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Let $X$ be a smooth, proper algebraic variety over a field $k$, of positive dimension.

Is it true that $X$ contains a smooth Zariski-closed curve?

If it is projective, this is true by Bertini. But is it true in general?

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Here is a strategy for showing that the answer is positive: blow up a proper smooth variety, call it $X$, to get a projective smooth $X'$ such that $X'\setminus D'$ is isomorphic to $X\setminus D$ with $D\subset X$ closed and $D'\subset X'$ a divisor with normal crossings; then take a curve in $X'$ that intersects transversally all the components of $D'$ and project back again. – algori Nov 16 '12 at 1:07
@algori : The problem with this strategy is that it is really possible to have a divisor $E$ on $X'$, exceptional over $X$, such that every curve meeting $E$ has singular image in $X$. This happens for instance if $X$ is a smooth surface, and $X'$ is constructed by blowing up a point, then blowing up a point on the exceptional divisor, then blowing up the intersection of the two exceptional divisors, and if you choose $E$ to be the third exceptional divisor. The problem is that the differential of $X'\to X$ is identically zero at every point of $E$. – Olivier Benoist Nov 16 '12 at 1:22
Olivier -- it's worse than you think: I'm not sure how to prevent the curve from intersecting itself. – algori Nov 16 '12 at 1:44
@algori : yes, indeed ! The answer is positive if $X$ is separably rationally connected (over an algebraically closed field). Indeed, if $X$ is of dimension $\leq 2$, it is projective so there is no problem. And if it is of dimension $\geq 3$ it contains (lots of) smooth rational curves by Theorem IV 3.9 in Koll'ar's book "Rational curves on algebraic varieties". – Olivier Benoist Nov 16 '12 at 3:14

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