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) ?

  • $\begingroup$ What do you mean by (locally) free morphism of schemes? $\endgroup$
    – Sasha
    Jul 26, 2021 at 10:24
  • 1
    $\begingroup$ 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. $\endgroup$
    – Aphelli
    Jul 26, 2021 at 10:25
  • $\begingroup$ @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. $\endgroup$ Jul 26, 2021 at 15:09

1 Answer 1


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.


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