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

Comment #6602 by Jonas Ehrhard on

Is that not just an application of Lemma 05BT to the morphism $\varphi: M \to M, x \mapsto y \cdot x$? From there we get a polynomial $P$ with $P(\varphi) = 0$, and then $P(y) = P(y) \cdot 1 = P(\varphi)(1) = 0$.

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