Lemma 114.5.2. Let $\varphi : R \to S$ be a ring map. Suppose $t \in S$ satisfies the relation $\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_ n) t^ n = 0$. Set $u_ n = \varphi (a_ n)$, $u_{n-1} = u_ n t + \varphi (a_{n-1})$, and so on till $u_1 = u_2 t + \varphi (a_1)$. Then all of $u_ n, u_{n-1}, \ldots , u_1$ and $u_ nt, u_{n-1}t, \ldots , u_1t$ are integral over $R$, and the ideals $(\varphi (a_0), \ldots , \varphi (a_ n))$ and $(u_ n, \ldots , u_1)$ of $S$ are equal.

Proof. We prove this by induction on $n$. As $u_ n = \varphi (a_ n)$ we conclude from Algebra, Lemma 10.123.1 that $u_ nt$ is integral over $R$. Of course $u_ n = \varphi (a_ n)$ is integral over $R$. Then $u_{n - 1} = u_ n t + \varphi (a_{n - 1})$ is integral over $R$ (see Algebra, Lemma 10.36.7) and we have

$\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_{n - 1})t^{n - 1} + u_{n - 1}t^{n - 1} = 0.$

Hence by the induction hypothesis applied to the map $S' \to S$ where $S'$ is the integral closure of $R$ in $S$ and the displayed equation we see that $u_{n-1}, \ldots , u_1$ and $u_{n-1}t, \ldots , u_1t$ are all in $S'$ too. The statement on the ideals is immediate from the shape of the elements and the fact that $u_1t + \varphi (a_0) = 0$. $\square$

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