The Stacks project

Proof. It suffices to prove (a) and (b) étale locally on $Y$, thus we may and do assume $Y$ is an affine scheme. By cohomology and base change (Lemma 75.25.4) the complex $E = Rf_*\mathcal{O}_ X$ is perfect and its formation commutes with arbitrary base change. In particular, for $y \in Y$ we see that $H^0(E \otimes ^\mathbf {L} \kappa (y)) = H^0(X_ y, \mathcal{O}_{X_ y}) = \kappa (y)$. Thus $\beta _0(y) \leq 1$ for all $y \in Y$ with notation as in Lemma 75.26.1. Apply Lemma 75.26.6 with $a = 0$ and $r = 1$. We obtain a universal closed subscheme $j : Z \to Y$ with $H^0(Lj^*E)$ invertible characterized by the equivalence of (4)(a), (b), and (c) of the lemma. Since formation of $E$ commutes with base change, we have

The morphism $\text{pr}_1 : X \times _ Y X$ has a section namely the diagonal morphism $\Delta $ for $X$ over $Y$. We obtain maps

in $D(\mathcal{O}_ X)$ whose composition is the identity. Thus $R\text{pr}_{1, *}\mathcal{O}_{X \times _ Y X} = \mathcal{O}_ X \oplus E'$ in $D(\mathcal{O}_ X)$. Thus $\mathcal{O}_ X$ is a direct summand of $H^0(Lf^*E)$ and we conclude that $X \to Y$ factors through $Z$ by the equivalence of (4)(c) and (4)(a) of the lemma cited above. Since $\{ X \to Y\} $ is an fppf covering, we have $Z = Y$. Thus $f_*\mathcal{O}_ X$ is an invertible $\mathcal{O}_ Y$-module. We conclude $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is an isomorphism because a ring map $A \to B$ such that $B$ is invertible as an $A$-module is an isomorphism. Since the assumptions are preserved under base change, we see that (a) is true.

Proof of (b). Above we have seen that for every $y \in Y$ the map $\mathcal{O}_ Y \to H^0(E \otimes ^\mathbf {L} \kappa (y))$ is surjective. Thus we may apply More on Algebra, Lemma 15.76.2 to see that in an open neighbourhood of $y$ we have a decomposition $Rf_*\mathcal{O}_ X = \mathcal{O}_ Y \oplus P$ $\square$