Lemma 10.136.4. Let $R \to S$ be a smooth ring map. Let $R \to R'$ be any ring map. Then the base change $R' \to S' = R' \otimes _ R S$ is smooth.

**Proof.**
Let $\alpha : R[x_1, \ldots , x_ n] \to S$ be a presentation with kernel $I$. Let $\alpha ' : R'[x_1, \ldots , x_ n] \to R' \otimes _ R S$ be the induced presentation. Let $I' = \mathop{\mathrm{Ker}}(\alpha ')$. Since $0 \to I \to R[x_1, \ldots , x_ n] \to S \to 0$ is exact, the sequence $R' \otimes _ R I \to R'[x_1, \ldots , x_ n] \to R' \otimes _ R S \to 0$ is exact. Thus $R' \otimes _ R I \to I'$ is surjective. By Definition 10.136.1 there is a short exact sequence

and the $S$-module $\Omega _{S/R}$ is finite projective. In particular $I/I^2$ is a direct summand of $\Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S$. Consider the commutative diagram

Since the right vertical map is an isomorphism we see that the left vertical map is injective and surjective by what was said above. Thus we conclude that $\mathop{N\! L}\nolimits (\alpha ')$ is quasi-isomorphic to $\Omega _{S'/R'} \cong S' \otimes _ S \Omega _{S/R}$. And this is finite projective since it is the base change of a finite projective module. $\square$

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## Comments (2)

Comment #720 by Keenan Kidwell on

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