## When does a vector bundle descend?

X and Y are irreducible curves, and f:X-->Y a morphism. Let E be a vector bundle over X.

When does there exist a vector bundle F over Y such that f*F=E and when will it be unique?

-
Regarding the last part of the question. $E$ needn't be unique. Take for example, $f$ to be multiplication by $n$ (prime to characteristic) on an elliptic curve $X=Y$. Anything in the kernel of $Pic(f)$, which is nontrivial, will pull back to $O_X$. – Donu Arapura Aug 8 2010 at 12:44
That should read "$F$ needn't be unique". – Donu Arapura Aug 8 2010 at 12:45
I see that there has been some back and forth in the edit history between the original author and Charles Matthews. Can I just say that "descent" is not usually recognized as a verb, as far as I know. – Yemon Choi Aug 9 2010 at 6:36
Let me expand my earlier comment a bit. When $Pic(f)$ is non-injective (which happens frequently) $F$ will never be unique when it exists: $f^*(F\otimes L)\cong E$if $L\in \ker Pic(f)$. – Donu Arapura Aug 9 2010 at 16:12

If $f$ does not factor through a point then $f$ is flat. So, you can use the usual descent condition --- let $p_1,p_2:X\times_Y X \to X$ be the projections. Then $E$ descends if there is an isomorphism $p_1^*E \cong p_2^*E$ satisfying the cocycle condition on the triple fiber product. Each such isomorphism gives a descent.

-
$f$ can fail to be flat if $Y$ is not normal! – James Borger Aug 8 2010 at 10:39
I think that you have descent for vector bundles even if $f$ is not flat, as long as $X$ and $Y$ are reduced. An example of this would be when $Y$ is a nodal curve and $X$ is the normalization; this would give you the classical description of vector bundles on a nodal curve as vector bundles on the normalization, together with isomorphisms of the fibers at the point that get glued. – Angelo Aug 8 2010 at 11:34
Good point! – James Borger Aug 8 2010 at 13:31
While this was not the original question, I'm pretty sure that descent of morphisms of vector bundles can fail for partial normalizations. I think it's possible to give subrings $A\subseteq B\subseteq k[x]$ such that $k[x]$ is integral over $A$ and such that $A$ is not the equalizer of the two maps $B\to B\otimes_A B$. So, in the non-flat case, I'm pretty sure there are some subtleties for full descent, though maybe one can descend objects. (I thought this example was in SGA3, but I haven't been able to find it...) – James Borger Aug 8 2010 at 22:34
Dear James, you are absolutely right, this is even false in very simple cases, like when $Y$ has an ordinary triple point and $X$ is the normalization. This is ironic, because some time ago I was explaining precisely this to a colleague. On the other hand, I believe that this is the only obstruction, that is, if $A$ is the equalizer of the two maps $B \to B\otimes_{A}B$, then descent works. – Angelo Aug 9 2010 at 9:42