Does a smooth projective variety over an algebraically closed field of negative Kodaira dimension contain a projective line?
I do not want to assume any conjectures.
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$\begingroup$ What is your precise definition of "projective line"? Just a smooth rational curve, or you also require that it has degree $1$ with respect to some fixed polarization? $\endgroup$– Francesco PolizziCommented Sep 2, 2020 at 18:24
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$\begingroup$ @FrancescoPolizzi smooth rational curve $\endgroup$– NguyenCommented Sep 2, 2020 at 19:26
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I remark that it is true in dimension at most $3$, over $\mathbb{C}$. (This probably isn't the easiest way to see it and is not a complete solution, but it was too short for a comment. )
In dimension $2$ it follows from the fact that all of the minimal surfaces with negative Kodaira dimension are covered by smooth rational curves (they are either $\mathbb{P}^1$-bundles or $\mathbb{P}^2$).
In dimension $3$ it follows from the existence and classification of extremal contractions for smooth $3$-folds due to Mori. Let $X$ be a smooth $3$-fold with Kodaira dimension $-\infty$ then $X$ is uniruled (Theorem 6.1.8 (FV)). Hence by a result of Miyoaka and Mori there is a non-empty Zariski open $U \subset X$ such that each point in $U$ is contained in an irreducible curve $C \subset X$ with $K_{X}.C <0$ (Theorem 6.1.4 (FV)), in particular $K_{X}$ is not nef.
(by (FV) I mean the book "Fano varieties" by Iskovskikh and Prokhorov).
Hence by the contraction theorem, there is some extremal contraction. You may consult the list of extremal contractions which can occur. In all cases; other than when the image is a point, some fibre of the extremal contraction clearly contains smooth rational curves (For conic bundles and del Pezzo bundles one needs that a general fibre is smooth, which is true). If the image is a point, $X$ is Fano $3$-fold with $b_{2}=1$. Then since Fano $3$-folds are classified we may go through each of the $17$ prime Fano $3$-folds and check there is a smooth rational curve in each.
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$\begingroup$ In dimension 3, a variety with negative Kodaira dimension is uniruled. $\endgroup$– abxCommented Sep 4, 2020 at 8:40
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$\begingroup$ I used that fact in my answer, although it is not clear to me that uniruledness immediately gives a smooth rational curve in $X$ as asked by the OP. $\endgroup$– Nick LCommented Sep 5, 2020 at 6:28
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$\begingroup$ @NickL Let's say that a proper variety is "covered by rational curves" if, for every proper closed subset $Z\subsetneq X$, there is a rational curve passing through a point in $X\setminus Z$. Note that this is clearly a birational invariant of the class of proper varieties. Moreover, if $X\to Z$ is a proper surjective morphism and $X$ is covered by rational curves, then so is $Z$, right? This is what abx probably had in mind (and is simpler than what you wrote). $\endgroup$ Commented Sep 7, 2020 at 10:54
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$\begingroup$ I am sorry, but I don't see any relationship between what you said and smooth rational curves. Could you explain what you mean? $\endgroup$– Nick LCommented Sep 11, 2020 at 7:27
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$\begingroup$ I wonder if a single uniruled parametrization may ever suffice in this question. Namely, do there exist uniruled parametrizations with images of all components of all fibers being singular rational curves? It's easy to arrange such parametrization with general curve being singular, but these typically break up to have smooth components. $\endgroup$ Commented Sep 11, 2020 at 8:19