Let $C,D$ curves (so $1$-dimensional proper $k$-schemes). Assume futhermore that they are also regular and $f:C \to D$ is a finite morphism.

It is known that in this case the pushforward functor $f_*: (QCoh-O_C-Mod) \to (QCoh-O_D-Mod)$ provides a category equivalence between quasi coherent modules of $O_C$ and $O_D$-modules with extra $f_*(O_C)$-structure. .

My question is if this functor also preserves local freeness for coherent modules: Namely if $F$ is a coherent locally free $O_C$-module does this also hold for $f_*F$?

My ideas:

The problem is local since $f$ is a affine morphism so by regularity we can assume that $C=Spec(A),D=Spec(R)$ with $A,R$ Dedekind-rings and $F$ is a free $A$-module $A^n$.

By classification of finitely generated modules over Dedekind rings every f.g. $A$-module $M$ has the shape $F= A^k \oplus T$ with free component $A^k$ and $T$ torsion.

My idea is firstly to observe that $f_*$ preserves (finite) direct sums. I guess that this follows from that as category equivalence of modules is preserves exact sequences?)

And then need to show that:

-$f_*A$ is a free $R$-module

-$f_*T$ is also torsion wrt $R$

Here I'm stuck. Why these two statements hold?

Futhermore does this category equivalence allow a "reverse" argument: namely if $f_*F$ is free then also $F$?

  • $\begingroup$ I dispute the claim that the pushforward is an equivalence of categories here. This would imply, for instance, that the category of modules over the ring of functions on an affine elliptic curve is equivalent to the category of modules over $k[x]$, but the former contains torsion-free non-free classes and the latter doesn't. $\endgroup$
    – hunter
    Jun 4, 2019 at 1:27
  • $\begingroup$ @hunter: I think I forgot a point: It should br CE between qc $O_C$-modules AND qc $O_D$ -modules WITH extra $f_*(O_C)$-structure. I will update my question. Thank you for remark. $\endgroup$
    – KarlPeter
    Jun 4, 2019 at 1:35

1 Answer 1


A module (finitely generated or not) over a Dedekind domain $R$ is flat if and only it is torsion-free.
So in your case of a finite morphism $R\to A$ the $R$-module $A$ is flat and finitely presented, hence projective.
We conclude that $A$, and all $A^n$, are locally free over $R$.
Geometrically this means that given a finite morphisms $f:C\to D$ between regular curves, the direct image $ f_*\mathcal F$ on $D$ of a locally free sheaf $\mathcal F$ on $C$ is a locally free sheaf on $D$.
This is always false if the target curve $D$ is not regular:
Indeed, in that case if $f:C\to D$ is the normalization, then the sheaf $f_*\mathcal O_C$ is never locally free on $D$.

  • $\begingroup$ Hi. Thank you a lot for your enlightening answer. One point stay unclear: Can we "go back" without loosing freeness? Namely if $F$ is a $A$-module and $f_*F$ is a free $R$-module then $F$ was already a free $A$-module? $\endgroup$
    – KarlPeter
    Jun 4, 2019 at 12:14
  • $\begingroup$ The answer to your question is NO: take for $A$ a finite algebra over $R=k[T]$ such that $Pic(A)\neq0$ and consider a non-trivial line bundle $F=\tilde L$ over $Spec(A)$ corresponding to the locally free but not free rank-one module $L$ over $A$ . Then the direct image $f_*(\tilde L)=\widetilde{_RL}$ (corresponding to $L$ seen as an $R$-module) is free, like all coherent torsion-free sheaves over $Spec (k[T])=\mathbb A^1_k$, but $F$ was not free. $\endgroup$ Jun 4, 2019 at 17:01
  • $\begingroup$ sorry for digging out this question again but lastly I noticed that it's not clear to me how you deduced that my morphism $R \to A$ is flat. By assumption it's finite between regular rings. does it imply that it is flat? $\endgroup$
    – KarlPeter
    Jun 13, 2019 at 16:40

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