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In a paper by F. Pop he claims the following fact-

Let $K$ be a field. The set (by which I believe he means the union) of all smooth $K$-curves passing through a smooth $K$-rational point of an integral $K$-variety is Zariski-dense.

Can anybody explain to me why this is the case or direct me to a proof? Pop does not give a reference for this and I have been trying to search it and prove it for myself for a while now. I only require to know the proof in the case of affine varieties really. Thanks.

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    $\begingroup$ A note: The statement, as written, seems to have a lot to do with rationality over $K$. If this is your intent, then the title is a bit deceptive, since it suggests you might, for instance, care primarily about what happens when $K$ is algebraically closed. $$ $$ It might also be helpful to state what paper you are looking at, and where in the paper this statement appears. $\endgroup$ Commented Aug 21, 2012 at 18:54
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    $\begingroup$ You can reduce to the affine case; then take hyperplane sections. $\endgroup$
    – Angelo
    Commented Aug 21, 2012 at 19:18
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    $\begingroup$ Dear Angelo, I'm confused slightly, do you need normality? Suppose that $X$ is a non-normal projective variety whose normalization is for example $\mathbb{P}^2_{\mathbb{C}}$. Suppose that the non-normal locus is a curve (I can do this with a pinch point). Then every hyperplane section will have a singularity, right? Of course, we can find tons of hyperplanes that are smooth in a neighborhood of the given point. $\endgroup$ Commented Aug 21, 2012 at 21:14
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    $\begingroup$ Do you want the curves to be closed in your variety? If so, then this will be difficult: for instance, if the variety itself is a singular curve, then you won't even find a single smooth curve! Otherwise, I agree with Angelo. $\endgroup$
    – M P
    Commented Aug 21, 2012 at 21:55
  • $\begingroup$ Oh that's true. If curves need not be closed then certainly things are fine. $\endgroup$ Commented Aug 22, 2012 at 5:35

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