
Lemma 10.143.1. Let $R$ be a ring. Let $f \in R[x]$ be a monic polynomial. Let $R \to B$ be a ring map. If $h \in B[x]/(f)$ is integral over $R$, then the element $f' h$ can be written as $f'h = \sum _ i b_ i x^ i$ with $b_ i \in B$ integral over $R$.

Proof. Say $h^ e + r_1 h^{e - 1} + \ldots + r_ e = 0$ in the ring $B[x]/(f)$ with $r_ i \in R$. There exists a finite free ring extension $B \subset B'$ such that $f = (x - \alpha _1) \ldots (x - \alpha _ d)$ for some $\alpha _ i \in B'$, see Lemma 10.134.9. Note that each $\alpha _ i$ is integral over $R$. We may represent $h = h_0 + h_1 x + \ldots + h_{d - 1} x^{d - 1}$ with $h_ i \in B$. Then it is a universal fact that

$f' h \equiv \sum \nolimits _{i = 1, \ldots , d} h(\alpha _ i) (x - \alpha _1) \ldots \widehat{(x - \alpha _ i)} \ldots (x - \alpha _ d)$

as elements of $B[x]/(f)$. You prove this by evaluating both sides at the points $\alpha _ i$ over the ring $B_{univ} = \mathbf{Z}[\alpha _ i, h_ j]$ (some details omitted). By our assumption that $h$ satisfies $h^ e + r_1 h^{e - 1} + \ldots + r_ e = 0$ in the ring $B[x]/(f)$ we see that

$h(\alpha _ i)^ e + r_1 h(\alpha _ i)^{e - 1} + \ldots + r_ e = 0$

in $B'$. Hence $h(\alpha _ i)$ is integral over $R$. Using the formula above we see that $f'h \equiv \sum _{j = 0, \ldots , d - 1} b'_ j x^ j$ in $B'[x]/(f)$ with $b'_ j \in B'$ integral over $R$. However, since $f' h \in B[x]/(f)$ and since $1, x, \ldots , x^{d - 1}$ is a $B'$-basis for $B'[x]/(f)$ we see that $b'_ j \in B$ as desired. $\square$

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