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.

“As we *alluded* to”

Of course. Thanks!