I want to prove the following statement:
For any two points $x$ and $y$ in an irreducible variety $X$, there is a one-dimensional, irreducible subvariety $C\subseteq X$ containing $x$ and $y$.
Both here and here, the same argument is outlined to prove this statement. In both cases, Bertini's theorem is applied to the exceptional divisors of the blow-up $\tilde X$ of $X$ in $x$ and $y$. The curve joining the exceptional divisors in $\tilde X$ is then mapped to a curve connecting $x$ and $y$ in $X$.
Question: I do not see why Bertini can be applied in the case where $x$ or $y$ are singular points: The exceptional divisor will not be smooth in general. Can you tell me why this works?
Another (but far less important) problem I have with the proof is the application of Chow's lemma – the variety is not required to be complete. This is irrelevant to me because I can assume $X$ to be quasi-projective. I was still wondering.
Edit. My error was in assuming that the theorem called Bertini's theorem in Hartshorne is the only Bertini theorem. I have a habit of prefering text-book references over papers (bear with me) and I found that in Shafarevich's book Basic Algebraic Geometry I, there are two Bertini theorems, and this one has no smoothness assumption, so it is probably the correct one:
Theorem. Let $X$ and $Y$ be irreducible varieties
defined over a field of characteristic $0$, and $f: X\to Y$ a regular map such that
$f(X)$ is dense in $Y$. Suppose that $X$ remains irreducible over the algebraic
closure of $k(Y)$. Then there exists an open dense set $U\subseteq Y$ such that all the fibres $f^{-1}(y)$ over $y\in U$ are irreducible.
I do not see immediately how to apply it, though. If someone could provide some help, I'd be very grateful.
Best Answer
Corollary 1.9 of http://www-math.mit.edu/~poonen/papers/bertini_irred.pdf proves your statement over an arbitrary field $k$, even if $k$ is finite. (It has "geometrically irreducible" in place of "irreducible", but this just makes the statement more difficult: the irreducible version follows by applying the geometrically irreducible version to the irreducible components of $X \times_k \overline{k}$ and then taking the scheme-theoretic image of the resulting $C$ under the projection $X \times_k \overline{k} \to X$. A nitpick: in your statement you need to assume that $\dim X \ge 1$ or that $x$ and $y$ are distinct!)