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](https://mathoverflow.net/questions/62843/path-connectedness-of-varieties/62883#62883) and [here](https://mathoverflow.net/questions/75320/connecting-points-on-a-variety-by-the-image-of-a-nonsingular-curve/75327#75327), 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.



Proving that any two points on a variety can be joined by a curve; why does Bertini apply?
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!)


