Lemma 29.34.15. Let

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

be a cartesian diagram of schemes. Let $W \subset X$, resp. $W' \subset X'$ be the open subscheme of points where $f$, resp. $f'$ is smooth. Then $W' = (g')^{-1}(W)$ if

$f$ is flat and locally of finite presentation, or

$f$ is locally of finite presentation and $g$ is flat.

**Proof.**
Assume first that $f$ locally of finite type. Consider the set

\[ T = \{ x \in X \mid X_{f(x)}\text{ is smooth over }\kappa (f(x))\text{ at }x\} \]

and the corresponding set $T' \subset X'$ for $f'$. Then we claim $T' = (g')^{-1}(T)$. Namely, let $s' \in S'$ be a point, and let $s = g(s')$. Then we have

\[ X'_{s'} = \mathop{\mathrm{Spec}}(\kappa (s')) \times _{\mathop{\mathrm{Spec}}(\kappa (s))} X_ s \]

In other words the fibres of the base change are the base changes of the fibres. Hence the claim is equivalent to Algebra, Lemma 10.137.19.

Thus case (1) follows because in case (1) $T$ is the (open) set of points where $f$ is smooth by Lemma 29.34.14.

In case (2) let $x' \in W'$. Then $g'$ is flat at $x'$ (Lemma 29.25.7) and $g \circ f$ is flat at $x'$ (Lemma 29.25.5). It follows that $f$ is flat at $x = g'(x')$ by Lemma 29.25.13. On the other hand, since $x' \in T'$ (Lemma 29.34.5) we see that $x \in T$. Hence $f$ is smooth at $x$ by Lemma 29.34.14.
$\square$

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