# Can a finite locally free morphism of integral affine schemes not be free?

Let $$f:X\to Y$$ be a finite locally free morphism of integral affine schemes. Is $$f$$ necessarily free?

Is there an easy counterexample to this (eg, a morphism of varieties over an algebraically closed field) ?

• What do you mean by (locally) free morphism of schemes? Jul 26, 2021 at 10:24
• That’s surely false. Otherwise, it means if $L/K$ is a finite extension of number fields, then $\mathcal{O}_L$ is a free $\mathcal{O}_K$-module. Jul 26, 2021 at 10:25
• @Sasha I mean that $f_* O_X$ is a finite locally free $O_Y$-module. So basically it's a ring extension $A\to B$ which makes $B$ into a finite locally free $A$-module. Jul 26, 2021 at 15:09

A simple geometric counterexample is the following. Let $$C$$ be a smooth projective curve, let $$L$$ be a non-trivial 2-torsion line bundle on $$C$$, and let $$P \in C$$ be a point. Let $$\tilde{C} = \mathrm{Spec}_C(\mathcal{O}_C \oplus L) \to C$$ be the etale double cover of $$C$$ associated with $$L$$. Set $$Y = C \setminus P$$ and $$X = \tilde{C} \times_C Y$$. Then $$f \colon X \to Y$$ is locally free, but not free. Indeed, $$f_*\mathcal{O}_X \cong \mathcal{O}_Y \oplus L\vert_Y$$ is locally free, but not free.