Let $X$ be a Del Pezzo surface over an algebraically closed field $k$, i.e. a projective surface with $-K_X$ ample.

I'm chiefly interested in the case that $k$ has characteristic $0$ and $X$ is smooth, so feel free to assume these if it simplifies things.

Here's the idea I have so far:

From the adjunction formula, we can see that any $-1$ curve is rational, since $K_X \cdot C < 0$ for any curve $C$. In fact, if $-K_X$ is very ample and $X$ is embedded in $\mathbb{P}^n$ via $-K_X$, this can be extended to show that lines are exactly the $-1$ curves, since $K_X \cdot C = \deg C$. In addition, $K_X^2 = \deg X$ as a subvariety of $\mathbb{P}^n$.

EDIT: As Pooh Bear pointed out in the comments, the idea below is not correct: blowing down a $-1$-curve increases $K_X^2$ by $1$. I'd still love to see a correct proof.

Now, I know Castelnuovo's Criterion shows that we can contract any of the $-1$ curves on $X$ to a point and that this decreases $K_X^2$ by $1$.

So if I knew:

  1. $-K_X$ is always very ample.
  2. The blow-down of a Del Pezzo surface is still Del Pezzo.
  3. Any Del Pezzo surface other than $\mathbb{P}^2$ has a $-1$ curve, or, any Del Pezzo surface has a rational curve.

Then I could keep contracting $-1$ curves and decreasing the degree until I find a Del Pezzo surface embedded in $\mathbb{P}^n$ as a degree $1$ subvariety, which should be a linear subspace (right??).

Does this work?

  • 1
    $\begingroup$ Unfortunately this strategy cannot work, because contracting at $(-1)$-curve increases $K_X^2$ by 1! $\endgroup$
    – Nefertiti
    Commented Apr 22, 2016 at 6:36
  • $\begingroup$ Ahh good catch! I misread the discussion of this in Hartshorne and got carried away... $\endgroup$
    – Dorebell
    Commented Apr 22, 2016 at 8:22
  • 1
    $\begingroup$ There is some nice historical discussion of this topic in Dolgachev's book Classical Algebraic Geometry, Chapter 8.The basic idea seems to be to embed $X$ using $|-K_X|$ as a surface of degree $d$ in $\mathbf P^d$, then project (birationally) to get a cubic surface in $\mathbf P^3$. $\endgroup$
    – Nefertiti
    Commented Apr 22, 2016 at 8:38

2 Answers 2


I would normally post this as a comment, but it is a bit too long. paf's argument works in characteristic zero to show $q(X) = h^1(K_X) = 0$ by Kodaira vanishing, but I think it's nice to know how you can still get vanishing in positive characteristic. We follow [Kollár, III.3].

Characteristic not 2 or 3

In this situation, we can use the following theorem describing how failure of Kodaira vanishing has strong geometric consequences. The proof of this is based on a construction due to Ekedahl, which I won't try to explain here.

Theorem [Kollár, II.6.2]. Let $X$ be a smooth, projective variety over a field of characteristic $p > 0$. Let $L$ be an ample line bundle on $X$ such that $H^1(X,-L) \ne 0$. Assume that $X$ is covered by a family of curves $\{D_t\}$ where $$\label{eq:a}(p-1)(L \cdot D_t) - (K_X \cdot D_t) > 0.\tag{$*$}$$ Then, through every point $x \in X$ there is a rational curve $C_x \subset X$ such that $$(p-1)(L \cdot C_x) - (K_X \cdot C_x) \le 1 + \dim X.$$

Now we apply this result to get the vanishing we want.

Proof of vanishing. Let $L = -K_X$, and suppose $H^1(X,-L) \ne 0$. In this case, the inequality \eqref{eq:a} holds for any covering family $\{D_t\}$ since $p > 3$. Then, by the Theorem we obtain a family of rational curves $C_x$ such that $$(p-1)(L \cdot C_x) - (K_X \cdot C_x) = p(L \cdot C_x) \le 3,$$ which implies $(L \cdot C_x) < 1$, a contradiction. $\blacksquare$

The proof of the theorem above can be adapted to apply to the characteristic 3 case as well [Kollár, II.6.8].

An argument that works in general

This argument is a nice application of the Albanese morphism.

Proof of vanishing. Suppose $h^1(K_X) = h^1(\mathcal{O}_X) > 0$, where the equality is by Serre duality. Since $h^2(\mathcal{O}_X) = h^0(K_X) = 0$ by amplitude of $-K_X$, by [Mumford, Lecture 27] we can use the Albanese construction as in characteristic zero to obtain a non-constant morphism $\mathrm{alb}\colon X \to \operatorname{Alb}(X)$, where $\operatorname{Alb}(X)$ is an abelian scheme.

Now let $H$ be an ample divisor on $\operatorname{Alb}(X)$. Then, there is a curve $D \subset X$ such that $(\mathrm{alb}^*H \cdot D) > 0$. Since $X$ is Fano, by [Kollár–Mori, Thm. 1.10] there exists a rational curve $C$ going through a point in the intersection of $\mathrm{alb}^*H$ and $D$, hence there exists a rational curve $C \subset X$ such that $(\mathrm{alb}^*H \cdot C) > 0$. On the other hand, this gives a nonconstant morphism $\mathbf{P}^1 \to C \to \operatorname{Alb}(X)$, which is impossible since $\operatorname{Alb}(X)$ is an abelian scheme, hence the morphism $\mathbf{P}^1 \to \operatorname{Alb}(X)$ must factor through the Jacobian of $\mathbf{P}^1$, which is a point. $\blacksquare$

  • $\begingroup$ Thanks a lot for the thorough answer, Takumi! I'll need to read about Albanese varieties a bit more to understand the last part. For the first argument, is it obvious that a covering family of curves exists? $\endgroup$
    – Dorebell
    Commented Apr 26, 2016 at 20:02
  • $\begingroup$ @Dorebell I guess you're okay with the fact that any covering family works, right? An easy way to construct a covering family is for each $x \in X$, choose a hyperplane $H$ that goes through $x$, and use the curve $H \cap X$. By the way, I think paf's answer deserves to be accepted, not mine… $\endgroup$ Commented Apr 27, 2016 at 19:59
  • $\begingroup$ Fair enough! I thought that a covering family needed to be parametrized by $\mathbb{P}^1$ or something - so any set of curves that covers $X$ works? $\endgroup$
    – Dorebell
    Commented Apr 27, 2016 at 21:52
  • 1
    $\begingroup$ Yes, if you read the proof in Kollár's book, you only need that every point in $X$ has some curve $D_t$ through it. $\endgroup$ Commented Apr 28, 2016 at 12:06

In addition to Pooh Bear's remark, this doesn't work because you forget the case of $\mathbb P^1\times \mathbb P^1$ which is a Del Pezzo surface non-isomorphic to $\mathbb P^2$ having no $-1$ curve (so your point 3. is false, but $\mathbb P^1\times \mathbb P^1$ is the only counterexample) and which can't be embedded in $\mathbb P^n$ as a degree 1 subvariety.

The only proof I know for rationality of Del Pezzo surfaces uses Castelnuovo's Theorem :

any complex surface such that $q$ and $P_2$ (the irregularity and second plurigenus) both vanish is rational.

In fact, by Nakai-Moishezon, one can prove that $P_2(X) = h^0(2K_X) = 0$ because $-K_X$ is ample and Kodaira vanishing Theorem shows that $q(X) = h^1(-(-K_X)) = 0$.


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