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 $X$ be a proper $k$ -scheme, and $Y$ a projective $k$ -scheme. If $f:X\to Y$ is finite, and $\mathscr{L}$ an ample line-bundle on $Y$ , then $f^\ast\mathscr{L}$ is ample on $X$ .

Proof: We use Serre’s ampleness criterion. Namely, let $\mathcal{F}$ be any coherent sheaf on $X$ . We need to show that there exists some $n\geqslant 0$ such that $H^i(X,\mathcal{F}\otimes f^\ast(\mathscr{L})^{\otimes n})=0$ , for $i>0$ . But, since $f$ is affine, and $\mathcal{F}\otimes f^\ast(\mathscr{L})^{\otimes n}$ quasicoherent, we know that

$H^i(X,\mathcal{F}\otimes f^\ast(\mathscr{L})^{\otimes n})=H^i(Y,f_\ast(\mathcal{F}\otimes f^\ast(\mathscr{L})^{\otimes n}))$ .

Applying the projection formula, which is applicable since $\mathscr{L}$ is locally free, we know that

$f_\ast(\mathcal{F}\otimes f^\ast(\mathscr{L})^{\otimes n})=f_\ast\mathcal{F}\otimes \mathscr{L}^{\otimes n}$ .

But, since $f$ is finite, we know that $f_\ast\mathcal{F}$ is coherent. So, by Serre’s criterion, there exists an $m$ such that

$H^i(Y,f_\ast\mathcal{F}\otimes\mathscr{L}^{\otimes m})=0$

for $i>0$ . Thus, the above calculation shows that taking $m=n$ allows us to conclude that $f^\ast\mathscr{L}$ is ample. $\blacksquare$

This is not particularly surprising. For example, suppose that $X$ and $Y$ are smooth curves over $k$ . Then, $\mathscr{L}$ on $Y$ (or $X$ ) is ample if and only if $\deg\mathscr{L}>0$ . But, positive degree is preserved by finite (equivalently non-constant) maps $f:X\to Y$ .

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 $f$ 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 $\mathscr{L}$ on $X$ is globally generated, then it gives us a map $\varphi_\mathscr{L}:X\to \mathbb{P}(V)$ , for the complete linear system $V=H^0(X,\mathscr{L})$ . Moreover, we know then that $\mathscr{L}$ will be $\varphi^\ast_\mathscr{L}(\mathcal{O}(1))$ . So, any globally generated line bundle is the pullback of a very ample line bundle. Thus, we need only produce an $X$ , and a globally generated line bundle $\mathscr{L}$ on $X$ which is not very ample, and for which the map $\varphi_\mathscr{L}$ is finite.

With an eye towards this second condition, we focus our attention on $X$ a (geometrically integral, smooth, projective) curve over $k$ . Indeed, if $\varphi:X\to Y$ is any $k$ -morphism, with $Y/k$ separated, which is not constant, then it is automatically finite. Indeed, by Zariski’s main theorem (or more specifically, Grothendieck’s version), $\varphi$ will be finite if and only if it’s proper and quasinite.

But, $\varphi$ is patently proper. This is by what Vakil calls the ‘Cancellation theorem’. Moreover, if $\varphi$ is not constant, then it must be quasifinite. Indeed, note that for any point $x\in X$ , which is not the generic point, $\varphi(x)$ is closed, thus so is $\varphi^{-1}(\varphi(x))$ . But, any infinite subset of $X$ is dense, and thus non-constancy forces $\varphi^{-1}(\varphi(x))$ to be finite as desired. Thus, we can remove finiteness from our list of conditions on $\mathscr{L}$ if we require $h^0(X,\mathscr{L})\geqslant 2$ , 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 $\deg\mathscr{L}\geqslant 2g+1$ , $g=g(X)$ the genus, then $\mathscr{L}$ is very ample, and if $\deg\mathscr{L}\geqslant 2g$ , then $\mathscr{L}$ is globally generated. Thus, if we want to easily guarantee that $\mathscr{L}$ is globally generated we should choose $\deg\mathscr{L}\geqslant 2g$ , but if we want a chance at it not being very ample, we need to choose $\deg\mathscr{L} <2g+1$ . This does not give a lot of options for $\deg\mathscr{L}$ . So, let us fix a line bundle $\mathscr{L}$ with $\deg\mathscr{L}=2g$ .

Let $D$ be the divisor associated to $\mathscr{L}$ . The Riemann-Roch formula then tells us:

$\ell(D)-\ell(K-D)=\deg\mathscr{L}+1-g$ .

Here $K$ is the canonical divisor. Our choices assure us then that $\deg K-D<0$ , so $\ell(K-D)=0$ . Thus, we see that $\ell(D)=h^0(X,\mathscr{L})$ is just $\deg\mathscr{L}+1-g$ . In particular, taking $\deg\mathscr{L}$ to be $2g$ , we see that

$\ell(D)=\deg\mathscr{L}+1-g=2g+1-g=g+1$ .

