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 $\mathbb{C}$. 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 $X$ and a linear system $\mathfrak{D}$ 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 $X$ such that the map $\phi:X\dashrightarrow \mathfrak{D}$ isn’t defined. So take a hyperplane in $\mathfrak{D}$, and look at its preimage in $X$, 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 $\{D_\lambda\}_{\lambda\in \mathbb{P}^1}$ 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 $\mathbb{C}$, taking its product with itself, and then taking the quotient topology) by $D_\lambda$ is the divisor of $f(z_1,\ldots,z_n)+\lambda g(z_1,\ldots,z_n)=0$.

We can always do this analytically, so this is a significant reduction that we’ve managed by sticking to characteristic zero. Now suppose that $P_\lambda$ is a singular point of $D_\lambda$ for $\lambda\neq 0,\infty$, which which isn’t in the pencil’s base locus. So we have $f(P_\lambda)+\lambda g(P_\lambda)=0$ and $\frac{\partial f}{\partial z_i}(P_\lambda)+\lambda\frac{\partial g}{\partial z_i}(P_\lambda)=0$ for $i=1,\ldots,n$.

Now, as $P_\lambda$ isn’t a base point of the pencil, we can’t have both $f,g$ vanishing, and so neither can, so we can write $\lambda=-\frac{f(P_\lambda)}{g(P_\lambda)}$ and so the equation from the derivatives becomes $\frac{\partial f}{\partial z_i}(P_\lambda)-\frac{f(P_\lambda)}{g(P_\lambda)}\frac{\partial g}{\partial z_i}(P_\lambda)=0$.

So now we look at the derivative of $f/g$ with respect to $z_i$ at $P_\lambda$, and we get $\frac{(\partial f/\partial z_i)(P_\lambda)-(f(P_\lambda)/g(P_\lambda))(\partial g/\partial z_i)(P_\lambda)}{g(P_\lambda)}=0$ because of the above.

So now, we call $V$ the locus of singular points of the divisors in the pencil. This is locally (again analytically) on $X\times\mathbb{P}^1$ cut by the equation $f+\lambda g=0$ and $\partial f/\partial z_i+\lambda \partial g/\partial z_i=0$ for all $i$. However, $f/g=-\lambda$ is locally constant on $V$ minus the base locus. Thus, $V$ can only meet finitely many divisors away from the base locus, because only finitely many values of $\lambda$ 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 $\mathbb{C}$. 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 $\bar{\mathbb{Q}}(x_1,\ldots,x_n)$ into the complex numbers. However, something here doesn’t feel quite right to me. People seem to be asserting that proving something over $\mathbb{C}$ 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? Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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9 Responses to Bertini’s Theorem

1. Walt says:

One version of the Lefschetz principle follows from mathematical logic: the set of first-order theorems over an algebraically closed field of characteristic zero does not depend on the underlying field.

2. Charles says:

I should have responded sooner. Anyway, first order logic is grossly insufficient for algebraic geometry, to my understanding (and recollection of logic, which is rusty). I’m fairly confident that the Nullstellensatz isn’t first order, nor is the statement that $R$ is Noetherian. It seems like there must be a stronger statement out there that cna be called the Lefschetz principle, but that first order statements are all ok doesn’t feel strong enough to me.

• steve says:

The nullstellensatz can be proved using first order methods, and this is well known. Just look up any book on the model theory of fields. For example here:
http://www.math.uic.edu/~marker/mtf.ps

Also the Lefschetz principle is precisely what Walt says: a statement about first order theorems in the theory of fields. This is clarification of the Lefschetz principle is due to Tarski.

As a heuristic: first order methods work when one considers sets defined by finite formulae (also called definable sets). Algebraic varieties, being defined by finitely many polynomial equations, are therefore definable in the model theory of fields. Many statements of algebraic geometry are therefore about definable sets as they involve algebraic varieties.

• Charles Siegel says:

That’s interesting…I don’t know much model theory, so probably I’d just heard something and remembered it incorrectly. Thanks for the correction!

3. Botong says:

I didn’t get how you can reduce to the case of a pencil. When the linear series change, the base locus changes.

4. Botong says:

Charles, just ignore the previous post. I think I get it. Thanks for sharing this.

5. Charles Staats says:

Is there no smoothness hypothesis on X?

• Charles Siegel says:

Ah, yeah, $X$ is smooth, or else we need to say union of the singular locus and the base locus.

6. Andrea says:

Hello Charles, can you please explain how you reduce to prove Bertini for a pencil? The base locus of any pencil in the system is BIGGER than the base locus of the system.
It’s probably trivial but I do not see it. Thank you, and many compliments for your nice blog.

Andrea