Ben Green has a nice clean writeup of the argument that there are no Siegel zeroes, found in a big OpenAI experiment a few weeks ago. Green’s writeup is itself a product of iterated back-and-forth with another OpenAI model. But it reveals something quite nice; that at the heart of the proof is the good old determinant method of Bombieri-Pila and Heath-Brown! (Except it crucially uses multiple primes at once, so I ought to mention Salberger and Walsh too, and probably some other people.)
The driving idea: if you have a bunch of low-height points on a variety X, and an equally-sized bunch of polynomials on X, you can make a big old matrix by evaluating each polynomial on each point, and then you can ask this matrix what its determinant is about. Because you’ve bounded the height of the points, that determinant can’t be too archimedeanly big. On the other hand, if p is a not-too-large prime, a lot of those points reduce to the same point mod p (or even higher powers of p), which means a lot of those rows agree mod p, which makes the determinant highly divisible by p. If that power of p is bigger than the archimedean bound, you have an integer at most X which is divislbe by a number bigger than X, which means it’s zero. But THAT means some even more massive polynomial vanishes at all your points while not actually being identically zero on X, and that means your low-height points are concentrated on a proper subvariety whose degree you can control. In the very popular case where X is a curve, this gives you a bound on the number of points of bounded height, which has the very nice feature of being independent of the coefficients of X. (This isn’t obvious from what I’ve said!)
Anyway, the Siegel zeroes proof runs this in the opposite direction. If chi is the quadratic character whose L-function you’re studying, X is the restriction of scalars of A^1 from the corresponding quadratic field K to Q. So it’s a surface whose integral points are in bijection with Z[d^{1/2}] (or with the ring of integers, whatever, where we’re going we don’t need constants.) And you take a bunch of low-height integral points and a bunch of polynomials, you get a determinant D which is archimedeanly bounded, and then you show that every p which is inert in K appears to a big power in D. But you’re not proving D is 0; the set of all low-height points on X is manifestly not contained in some low-degree curve, so D isn’t zero, which means you just can’t have too many of those problematic (i.e. inert) primes p, and that is exactly what one is trying to prove.
Of course it’s a bit trickier than that! In traditional determinant method your bunch of polynomials (at least in the cases I’m familiar with) is basically just “all polynomials of degree below some bound.” Here, the polynomials have to be chosen more carefully. The idea is that we can think of X(Q) as parametrized by pairs (z,z’) of Galois-conjugate pairs of elements in K; when p is inert, the rational points X(F_p) all lie on the curve z^p = z’, and this curve is inseparable on projection to the z’ coordinate. So you choose your polynomials in a careful way that breaks the z/z’ symmetry and gives you tons of p-adic vanishing precisely for these inert primes, where the points are concentrated on this curve. (And I should say that a bit of back and forth with Claude Fable was helpful in pinning down what it was about this particular bunch of polynomials that makes inert primes especially apt to divide the determinant with large multiplicity!)
