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I would like a reference for the following (easy/classical?) result:

Let $X$ be a quasi-projective irreducible algebraic variety of dimension $\ge 1$, defined over an algebraically closed field $k$ (of any characteristic), let $U\subset X$ be a non-empty (dense) open subset and $p\in X$ be a point. Then, there exists a closed curve $\Gamma\subset X$, passing through $p$ and meeting $U$.

One proof that I know is is page 262 of "Geometrische Methoden in dar Invariantentheorie" from Hanspeter Kraft, but everything is stated over $\mathbb{C}$. It seems to me that the techniques work over any algebraically closed field, but I would be happy with a reference which does explicitly the general case.

PS: When $X$ is smooth, a quick proof is given by Bertini's theorem, taking a general hyperplane section through $p$ and applying induction.

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  • $\begingroup$ Maybe I'm missing something, but why do you need the smoothness of $X$? It seems to me that, in any case, the hyperplane argument gives a curve $\Gamma$ with the required properties. Just take a projective closure $\bar X$ of $X$, embed it in a projective space and start cutting with general hyperplanes through $p$. Since $U$ is dense in $X$, and hence in $\bar X$, these hyperplanes will meet $U$, so the same holds for their intersection. Maybe you want $\Gamma$ smooth? $\endgroup$
    – Francesco Polizzi
    Commented Oct 7, 2015 at 7:58
  • $\begingroup$ I need $\Gamma$ irreducible (a priori the component touching $p$ could be outside of $U$). When $X$ is smooth, this follows from Bertini. In general, is it true also? $\endgroup$
    – Jérémy Blanc
    Commented Oct 7, 2015 at 8:01
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    $\begingroup$ There is a reference near the beginning of Chapter II, "Algebraic Theory via Varieties", of Mumford's "Abelian Varieties", the Lemma on p. 56. This is one step in the proof of the Theorem of the Cube. $\endgroup$
    – Jason Starr
    Commented Oct 7, 2015 at 9:13
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    $\begingroup$ mathoverflow.net/questions/62843/… the actual reference to a paper of C.P. Ramanujam is in a comment to an answer. $\endgroup$
    – roy smith
    Commented Oct 8, 2015 at 15:07

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