Lemma 10.161.9. Let $\varphi : R \to S$ be a ring map. Assume

$\varphi $ is smooth,

$R$ is normal.

Then $S$ is normal.

Lemma 10.161.9. Let $\varphi : R \to S$ be a ring map. Assume

$\varphi $ is smooth,

$R$ is normal.

Then $S$ is normal.

**Proof.**
Observe that $R \to S$ is flat with regular fibres (see the list of results on smooth ring maps in Section 10.141). In particular, the fibres are normal. Thus if $R$ is Noetherian, then $S$ is Noetherian and we get the result from Lemma 10.161.8.

The general case. First note that $R$ is reduced and hence $S$ is reduced by Lemma 10.161.7. Let $\mathfrak q$ be a prime of $S$ and let $\mathfrak p$ be the corresponding prime of $R$. Note that $R_{\mathfrak p}$ is a normal domain. We have to show that $S_{\mathfrak q}$ is a normal domain. To do this we may replace $R$ by $R_{\mathfrak p}$ and $S$ by $S_{\mathfrak p}$. Hence we may assume that $R$ is a normal domain.

Assume $R \to S$ smooth, and $R$ a normal domain. We may find a finitely generated $\mathbf{Z}$-subalgebra $R_0 \subset R$ and a smooth ring map $R_0 \to S_0$ such that $S \cong R \otimes _{R_0} S_0$, see remark (10) in Section 10.141. As $R_0$ is a Nagata domain (see Proposition 10.160.16) we see that its integral closure $R_0'$ is finite over $R_0$. Moreover, as $R$ is a normal domain it is clear that $R_0' \subset R$. Hence we may replace $R_0$ by $R_0'$ and $S_0$ by $R_0' \otimes _{R_0} S_0$ and assume that $R_0$ is a normal Noetherian domain. By the first paragraph of the proof we conclude that $S_0$ is a normal ring (it need not be a domain of course). In this way we see that $R = \bigcup R_\lambda $ is the union of normal Noetherian domains and correspondingly $S = \mathop{\mathrm{colim}}\nolimits R_\lambda \otimes _{R_0} S_0$ is the colimit of normal rings. This implies that $S$ is a normal ring. Some details omitted. $\square$

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