**Proof.**
Let $S = R[x]_ g/(f)$ be a presentation of $S$ as in Definition 10.144.1. Write $f = x^ n + a_1 x^{n - 1} + \ldots + a_ n$ with $a_ i \in R$. By Lemma 10.136.14 there exists a finite locally free and faithfully flat ring map $R \to S'$ such that $f = \prod (x - \alpha _ i)$ for certain $\alpha _ i \in S'$. Hence $R \to S'$ satisfies conditions (1), (2). Let $\mathfrak q \subset R[x]/(f)$ be a prime ideal with $g \not\in \mathfrak q$ (i.e., it corresponds to a prime of $S$). Let $\mathfrak p = R \cap \mathfrak q$ and let $\mathfrak q' \subset S'$ be a prime lying over $\mathfrak p$. Note that there are $n$ maps of $R$-algebras

\begin{eqnarray*} \varphi _ i : R[x]/(f) & \longrightarrow & S' \\ x & \longmapsto & \alpha _ i \end{eqnarray*}

To finish the proof we have to show that for some $i$ we have (a) the image of $\varphi _ i(g)$ in $\kappa (\mathfrak q')$ is not zero, and (b) $\varphi _ i^{-1}(\mathfrak q') = \mathfrak q$. Because then we can just take $g' = \varphi _ i(g)$, and $\varphi = \varphi _ i$ for that $i$.

Let $\overline{f}$ denote the image of $f$ in $\kappa (\mathfrak p)[x]$. Note that as a point of $\mathop{\mathrm{Spec}}(\kappa (\mathfrak p)[x]/(\overline{f}))$ the prime $\mathfrak q$ corresponds to an irreducible factor $f_1$ of $\overline{f}$. Moreover, $g \not\in \mathfrak q$ means that $f_1$ does not divide the image $\overline{g}$ of $g$ in $\kappa (\mathfrak p)[x]$. Denote $\overline{\alpha }_1, \ldots , \overline{\alpha }_ n$ the images of $\alpha _1, \ldots , \alpha _ n$ in $\kappa (\mathfrak q')$. Note that the polynomial $\overline{f}$ splits completely in $\kappa (\mathfrak q')[x]$, namely

\[ \overline{f} = \prod \nolimits _ i (x - \overline{\alpha }_ i) \]

Moreover $\varphi _ i(g)$ reduces to $\overline{g}(\overline{\alpha }_ i)$. It follows we may pick $i$ such that $f_1(\overline{\alpha }_ i) = 0$ and $\overline{g}(\overline{\alpha }_ i) \not= 0$. For this $i$ properties (a) and (b) hold. Some details omitted.
$\square$

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