Fix an algebraically closed field $k$ (arbitrary characteristic), all schemes will be of finite type over $k$.

(Property *): I'm interested in (classes of) examples of schemes $X$ (irreducible, of dimension $n$) so that any morphism of schemes $\phi: X \rightarrow Y$ with $\dim Y < n$ is constant.

There are two examples I know of. Projective spaces $\mathbb{P}^n_k$ have this property and simple abelian varieties do too. (One may also put arbitrary non-reduced structures on these, see below).

$\textbf{Claim}$: More generally, if $X$ is a proper irreducible scheme so that every effective divisor is ample (so proper=projective), then $X$ has property (*). (Projective spaces have this property by definition, and simple abelian varieties do too by a general result from Mumford's book.)

Proof: (Eisenbud-Harris give a similar argument for the case of projective space). Let $X$ be as in the theorem and $\phi: X \rightarrow Y$ be a morphism with $Y$ of smaller dimension than $n = \dim X$. Without loss of generality, we may assume $\phi$ is surjective (hence we can pullback cartier divisors). Choose an effective Cartier divisor $D$ and a point $p \not \in |D|$ the support of $D$, but in the image of $\phi$ (since $\phi$ surjective). The pullback of an effective Cartier divisor is also effective, Cartier - hence ample. The pullback of the point will contain a complete curve (hence these two subschemes of $X$ meet by the Nakai-Moishezon criteria - contradicting that $p \not \in |D|$. $\square$

Easy observations

(1) Having property $*$ is not stable under blowing-up (Blow up of $\mathbb{P}^2$ is $\mathbb{P}^1 \times \mathbb{P}^1$.)

(2) If a scheme $X$ satisfies the claim, then by definition, so does $X_{red}$. Further, any thickening of $X$ has property $*$.

$\textbf{Proof: }$ To check that an $X$ satisfying the claim satisfies $*$, we did calculations in the intersection ring. This is invariant under changing the non-reduced structure. $\square$


Main question: Are there other (families of?) examples of schemes satisfying (*)?

(1) Does every scheme (no finiteness conditions!) have a dense affine open subset? This came up when I was thinking about this, and I realized I can't prove it offhand. Certainly it is true for irreducible schemes, and suffices to show it for connected schemes.

(2) Do you suspect that the only examples also satisfy the claim above? That is, have every effective divisor ample?

(3) Certainly all examples of schemes satisfying $*$ must be connected. Are there connected, but not irreducible examples?

I thought this was a little interesting (and admittidenly, I have no applications in mind.)

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    $\begingroup$ Dear LMN, Regarding question (1), you will need some finiteness conditions. For example, if $X$ is the disjoint union of countably many copies of Spec $k$ (for some field $k$), then any affine subscheme will be finite, and so $X$ has no dense affine open subscheme. (This is related to the fact that an affine open is quasi-compact, so its closure will tend to want to be quasi-compact too, although I'm not sure in what generality that will literally be true.) Regards, $\endgroup$
    – Emerton
    Dec 15, 2012 at 0:44
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    $\begingroup$ "Curves" is another such family! And gives examples of (3). $\endgroup$ Dec 15, 2012 at 2:12
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    $\begingroup$ If a K3 surfaces has no elliptic pencils then it has (*). Indeed, there are no maps to curves of positive genus since pullbacks of 1-forms are zero. $\endgroup$
    – Sasha
    Dec 15, 2012 at 3:03
  • $\begingroup$ Your statement about blowing up $\mathbb{P}^2$ needs a little more care (although the basic point stands). $\endgroup$
    – S. Carnahan
    Dec 15, 2012 at 3:33
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    $\begingroup$ LMN: $H^{1,0}$ of a K3 surface vanishes. Also, you can't get $\mathbb{P}^1 \times \mathbb{P}^1$ just by blowing up $\mathbb{P}^2$. You have to blow down after two blow-ups. $\endgroup$
    – S. Carnahan
    Dec 15, 2012 at 6:38

1 Answer 1


The point is not that every effective divisor should be ample, but that every semi-ample (=some multiple is basepoint-free) should have maximal Kodaira-Iitaka dimension. Actually this is an equivalent condition, but it is really just a reformulation of the condition you are looking for. Semi-ample divisors that are not ample lie on the boundary of the ample cone, so a sufficient condition would be something along the lines that the boundary of the ample cone does not contain any (integral) non-zero divisors.

One probably has to make some assumptions so this would make sense. Also, this is sufficient, but not necessary. That is, if this condition on the cone holds, then your desired condition holds, but your desired condition may hold even without this. In particular, Mumford (first and then others) gave examples of nef but not semi-ample divisors. Those divisors would lie on the boundary as well. Anyway, I don't have time to add more details right now, but I will try to do it later. Alternatively, you can consider this a hint and work it out yourself. (Mumford's example is a ruled surface, so that does not work, but I would expect that one should be able to construct an example that satisfies your condition and still has a nef but not semi-ample divisor).

Sasha's example of a K3 surface falls into this category. If there are no elliptic pencils, which is equivalent to there being no smooth elliptic curve on the surface, then the only non-trivial morphisms of the surface are the contraction of some $(-2)$-curves, but those morphisms are birational, so the target has the same dimension.

For a reducible example take two varieties of the same dimension, both of which has Picard number one and who intersect each other. For instance two intersecting planes. If the morphism maps to something lower dimensional, it has to be constant on both planes but they have to map to the same point.

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    $\begingroup$ Isn't picard number $1$ more a situation where the boundary is just the origin, or trivial, rather than the entire cone? $\endgroup$
    – Will Sawin
    Dec 15, 2012 at 6:23
  • $\begingroup$ Will, you're absolutely right. I corrected the relevant sentence. Thanks! $\endgroup$ Dec 15, 2012 at 15:29

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