Lemma 10.122.9. Suppose $R\subset S$ is an inclusion of domains and let $x \in S$. Assume $x$ is (strongly) transcendental over $R$ and that $S$ is finite over $R[x]$. Then $R \to S$ is not quasi-finite at any prime of $S$.

**Proof.**
As a first case, assume that $R$ is normal, see Definition 10.36.11. By Lemma 10.36.14 we see that $R[x]$ is normal. Take a prime $\mathfrak q \subset S$, and set $\mathfrak p = R \cap \mathfrak q$. Assume that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite. This would be the case if $R \to S$ is quasi-finite at $\mathfrak q$. Let $\mathfrak r = R[x] \cap \mathfrak q$. Then since $\kappa (\mathfrak p) \subset \kappa (\mathfrak r) \subset \kappa (\mathfrak q)$ we see that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak r)$ is finite too. Thus the inclusion $\mathfrak r \supset \mathfrak p R[x]$ is strict. By going down for $R[x] \subset S$, see Proposition 10.37.7, we find a prime $\mathfrak q' \subset \mathfrak q$, lying over the prime $\mathfrak pR[x]$. Hence the fibre $\mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p))$ contains a point not equal to $\mathfrak q$, namely $\mathfrak q'$, whose closure contains $\mathfrak q$ and hence $\mathfrak q$ is not isolated in its fibre.

If $R$ is not normal, let $R \subset R' \subset K$ be the integral closure $R'$ of $R$ in its field of fractions $K$. Let $S \subset S' \subset L$ be the subring $S'$ of the field of fractions $L$ of $S$ generated by $R'$ and $S$. Note that by construction the map $S \otimes _ R R' \to S'$ is surjective. This implies that $R'[x] \subset S'$ is finite. Also, the map $S \subset S'$ induces a surjection on $\mathop{\mathrm{Spec}}$, see Lemma 10.35.17. We conclude by Lemma 10.121.6 and the normal case we just discussed. $\square$

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