
Lemma 10.37.3. 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 10.37.2. 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 10.35.7. $\square$

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).