Bertini’s Theorem

Today we’re going to prove a theorem of Bertini’s. Next time, we’ll focus on using the theorem to do something that’s even more geometric. For now, we’re only going to prove the theorem in characteristic zero. In fact, we’re going to work over . Why? Because there’s a long, proud history of transcendental methods in the subject, and using differential geometry and legitimate calculus (as opposed to formal) to simplify proofs, and we’re about to hit some of that. Don’t worry, though, positive characteristic is still out there, and we’ll have to use it if I ever decide to make a serious go at Mori’s Bend-and-Break and some of its nicer consequences.

So today we’re going to work on Bertini’s Theorem. We’ll start out just stating it.

Bertini’s Theorem: The generic element of a linear system is smooth away from the base locus of the system.

So, what’s this mean? Well, take a projective variety and a linear system on it. The linear system is a projective space, so we’ve got the Zariski topology on it. By a generic element, we mean that there is an open dense set of these elements in the linear system. We remember the base locus is the subset of such that the map isn’t defined. So take a hyperplane in , and look at its preimage in , and remember to avoid the base locus. Then what this says is that there’s an open subset of these hyperplanes such that the inverse image is smooth.

Proof: We’ll go by contradiction. If a generic element of a linear system is singular away from the base locus, then it works for a pencil, that is, a linear system of dimension 1. So now let be a pencil given locally (in the analytic topology, this time, which means we’ve got MUCH smaller open sets. The analytic topology is obtained by taking the usual topology on , taking its product with itself, and then taking the quotient topology) by is the divisor of .

We can always do this analytically, so this is a significant reduction that we’ve managed by sticking to characteristic zero. Now suppose that is a singular point of for , which which isn’t in the pencil’s base locus. So we have and for .

Now, as isn’t a base point of the pencil, we can’t have both vanishing, and so neither can, so we can write and so the equation from the derivatives becomes .

So now we look at the derivative of with respect to at , and we get because of the above.

So now, we call the locus of singular points of the divisors in the pencil. This is locally (again analytically) on cut by the equation and for all . However, is locally constant on minus the base locus. Thus, can only meet finitely many divisors away from the base locus, because only finitely many values of appear for such singular points, and so a generic element of a pencil is smooth away from the base locus.

Thus, by the beginning remark, it must be that for any linear system, Bertini holds. QED.

Notice how intrinsically but subtly the proof relies on the complex numbers. All we ever needed was that we could work locally in the analytic topology. This theorem DOES hold for arbitrary varieties over arbitrary base fields, but to prove it in general requires more sophisticated machinery from schemes, and I prefer to minimize their use, because these methods are much more algebraic and hide the geometry.

This is often the case. Now, one caveat, and its something that bothers me. When working in characteristic zero, algebraic geometers often assert that “by the Lefshetz Principle” they can assume that they are working over . Now, my understanding of the reasoning is that any variety is determined by finitely many elements of whatever field you’re working over, and we can always embed into the complex numbers. However, something here doesn’t feel quite right to me. People seem to be asserting that proving something over proves it over other fields of characteristic zero, but I’m really not seeing it. This seems more like a rule of thumb than a rigorous principle. Anyone know this better?