The Stacks project

Proof. Let $L/K$ be a field extension such that we can write $x^ m + b_{m - 1} x^{m - 1} + \ldots + b_0 = \prod _{i = 1}^ m (x - \beta _ i)$ with $\beta _ i \in L$. See Fields, Section 9.16. Each $\beta _ i$ is integral over $R_0$. Since each $a_ i$ is a homogeneous polynomial in $\beta _1, \ldots , \beta _ m$ we deduce the same for the $a_ i$ (use Lemma 10.36.7). This proves (1).

Let $R$ be as in (2). Choose $c_0, \ldots , c_{m - n - 1} \in K$ such that

This equation implies

Thus $c_ j \in R$ for all $j$. Dividing out the radical $\sqrt{(b_0, \ldots , b_{m - 1})}$ we get a reduced ring $\overline{R}$. We have to show that the images $\overline{a}_ i \in \overline{R}$ are zero. And in $\overline{R}[x]$ we have the relation

It is easy to see that this implies $\overline{a}_ i = 0$ for all $i$. Indeed by Lemma 10.25.1 the localization of $\overline{R}$ at a minimal prime $\mathfrak {p}$ is a field and $\overline{R}_{\mathfrak p}[x]$ a UFD. Thus $f = x^ n + \sum \overline{a}_ i x^ i$ is associated to $x^ n$ and since $f$ is monic $f = x^ n$ in $\overline{R}_{\mathfrak p}[x]$. Then there exists an $s \in \overline{R}$, $s \not\in \mathfrak p$ such that $s(f - x^ n) = 0$. Therefore all $\overline{a}_ i$ lie in $\mathfrak p$ and we conclude by Lemma 10.25.2. $\square$