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Let $k$ be a not-necessarily algebraically closed field of characteristic zero. Let $X$ be a positive-dimensional projective variety over $k$. Let $x$ be a closed point on $X$. Does there exist a curve over $k$ on $X$ which contains this point?

Variety = geometrically integral quasi-projective $k$-scheme

Curve = $1$-dimensional variety.

What about the special case where $x$ is a $k$-rational point?

I can blow-up $X$ at $x$ and take the image in $X$ of an effective ample Cartier divisor via this blow-up and reason by induction. But I'm afraid this doesn't give me a geometrically connected curve passing through $x$.

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There is a result of Katz showing that there is a whole Lefschetez pencil defined over $k$ and going through $x$, if ${\rm dim}(X)>1$. If you allowed positive characteristic, then there would be a problem in the case where $k$ is finite or in the case where the residue field of $x$ is not separable over $k$. See web.math.princeton.edu/~nmk/newlefapp12.pdf –  Damian Rössler Nov 10 '12 at 8:57
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Sorry, you also have to suppose that $X$ is smooth for Katz's result to apply. So all in all, the answer to you question is yes, since you have resolution of singularities in char. $0$ and you may apply Katz's result several times to lower the dimension. –  Damian Rössler Nov 10 '12 at 9:12

2 Answers 2

Your idea is good. Let $X'\to X$ be the blowup along $x$. Then $X'$ is projective, geometrically integral and of dimension $\dim X>1$. Embed $X'$ in some $\mathbb P^n_k$.

When $k$ is infinite, by Jouanolou, "Théorèmes de Bertini et applications" (Progress in Maths), Corollaire 6.11 (2)+(3), there exists a hyperplane $H$ such that $H\cap X'$ is geometrically integral.

When $k$ is finite, the existence of such a hypersurface $H$ is proved in Poonen "Bertini theorem over finite fields", Ann. Math. (2000), Proposition 2.7.

Now the image of $H\cap X'$ in $X$ is a geometrically integral closed subscheme of $X$ passing throught $x$ of dimension $<\dim X$. By induction we find a geometrically integral curve in $X$ passing through $x$.

Edit In fact through any closed finite subset $Z$ of $X$, it passes a geometrically integral curve in $X$. The proof is the same, we just blowup $X$ along $Z$ instead of $x$.

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See Liu, Exercise 8.1.5. If a connected $k$-variety has a $k$-rational point, it is geometrically connected (Liu, Exercise 3.2.11).

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@Timo: The OP does not specify that the closed point is a rational point. –  Jason Starr Nov 9 '12 at 19:28
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@Timo: Aha, I see now that you are answering the special case! –  Jason Starr Nov 9 '12 at 19:33

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