In this post we give a counterexample to the claim that the pullback of a very ample sheaf under a finite map is very ample, and prove that the result is true with ‘very ample’ replaced by ‘ample’.
Pullback of ample is ample
Let us first prove that the result is true for ample:
Theorem: Let
be a proper
-scheme, and
a projective
-scheme. If
is finite, and
an ample line-bundle on
, then
is ample on
.
Proof: We use Serre’s ampleness criterion. Namely, let be any coherent sheaf on
. We need to show that there exists some
such that
, for
. But, since
is affine, and
quasicoherent, we know that
.
Applying the projection formula, which is applicable since is locally free, we know that
.
But, since is finite, we know that
is coherent. So, by Serre’s criterion, there exists an
such that
for . Thus, the above calculation shows that taking
allows us to conclude that
is ample.
This is not particularly surprising. For example, suppose that and
are smooth curves over
. Then,
on
(or
) is ample if and only if
. But, positive degree is preserved by finite (equivalently non-constant) maps
.
Doesn’t work for very ample
Of course the above proof cannot be adapted to work for ‘ample’ replaced by ‘very ample’. In particular, there is not cohomological criterion for very ampleness that we can leverage using the fact that is affine (and so, in particular, preserves cohomology). But, what’s a little less obvious is an example of a counterexample. The below could have been made much shorter, but I tried to give you an idea of how one might cook up such an example by themselves.
Here’s how we can build an example. We know that if on
is globally generated, then it gives us a map
, for the complete linear system
. Moreover, we know then that
will be
. So, any globally generated line bundle is the pullback of a very ample line bundle. Thus, we need only produce an
, and a globally generated line bundle
on
which is not very ample, and for which the map
is finite.
With an eye towards this second condition, we focus our attention on a (geometrically integral, smooth, projective) curve over
. Indeed, if
is any
-morphism, with
separated, which is not constant, then it is automatically finite. Indeed, by Zariski’s main theorem (or more specifically, Grothendieck’s version),
will be finite if and only if it’s proper and quasinite.
But, is patently proper. This is by what Vakil calls the ‘Cancellation theorem’. Moreover, if
is not constant, then it must be quasifinite. Indeed, note that for any point
, which is not the generic point,
is closed, thus so is
. But, any infinite subset of
is dense, and thus non-constancy forces
to be finite as desired. Thus, we can remove finiteness from our list of conditions on
if we require
, and choosing a basis, for then we’re mapping into a positive dimensional projective space, with independent sections.
That said, it is a common theorem that if ,
the genus, then
is very ample, and if
, then
is globally generated. Thus, if we want to easily guarantee that
is globally generated we should choose
, but if we want a chance at it not being very ample, we need to choose
. This does not give a lot of options for
. So, let us fix a line bundle
with
.
Let be the divisor associated to
. The Riemann-Roch formula then tells us:
.
Here is the canonical divisor. Our choices assure us then that
, so
. Thus, we see that
is just
. In particular, taking
to be
, we see that
.
So, as long as , we have produced a line bundle with the desired qualities.
Just to recap, the above tells us that if we take a genus curve
, and a line bundle
of degree
, then
. Thus, by global generation, we obtain a map
which, since it’s non-constant, is finite. Moreover, as we have already remarked
. So we’re done!…not really.
As we alluded to above, while we included the bound so that
isn’t guaranteed to very ample, it’s not for sure true that
isn’t still ample. Thus, we need some means of guaranteeing that
is not a closed embedding.
Now, while we could attack this by trying to give an even finer structural analysis of , we could instead approach it from a different angle. Namely, if we knew what type of (smooth) curves lived inside of
, then we could perhaps discount
from being a closed embedding by saying that our curve doesn’t show up on the list of curves inside of
. But, there is a pretty natural
for which we know all the curves in
…
! The only curve inside of
is
itself. Thus, it seems like it would be prudent for us to take
.
Thus, the above informs explicitly what our counterexample should be. Trying to keep the complicatedness of our curve to a minimum, we should take a genus curve
and a line bundle
on
with
.
Let’s just go through the above, and check that this does, in fact, work. Namely, let be a smooth, genus
curve with a
-point
(i.e. an elliptic curve). Let
. Since
, we know that
is globally generated. We know from Riemman-Roch that
. Thus, we see that
has negative degree, and so
. Thus, the Riemann-Roch formula tells us that
.
In particular, is a
-dimensional space and so the map
provided by a basis is non-constant and so (by previous discussion) finite. This map is not a closed embedding though, else it would be an isomorphism, and
has genus
! Thus, we see that
, and since
is finite,
not very ample, and
very ample, this gives us the desired contradiction.
Note that in the above analysis, save the last part where we show that is not a closed embedding, there was no need to take
. Namely, we could take
, and pick a line bundle
with degree
. Then we obtain another counterexample as long as we can show that the map
is not a closed embedding. While this is not as immediately obvious as the case
, there is a trick we can use. Namely, we know what all the curves in
look like–they are of the form
. Moreover, the genus of such a curve is
, where
. In particular, we see that genus
curves aren’t of the form
, and so don’t show up in
. Thus,
is not a closed embedding, and so we have produced another counterexample.
While we could try and produce counterexamples with higher genus curves, we can no longer use mess-free tricks to check that is not a closed embedding. Namely, there is no silly issue with embedding curves into
, for
, as there was for
, and there is no list of curves contained in
, as there were for
.
Nice post. Another example you might like is the canonical morphism of a hyperelliptic curve. This is never an embedding (in fact it is always a double cover onto its image, which turns out to be a
embedded in
by the Veronese map). Details in Hartshorne IV.3.
Typo alert: quasinite
“As we *alluded* to”
Of course. Thanks!