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

Question

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.

Answer 1

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!)

Answer 2

You can see this using Bertini's Theorem (Theorem 8.18 in Hartshorne). Let $X$ be an irreducible projective variety of dimension $dim(X)\geq 2$. If $char(k) = 0$ by Hironaka Theorem we can find a resolution of singularities $f:\widetilde{X}\rightarrow X$. Let $x,y\in X$ be two distinct points, and $\tilde{x}\in f^{-1}(x),\tilde{y}\in f^{-1}(y)$. Note that $f(\tilde{x})\neq f(\tilde{y})$.

Since $X$ is projective $\widetilde{X}$ is projective as well. Consier an embedding $\widetilde{X}\subset\mathbb{P}^{n}$. Now, the general hyperplane containing $\tilde{x},\tilde{y}$ intersect $\widetilde{X}$ in a smooth divisor (Theorem 8.18). Furthermore, since $H$ is ample the intersection $\widetilde{X}\cap H$ is connected and hence irreducible. Now consider $\widetilde{X}\cap H\subset H$ instead of $\widetilde{X}$, and take a general hyperplane section containing $\tilde{x},\tilde{y}$. Proceeding in this way at the last step you end up with a surface, and taking a general hyperplane section of this surface you find a smooth irreducible curve $\widetilde{C}$ in $\widetilde{X}$ passing through $\tilde{x},\tilde{y}$.

Finally, consider $C = f(\widetilde{C})$. Clearly $f$ does not contract $\widetilde{C}$. Furthermore $\widetilde{C}$ irreducible implies $C$ irreducible. Note that in general $C$ is not smooth (For instance take $X\subset\mathbb{P}^3$ an irreducible cubic surface with a double line, and two general points $x,y\in X$. Then an hyperplane containing $x,y$ intersects $X$ in a nodal plane cubic). However $C$ is an irrerucible curve in $X$ joining $x$ and $y$.