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Let $X$ be the blow-up of $\mathbb{P}^2$ at three general points $p_1,p_2,p_3$, that is a del Pezzo surface of degree six, and let $\pi_i:X\rightarrow\mathbb{P}^1$ be the morphism induced by the projection from $p_i$.

Does anyone know a reference for the following classical fact?

Any morphism $f:X\rightarrow\mathbb{P}^1$ factors through $\pi_i$ for some $i = 1,2,3$.

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    $\begingroup$ Why do you think this is true? $\endgroup$ Sep 7, 2017 at 19:36
  • $\begingroup$ To prove the result it suffices to show the following. For $i \in \{1,2,3\}$ let $C_i$ be the strict transform of some line in the linear system given by the lines in $\mathbb{P}^2$ which contain $p_i$. (The linear systems $|C_i|$ define the conic bundle structures $\pi_i$ on $X$; these have $2$ singular fibres over $\bar{k}$.) Let $D$ be a divisor on $X$. If $\dim H^0(X,D) = 2$ then it suffices to show that $D \in |C_i|$ for some $i$. Probably the best way to approach this is to use the description of $X$ as a toric variety. $\endgroup$ Sep 7, 2017 at 19:58
  • $\begingroup$ There is one thing I'm still confused about. Take a general pencil $P$ of hyperplanes in $\mathbb{P}^6$. (We view $X \subset \mathbb{P}^6$ with respect to the anticanonical embedding). Then (presumably) $P$ defines a morphism $f: X \to \mathbb{P}^1$. This does not look like it factors through one of the morphisms $\pi_i$ as the smooth fibres of $f$ are curves of genus $1$, whereas the morphisms $\pi_i$ are conic bundles. $\endgroup$ Sep 7, 2017 at 20:02
  • $\begingroup$ @DanielLoughran: Daniel, a pencil of hyperplanes has a base locus (6 points), so it only defines a rational map to $\mathbb{P}^1$. To get a regular map one needs a divisor $D$ with $D^2 = 0$ and two global sections. $\endgroup$
    – Sasha
    Sep 7, 2017 at 20:12

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Let me sketch a proof. Let $D$ be the divisor class giving the map $f$. Then $D^2 = 0$ (because $D$ is the pullback of a point on $\mathbb{P}^1$). By adjunction formula $\deg(K_D) = K_X \cdot D$ is negative since $-K_X$ is ample. Therefore, general fiber of $f$ is a rational curve, hence $K_X \cdot D = -2$, hence $D$ is a conic (with respect to the anticanonical embedding of $X$). This means that $D$ is linearly equivalent to one of $C_i$, hence $f$ coincides with one of $\pi_i$.

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  • $\begingroup$ How does your argument take care of the divisor classes $D = n C_i$ for some $n > 1$? These also satisfy $D^2=0$, but $K_X \cdot D = -2n$. $\endgroup$ Sep 8, 2017 at 9:13
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    $\begingroup$ @DanielLoughran: Daniel, you are right; in fact I skipped one step. Actually, the general fiber doesn't have to be connected; and if it is not one should use the Stein factorization. So, the correct statement is that $f$ decomposes as $g \circ \pi_i$, where $g \colon \mathbb{P}^1 \to \mathbb{P}^1$ is a finite covering. $\endgroup$
    – Sasha
    Sep 8, 2017 at 9:45
  • $\begingroup$ Thank you very much. I agree with your argument. Do you know if this is written somewhere? I mean in a paper that one can cite. It seems to me that this should be very classical. $\endgroup$
    – user114198
    Sep 11, 2017 at 16:47
  • $\begingroup$ @WLC: I don't know a reference, but I think you can start searching from "Classical Algebraic Geometry" by Igor Dolgachev. $\endgroup$
    – Sasha
    Sep 12, 2017 at 17:46

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