s \in S$ is integral over $R$, and $s'$ is integral over $I$, then $ss'$ is integral over $I$.
Proof. Closure under addition is clear from the characterization of Lemma 10.38.2. Any element $s \in S$ which is integral over $R$ corresponds to the degree $0$ element $s$ of $S[x]$ which is integral over $A$ (because $R \subset A$). Hence we see that multiplication by $s$ on $S[x]$ preserves the property of being integral over $A$, by Lemma 10.36.7. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.