We start with a semistable elliptic curve $E/\mathbb Q$ such that such that its mod 3 representation $\overline{\rho}_{E,3}$ is reducible. We wish to show that $\overline{\rho}_{E,5}$ is modular. This follows immediately from the following lemma from the DDT notes:
Lemma 3.49: There is a semistable elliptic curve $A/\mathbb Q$ such that
- $A[5] \simeq E[5]$ as $G_\mathbb Q$-modules.
- $A[3]$ is an irreducible $G_\mathbb Q$-module.
Proof: Let $Y'(5)$ be the $\mathbb Q$-curve classifying elliptic curves $A$ together with an isomorphism $E[5] \simeq A[5]$ compatible with Weil pairings. Elliptic curves over $\mathbb Q$ satisfying 1) correspond to points in $Y'(5)(\mathbb Q)$. Adjoining finitely many points to $Y'(5)$ gives its compactification $X'(5)$ which is a twist of $X(5)$. We know that $X(5)$ has genus $0$ as a $\mathbb C$-curve. Since $X'(5)$ has a point $x_0$ over $\mathbb Q$ corresponding to $E$, it is isomorphic to $\mathbb P^1$ over $\mathbb Q$. The rational points of $Y'(5)$ therefore give a large supply of elliptic curves satisfying 1). Consider the curve $Y'(5,3)$ classifying elliptic curves $A$ satisfying 1) and equipped with a subgroup of order 3. One quickly checks that its compactification $X'(5,3)$ has genus $> 1$ and hence has only finitely many rational points by Faltings' Theorem. Hence only finitely many points in $Y'(5)(\mathbb Q)$ are in the image of $Y'(5,3)(\mathbb Q)$ under the natural map $Y'(5,3) \to Y'(5)$. Hence for all but finitely many $x \in Y'(5)(\mathbb Q)$, the corresponding $A$ satisfies 2) since it has no rational subgroup of order 3. Choose $x$ arbitrarily close in the 5-adic topology to $x_0$ to find an elliptic curve $A$ (associated to $x$) which is semistable and satisfying 1) and 2). Hence the Lemma.
I fail to understand the last line of the proof: why does choosing $x$ 5-adically close to $x_0$ give us a semistable elliptic curve $A$ with the desired properties? What does the 5-adic topology have to do with anything? Correct me if I am wrong here: it seems we have established a) and b) in the preceding lines of the proof, so it seems to me that this 5-adic topology argument must be being used to establish semistability. Is this true? If so, how/why is this true? If not, where am I going wrong?
I have seen this exact argument given in other sources, and I feel it bears at least a little explanation.
Best Answer
As far as I can tell, there are two distinct steps to the argument: semistability outside $5$ and semistability at $5$.
The first step is automatic for any elliptic curve represented by some $x \in Y’(5)(\mathbb{Q})$ – see Lemma 1.
The second step requires the $5$-adic approximation. The sketch of the argument is as follows (see below for details): if the “new point” is close $5$-adically to the original curve, then they can be represented by arbitrarily $5$-adically close minimal Weierstrass equations. In particular, they have the same reduction type mod $5$.
Lemma 1: let $p,\ell$ be distinct primes with $\ell>3$. Let $E$ be a semistable elliptic curve over $\mathbb{Q}_p$, and $F/\mathbb{Q}_p$ be another elliptic curve such that $E[\ell]$ and $F[\ell]$ are isomorphic as Galois-modules. Then $F$ is semistable.
Proof: let $I_p$ be the inertia group at $p$ and $I_p^+$ be the wild inertia group. The goal is to show that $1$ is the only eigenvalue of an element of $I_p$ acting on $T_{\ell}F$.
Note first that $I_p^+$ acts trivially on $E[\ell]$, hence on $F[\ell]$; so its action on $T_{\ell}F$ is contained in $I_2+\ell\mathcal{M}_2(\mathbb{Z}_{\ell})$, which is a pro-$\ell$-group. Since $I_p^+$ is a pro-$p$-group, $I_p^+$ acts trivially.
Next, let $M \in GL_2(\mathbb{Z}_{\ell})$ be a matrix in the image of $I_{p}$ on $T_{\ell}F$. We know (because $F[\ell] \cong E[\ell]$ and $E$ is semistable) that $M$ is (trivial or) unipotent mod $\ell$, so that its two eigenvalues are congruent to $1$ mod $\ell$.
Moreover, since the Frobenius acts by $p$-th power on $I_p/I_p^+$, $M$ and $M^p$ are similar. So let $u,v$ be the two (generalized) eigenvalues of $M$: then $uv=1$, and $\{u,v\}=\{u^p,v^p\}$. Thus $u^{p \pm 1}=1$, so $u$ is a root of unity.
If $u \neq 1$, then the order of $u$ is divisible by $\ell$. Hence $\ell-1 \leq [\mathbb{Q}_{\ell}(u):\mathbb{Q}_{\ell}] \leq 2$, a contradiction. QED.
Now, for semistability at $5$. This one is actually a bit more mysterious and I wonder how necessary $5$-adic approximation is.
We use Prop. 11 in Chapter XVI of Cornell-Silverman-Stevens: it states that there is an open subset $U \subset \mathbb{P}^1_{\mathbb{Q}}$, an isomorphism $\psi: U \rightarrow Y’(5)$, and polynomials $f,g \in \mathbb{Q}[t]$ with respective degrees $20,30$ such that for $u \in U(K)$, $\psi(u)$ is represented by the elliptic curve $y^2=x^3+f(u)x+g(u)$.
Up to translating, we may assume that $0 \in U(\mathbb{Q})$ represents the original elliptic curve. Up to rescaling $f,g$, we may assume that $y^2=x^3+f(0)x+g(0)$ is a minimal equation at $5$ and has (at worst) a node mod $5$.
Well, if $u \in \mathbb{Q}$ is close enough $5$-adically to $0$, then $u \in U(\mathbb{Q})$, $f(u),g(u) \in \mathbb{Z}_5$ and $(f(u),g(u)) = (f(0),g(0)) \pmod{5}$, so that $y^2=x^3+f(u)x+g(u)$ (which represents the elliptic curve $\psi(u)$) is a minimal equation which also has (at worst) a node mod $5$. Thus we’re done.