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Mumford's book "Abelian Varieties" contains a proof of the following statement: given two points $x$ and $y$ on a variety, there is an irreducible curve containing both (Lemma on p. 56 in the section on the Theorem of the Cube). The normalization of the curve is non-singular, so this yields the desired result.

The proof is roughly what Artie said. Arguing my by dimension, it is enough to produce an irreducible codimension $1$ subvariety containing the points. Pick a blow-up $f: X' \to X$ such that $X'$ is projective and the fibers $f^{-1}(x)$, $f^{-1}(y)$ are positive dimensional. Now fix a projective embedding of $X'$ and take a general hyperplane section $H$. This section is irreducible (Bertini) and meets the fibers $f^{-1}(x)$ and $f^{-1}(y)$ (for dimensional reasons). The image of $H$ under $f$ is the desired subvariety.

Question for the experts: What's an example where it is impossible to take the curve to be smooth?

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Mumford's book "Abelian Varieties" contains a proof of the following statement: given two points $x$ and $y$ on a variety, there is an irreducible curve containing both (Lemma on p. 56 in the section on the Theorem of the Cube). The normalization of the curve is non-singular, so this yields the desired result.

The proof is roughly what Artie said. Arguing my dimension, it is enough to produce an irreducible codimension $1$ subvariety containing the points. Pick a blow-up $f: X' \to X$ such that $X'$ is projective and the fibers $f^{-1}(x)$, $f^{-1}(y)$ are positive dimensional. Now fix a projective embedding of $X'$ and take a general hyperplane section $H$. This section is irreducible (Bertini) and meets the fibers $f^{-1}(x)$ and $f^{-1}(y)$ (for dimensional reasons). The image of $H$ under $f$ is the desired subvariety.

Question for the experts: What's an example where it is impossible to take the curve to be smooth?

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Mumford's book "Abelian Varieties" contains a proof of the following statement: given two points $x$ and $y$ on a variety, there is an irreducible curve containing both (Lemma on p. 56 in the section on the Theorem of the Cube). The normalization of the curve is non-singular, so this yields the desired result.

The proof is roughly what Artie said.