Lemma 10.123.3 (00PV)—The Stacks project

Lemma 10.123.3. Let $R$ be a ring. Let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. Assume $t$ is integral over $R[x]$. Let $p \in R[x]$, $p = a_0 + a_1x + \ldots + a_ k x^ k$ such that $t \varphi (p) \in \mathop{\mathrm{Im}}(\varphi )$. Then there exists a $q \in R[x]$ and $n \geq 0$ such that $\varphi (a_ k)^ n t - \varphi (q)$ is integral over $R$.

Proof. Let $R'$ and $S'$ be the localization of $R$ and $S$ at the element $a_ k$. Let $\varphi ' : R'[x] \to S'$ be the localization of $\varphi $. Let $t' \in S'$ be the image of $t$. Set $p' = p/a_ k \in R'[x]$. Then $t' \varphi '(p') \in \mathop{\mathrm{Im}}(\varphi ')$ since $t \varphi (p) \in \mathop{\mathrm{Im}}(\varphi )$. As $p'$ is monic, by Lemma 10.123.2 there exists a $q' \in R'[x]$ such that $t' - \varphi '(q')$ is integral over $R'$. We may choose an $n \geq 0$ and an element $q \in R[x]$ such that $a_ k^ n q'$ is the image of $q$. Then $\varphi (a_ k)^ n t - \varphi (q)$ is an element of $S$ whose image in $S'$ is integral over $R'$. By Lemma 10.36.11 there exists an $m \geq 0$ such that $\varphi (a_ k)^ m(\varphi (a_ k)^ n t - \varphi (q))$ is integral over $R$. Thus $\varphi (a_ k)^{m + n}t - \varphi (a_ k^ m q)$ is integral over $R$ as desired. $\square$

Comments (2)

Comment #3631 by Brian Conrad on October 19, 2018 at 21:40

Why not localize at $a_k$ throughout from the start (harmless for the proof), to instantly reduce to the case $a_k=1$ handled by the Lemma invoked near the end of the argument and thereby avoid all of the futzing around here?

Comment #3730 by Johan on October 23, 2018 at 18:58

Great! This also means I can move Lemma 115.5.1 to the obsolte chapter as it is now no longer used. Changes are here.

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