# Is each rationally chain connected surface rational?

Consider a 2-dimensional smooth projective algebraic surface S over complex numbers. Could you recommend any exact references to the proofs of the following assertions (of course, if they are true):

1. If S is rationally chain connected then S is rational.

2. If through each point of S one can draw a rational curve then S is uniruled.

3. If S is uniruled then S is ruled.

Recall that a surface is rationally chain connected, through each generic pair of points can be joined by a chain of rational curves on the surface. A uniruled surface is the one which admits a dominant map from $X\times \mathbb{P}^1$ for some curve $X$. A ruled surface is the one which admits a birational map from $X\times \mathbb{P}^1$ for some curve $X$.

Assertion 1 = [1,Proposition IV.3.6] + [1,Theorem IV.3.10] + [1,Proposition IV.3.3.1] + [1,Excercise IV.3.3.5], the latter given without proof. Cf. [1, Excercise IV.3.12.2]. Although this approach seems to be an overkill.

[1] J. Kollar, Rational curves on algebraic varieties, Springer-Verlag, Berlin–Heidelberg, 1996.

Edit

[2] A.Beauville, Complex algebraic surfaces,London Math. Soc. Student Texts 34, 2nd edition, Cambridge Univ. Press, 1996, 132p.

• You need three nontrivial results : rationally chain connected $\ \Rightarrow\$ rationally connected (see Koll\'ar's book), the Noether-Enriques theorem which says that any surface fibered over a curve with rational general fibers is ruled, and the Castelnuovo theorem which says that a surface dominated by $\mathbb{P}^1\times \mathbb{P}^1$ is rational. The last two can be found in any book on surfaces. The rest is an exercise. – abx Sep 3 '14 at 6:48
• Thank you very much. It seems that the resulting excercise requires further nontrivial results (like the Enriques-Kodaira classification of surfaces) to construct the fibration with rational fibers and relate it to the initial surface. That is why a reference is preferrable. – Mikhail Skopenkov Sep 3 '14 at 7:31
• No. Once you know you have a family of rational curves through each point, just take a general curve in the parameter space of the family and you have a surface fibered with rational curves dominating your surface. – abx Sep 3 '14 at 7:33
• Thanx! Then, by Noether-Enriques theorem, the initial surface is dominated by $X\times \mathbb{P}^1$ (not $\mathbb{P}^1\times \mathbb{P}^1$). In other words, the initial surface is uniruled, not yet unirational and not yet ruled, cf. assertion 3 in the question. How can I apply the Castelnuovo theorem? – Mikhail Skopenkov Sep 3 '14 at 8:16
• Actually you don't need Castelnuovo, sorry. Once you know your surface is ruled (and rationally connected), you get plenty of rational curves which map onto your base curve $X$. – abx Sep 3 '14 at 9:11

## 1 Answer

By the comments of abx you only need an answer to question $3$. You get easily an answer if you assume the results in the classical book of Beauville on algebraic surfaces.

Namely, if $S$ is uniruled, then you can solve the indeterminacy (Theorem II.7) and get a dominant morphism $X \rightarrow S$ where $X$ is birational to a product $C \times {\mathbb P}^1$. It follows, by Proposition III.20, that (being birational to a surface with vanishing plurigenera) all plurigenera of $X$ vanish. The argument of the proof of the same proposition applied to the dominant map $X \rightarrow S$ show that $P_n(S) \leq P_n(X)$ and therefore all plurigenera of $S$ vanish too.

Then, if $q=0$, the result follows by Castelnuovo rationality criterion. For the irregular ($q>0$) case, Beuville shows, (Proposition VI.15.(1)) that, if the surface is not ruled, then either $P_4 \neq 0$ or $P_6 \neq 0$.

Of course all the mentioned results are highly non trivial, but all proofs are in Beauville's book.

• Thank you very much, this completes the proof. Indeed, once all plurigenera vanish, the surface is ruled by the Enriques theorem [2,Theorem VI.17]. However, the assertion if a surface S is uniruled, all plurigenera vanish'' requires some work for the proof. E.g., we cannot just take a pullback of a pluricanonical section because the dominant map $X\times \mathbb{P}^1\to S$ may not be defined everywhere. Thus we need to start with eliminating indeterminacy using [2,Theorem II.7] and so on. That is why a reference to the assertions is preferable. – Mikhail Skopenkov Sep 3 '14 at 17:31
• I edited my post explaining this point, I hope is ok now. Really, i thik you could avoid solving the indeterminacy, as this is just a finite set (see Beauville's proof of Proposition III.20). Anyway, I solved them to be safe: then you pull-back holomorphic forms, and this gives the injective map we need. – Roberto Pignatelli Sep 3 '14 at 21:55