So, as long as $g\geqslant 1$ , we have produced a line bundle with the desired qualities.

Just to recap, the above tells us that if we take a genus $g\geqslant 1$ curve $X/k$ , and a line bundle $\mathscr{L}$ of degree $2g$ , then $h^0(X,\mathscr{L})=g+1\geqslant 2$ . Thus, by global generation, we obtain a map $\varphi_\mathscr{L}:X\to \mathbb{P}^{g}$ which, since it’s non-constant, is finite. Moreover, as we have already remarked $\varphi_\mathscr{L}^\ast(\mathcal{O}(1))=\mathscr{L}$ . So we’re done!…not really.

As we alluded to above, while we included the bound $\deg\mathscr{L}<2g+1$ so that $\mathscr{L}$ isn’t guaranteed to very ample, it’s not for sure true that $\mathscr{L}$ isn’t still ample. Thus, we need some means of guaranteeing that $\varphi_\mathscr{L}$ is not a closed embedding.

Now, while we could attack this by trying to give an even finer structural analysis of $\mathscr{L}$ , we could instead approach it from a different angle. Namely, if we knew what type of (smooth) curves lived inside of $\mathbb{P}^g$ , then we could perhaps discount $\varphi_\mathscr{L}$ from being a closed embedding by saying that our curve doesn’t show up on the list of curves inside of $\mathbb{P}^g$ . But, there is a pretty natural $g$ for which we know all the curves in $\mathbb{P}^g$ … $g=1$ ! The only curve inside of $\mathbb{P}^1$ is $\mathbb{P}^1$ itself. Thus, it seems like it would be prudent for us to take $g=1$ .

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 $1$ curve $X$ and a line bundle $\mathscr{L}$ on $X$ with $\deg\mathscr{L}=2(1)=2$ .

Let’s just go through the above, and check that this does, in fact, work. Namely, let $X/k$ be a smooth, genus $1$ curve with a $k$ -point $p$ (i.e. an elliptic curve). Let $\mathscr{L}=\mathcal{O}(2p)$ . Since $\deg\mathscr{L}=2\geqslant 2g$ , we know that $\mathscr{L}$ is globally generated. We know from Riemman-Roch that $\deg K=2(1)-2=0$ . Thus, we see that $K-2p$ has negative degree, and so $\ell(K-2p)=0$ . Thus, the Riemann-Roch formula tells us that

$2=\deg(2p)+1-1=\ell(K-2p)+\ell(2p)=\ell(2p)$ .

In particular, $H^0(X,\mathscr{L})$ is a $2$ -dimensional space and so the map $\varphi_\mathscr{L}:X\to\mathbb{P}^1$ 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 $\mathbb{P}^1$ has genus $0$ ! Thus, we see that $\mathscr{L}=\varphi_\mathscr{L}^\ast(\mathcal{O}(1))$ , and since $\varphi_\mathscr{L}$ is finite, $\mathscr{L}$ not very ample, and $\mathcal{O}(1)$ very ample, this gives us the desired contradiction.

Note that in the above analysis, save the last part where we show that $\varphi_\mathscr{L}$ is not a closed embedding, there was no need to take $g=1$ . Namely, we could take $g=2$ , and pick a line bundle $\mathscr{L}$ with degree $2g=4$ . Then we obtain another counterexample as long as we can show that the map $\varphi_\mathscr{L}:X\to\mathbb{P}^2$ is not a closed embedding. While this is not as immediately obvious as the case $g=1$ , there is a trick we can use. Namely, we know what all the curves in $\mathbb{P}^2$ look like–they are of the form $V(f)$ . Moreover, the genus of such a curve is $\frac{(d-1)(d-2)}{2}$ , where $d=\deg(f)$ . In particular, we see that genus $2$ curves aren’t of the form $\frac{(d-1)(d-2)}{2}$ , and so don’t show up in $\mathbb{P}^2$ . Thus, $\varphi_\mathscr{L}$ 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 $\varphi_\mathscr{L}$ is not a closed embedding. Namely, there is no silly issue with embedding curves into $\mathbb{P}^g$ , for $g\geqslant 3$ , as there was for $\ell=1$ , and there is no list of curves contained in $\mathbb{P}^g$ , as there were for $g=2$ .

4 comments

levinsonjake says:

November 26, 2014 at 5:11 pm

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 $\mathbb{P}^1$ embedded in $\mathbb{P}^{g-1}$ by the Veronese map). Details in Hartshorne IV.3.

PrimeRibeyeDeal says:

November 21, 2018 at 10:43 am

“As we *alluded* to”

1. alexyoucis says:
  
  October 30, 2019 at 3:17 pm
  
  Of course. Thanks!
  
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Weird Example: Pullback of Very Ample by finite is not Very Ample

Pullback of ample is ample

Doesn’t work for very ample

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Pullback of ample is ample

Doesn’t work for very ample

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