Lemma 10.123.8. Suppose $R \subset S$ is an inclusion of reduced rings and suppose that $x \in S$ is strongly transcendental over $R$. Let $\mathfrak q \subset S$ be a minimal prime and let $\mathfrak p = R \cap \mathfrak q$. Then the image of $x$ in $S/\mathfrak q$ is strongly transcendental over the subring $R/\mathfrak p$.

Proof. Suppose $u(a_0 + a_1x + \ldots + a_ k x^ k) \in \mathfrak q$. By Lemma 10.25.1 the local ring $S_{\mathfrak q}$ is a field, and hence $u(a_0 + a_1x + \ldots + a_ k x^ k)$ is zero in $S_{\mathfrak q}$. Thus $uu'(a_0 + a_1x + \ldots + a_ k x^ k) = 0$ for some $u' \in S$, $u' \not\in \mathfrak q$. Since $x$ is strongly transcendental over $R$ we get $uu'a_ i = 0$ for all $i$. This in turn implies that $ua_ i \in \mathfrak q$. $\square$

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