Lemma 10.135.13. Let $R \to S$ be a ring map. Then $R \to S$ is smooth if and only if $R \to S$ is smooth at every prime $\mathfrak q$ of $S$.

** A ring map is smooth if and only if it is smooth at all primes of the target **

**Proof.**
The direct implication is trivial. Suppose that $R \to S$ is smooth at every prime $\mathfrak q$ of $S$. Since $\mathop{\mathrm{Spec}}(S)$ is quasi-compact, see Lemma 10.16.10, there exists a finite covering $\mathop{\mathrm{Spec}}(S) = \bigcup D(g_ i)$ such that each $S_{g_ i}$ is smooth. By Lemma 10.22.3 this implies that $S$ is of finite presentation over $R$. According to Lemma 10.132.13 we see that $\mathop{N\! L}\nolimits _{S/R} \otimes _ S S_{g_ i}$ is quasi-isomorphic to a finite projective $S_{g_ i}$-module. By Lemma 10.77.2 this implies that $\mathop{N\! L}\nolimits _{S/R}$ is quasi-isomorphic to a finite projective $S$-module.
$\square$

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## Comments (1)

Comment #3626 by Herman Rohrbach on