The mod $p$ Galois representation of the Frey curve is unramified away from $2, p$

Question

Given a hypothetical solution to Fermat's last theorem for $p \ge 5$ $$a^p + b^p + c^p = 0$$with $a \equiv -1 \pmod 4$, $b$ even, we can write down the Frey Curve$$E: y^2 = x(x-a^p)(x+b^p)$$which has discriminant $\Delta_E = -2^{-8}(abc)^{2p}$.

Letting $G_\mathbb Q =\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ act on the $p$-torsion points of $E$, we obtain a Galois representation $$\rho:G_\mathbb Q \to \mathrm{Aut}(E[p])\cong \mathrm{GL}_2(\mathbb F_p)$$ I've seen in numerous texts that

$\rho$ is unramified at all primes $\ell \ne 2, p$

but most of these texts just say that this is "easy to check".

Is there an easy way to see why this is true? I'm guessing I'm missing something obvious!

Answer 1

Well, it was easy for Serre to check ;D

This is not obvious, but once you know the right tools to use, it does become an easy exercise. A great resource for learning about the proof of FLT is this expository paper by Darmon, Diamond, and Taylor: https://www.math.wisc.edu/~boston/ddt.pdf

In particular, see Theorem 2.15 (b). The proof refers you to other non-trivial propositions, but you'll see that, with the relevant references in hand, you can indeed check that the representation is unramified away from $2,p$. To make this a worthy post, I'll sketch the idea below.

You've set things up so that the discriminant of $E$ is $\Delta = 2^{-8}(abc)^{2p}$. For any elliptic curve, the only primes at which ramification is possible are those which divide the discriminant. In particular, for the Frey curve, the only primes at which $\rho$ can ramify are $2,p$, and the primes $\ell$ dividing $abc$.

Our situation is even better, because this discriminant is minimal. In fact, if $\ell \mid abc$, then $\rho$ is unramified at $\ell$ if and only if $p \mid v_{\ell}(\Delta)$ (where $v_\ell$ is the $\ell$-adic valuation). This is Proposition 2.12(c) at the link I provided. So in fact $\rho$ is unramified at all those primes $\ell$ which divide $abc$.

So the only primes left at which our representation can ramify are $2$ and $p$.

Answer 2

To elaborate a little more on the existing answer:

There is a criterion (Neron--Ogg--Shafarevic) whichs says that $E$ has good reduction at $\ell \neq p$ iff the Galois action on the $p$-adic Tate module is unramified.

So if $\ell$ divides the (minimal) discriminant, and so is a prime of bad reduction, then we know that the Galois action on the $p$-adic Tate module will be ramified.

Recall that the $p$-adic Tate module is built as the inverse limit of $E[p^n]$ over all $n$. This suggests the more subtle question, namely: how large a value of $n$ do we have to take in order to start seeing ramification at $\ell$.

It turns out that, in the case of semi-stable reduction, this is related to the power of $\ell$ that divides the discriminant: for example, the value $n =1$ is not large enough (i.e. the Galois action on the $p$-torsion is unramified at $\ell$, despite $\ell$ being a prime of bad reduction) iff the power of $\ell$ dividing the discriminant is a multiple of $p$.

The usual way to see this is via the theory of Tate curves. The rough idea is that having semistable reduction means that the elliptic curve becomes a nodal curve mod $\ell$, and the group law on the nonsingular points of a nodal cubic is isomorphic to the multiplicative group $\mathbb G_m$. The $p^n$-torsion on here is equal to the $p^n$th roots of $1$, and hence has size $p^n$, rather than $p^{2n}$ (the size of $E[p^{2n}]$). Thus a lot of the $p^n$-torsion on $E$ ``disappears'' somehow upon reduction mod $\ell$, and this is reflect in the appearance of ramification.

However, if the power of $\ell$ in the discriminant is a multiple of $p$, then the reduction of $E$ at $\ell$ is not actually a nodal curve --- rather, it is the union of $p$ lines, each meeting at a point, arranged in a $p$-gon (actually a $p^r$-gon if $p^r$ is the power of $\ell$ dividing $\Delta,$ but let me write as if $r = 1$). (To see the relationship with a nodal curve, think of a nodal curve as a single line meeting itself in a point --- so a $1$-gon.) If we remove the singular points, we get not just $\mathbb G_m$, but $\mathbb G_m \times (\mathbb Z/p)$, and this has $p^2$ $p$-torsion points; thus there is ``enough room'' for all the $p$-torsion on $E$ to reduce nicely mod $\ell$, and so we don't get any ramification at $\ell$ on the $p$-torsion.