The Stacks project

Proof. Write $t \varphi (p) = \varphi (r)$ for some $r \in R[x]$. Using euclidean division, write $r = qp + r'$ with $q, r' \in R[x]$ and $\deg (r') < \deg (p)$. We may replace $t$ by $t - \varphi (q)$ which is still integral over $R[x]$, so that we obtain $t \varphi (p) = \varphi (r')$. In the ring $S_ t$ we may write this as $\varphi (p) - (1/t) \varphi (r') = 0$. This implies that $\varphi (x)$ gives an element of the localization $S_ t$ which is integral over $\varphi (R)[1/t] \subset S_ t$. On the other hand, $t$ is integral over the subring $\varphi (R)[\varphi (x)] \subset S$. Combined we conclude that $t$ is integral over the subring $\varphi (R)[1/t] \subset S_ t$, see Lemma 10.36.6. In other words there exists an equation of the form

in $S_ t$ with $r_{i, j} \in R$. This means that $t^{d + N} + \sum _{i < d} \sum _{j = 0, \ldots , n_ i} \varphi (r_{i, j}) t^{i + N - j} = 0$ in $S$ for some $N$ large enough. In other words $t$ is integral over $R$. $\square$