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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.

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If you accept to have your $X$ quasi-projective, you can just apply directly Bertini's theorem to the linear system of hyperplanes passing through $x$ and $y$. If $X$ contains the line $\langle x,y\rangle$ you are done, otherwise Bertini tells you that a general divisor in the system is irreducible, and you get the result by induction on the dimension. – abx Apr 7 '14 at 7:28
But here we are back to the original problem: Why can I apply Bertini when $X$ is not smooth? – Jesko Hüttenhain Apr 7 '14 at 7:46
There is no smoothness assumption in the second Bertini theorem, see for instance this paper, Theorem 5.3. – abx Apr 7 '14 at 7:54
If you are in characteristic $0$, you can resolve the singularities, apply the statement to any pair of points in the respective exceptional divisors, and then map the curve back to $X$. – Alex Degtyarev Apr 7 '14 at 12:21
@JeskoHüttenhain: Are you reading the proof of the theorem of the cube in Mumford's book on abelian varieties? An excellent characteristic-free reference on Bertini theorems (of many flavors!) is the book by Jouanolou (just Google that name and "Bertini"). And Chow's Lemma is valid with separatedness rather than completeness (even though Hartshorne's textbook exercise imposes completeness); e.g., look in EGA II, 5.6. – user76758 Apr 7 '14 at 12:41

2 Answers 2

up vote 11 down vote accepted

Corollary 1.9 of 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!)

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That's just perfect. Thanks =) – Jesko Hüttenhain Nov 11 '14 at 7:49

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 $y$.

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