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

Lemma 10.137.13. 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 $S$.

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 10.17.10, there exists a finite covering $\mathop{\mathrm{Spec}}(S) = \bigcup D(g_ i)$ such that each $S_{g_ i}$ is smooth. By Lemma 10.23.3 this implies that $S$ is of finite presentation over $R$. According to Lemma 10.134.13 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 10.78.2 this implies that $\mathop{N\! L}\nolimits _{S/R}$ is quasi-isomorphic to a finite projective $S$-module. $\square$

## Comments (3)

Comment #3626 by Herman Rohrbach on

Suggested slogan: "A ring map is smooth if and only if it is smooth at all primes of the target."

Comment #4736 by Andy on

Hi, could you elaborate on why being quasi-isomorphic to finite projective module in degree $0$ on cover implies it is true over the entire ring? It's not clear to me that 00NX as stated applies?

Comment #4820 by on

The lemma applies to the cohomology module.

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