Lemma 37.72.6. Consider a commutative diagram
of morphisms of schemes. Assume $X \to S$ is flat, $f$ is proper, $\dim (X_ y) \leq 1$ for $y \in Y$, and $R^1f_*\mathcal{O}_ X = 0$. Then $f_*\mathcal{O}_ X$ is $S$-flat and formation of $f_*\mathcal{O}_ X$ commutes with arbitrary base change $S' \to S$.
Proof. We may assume $Y$ and $S$ are affine, say $S = \mathop{\mathrm{Spec}}(A)$. To show the quasi-coherent $\mathcal{O}_ Y$-module $f_*\mathcal{O}_ X$ is flat relative to $S$ it suffices to show that $H^0(X, \mathcal{O}_ X)$ is flat over $A$ (some details omitted). By Lemma 37.72.3 we have $H^1(X, \mathcal{O}_ X \otimes _ A M) = 0$ for every $A$-module $M$. Since also $\mathcal{O}_ X$ is flat over $A$ we deduce the functor $M \mapsto H^0(X, \mathcal{O}_ X \otimes _ A M)$ is exact. Moreover, this functor commutes with direct sums by Cohomology, Lemma 20.19.1. Then it is an exercise to see that $H^0(X, \mathcal{O}_ X \otimes _ A M) = M \otimes _ A H^0(X, \mathcal{O}_ X)$ functorially in $M$ and this gives the desired flatness. Finally, if $S' \to S$ is a morphism of affines given by the ring map $A \to A'$, then in the affine case just discussed we see that
This shows that formation of $f_*\mathcal{O}_ X$ commutes with any base change $S' \to S$. Some details omitted. $\square$
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