Lemma 10.161.7. Let $\varphi : R \to S$ be a ring map. Assume

$\varphi $ is smooth,

$R$ is reduced.

Then $S$ is reduced.

Lemma 10.161.7. Let $\varphi : R \to S$ be a ring map. Assume

$\varphi $ is smooth,

$R$ is reduced.

Then $S$ is reduced.

**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 reduced. Thus if $R$ is Noetherian, then $S$ is Noetherian and we get the result from Lemma 10.161.6.

In the general case 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. Now, if $x \in S$ is an element with $x^2 = 0$, then we can enlarge $R_0$ and assume that $x$ comes from an element $x_0 \in S_0$. After enlarging $R_0$ once more we may assume that $x_0^2 = 0$ in $S_0$. However, since $R_0 \subset R$ is reduced we see that $S_0$ is reduced and hence $x_0 = 0$ as desired. $\square$

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