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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 .

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

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