Smoothness is preserved under base change

Lemma 10.137.4. 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 10.137.1 there is a short exact sequence

$0 \to I/I^2 \to \Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S \to \Omega _{S/R} \to 0$

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

$\xymatrix{ R' \otimes _ R (I/I^2) \ar[r] \ar[d] & R' \otimes _ R (\Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S) \ar[d] \\ I'/(I')^2 \ar[r] & \Omega _{R'[x_1, \ldots , x_ n]/R'} \otimes _{R'[x_1, \ldots , x_ n]} (R' \otimes _ R S) }$

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. $\square$

## Comments (2)

Comment #720 by Keenan Kidwell on

In the fourth line of the proof, $R^\prime\otimes_rS$ should be $R^\prime\otimes_RS$.

Comment #4883 by Rankeya on

Suggested slogan: Smoothness is preserved under base change.

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