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Let $\mathbf{T}$ be the reduced nearly ordinary Hecke algebra of level $N$ of Hida theory for $\operatorname{GL}_{2}$ over $\mathbb{Q}$ (or more generally over a totally real field $F$). Then $\mathbf{T}$ is finitely generated over a regular ring $\Lambda$ of dimension 3. Let $\mathfrak{m}$ be a maximal non-Eisentein ideal of $\mathbf{T}$.

By patching pseudo-representations attached to algebraic modular forms, Wiles (and Hida) have constructed a two-dimensional $G_{\mathbb{Q}}$-representation $(V,\rho)$ with coefficients in $\mathbf{T}_{\mathfrak{m}}\otimes_{\Lambda}\operatorname{Frac}(\Lambda)$. This representation admits a 1-dimensional sub-space $V^{+}$ and a free 1-dimensional quotient $V^{-}$ both stable under the action of $G_{\mathbb{Q}_{p}}$. Because $\mathfrak{m}$ is non-Eisenstein, there exists a choice of basis of $V$ such that $\rho$ has values in $\operatorname{GL}_{2}(\mathbf{T}_{\mathfrak{m}})$. The lattice $L\subset V$ corresponding to this choice of basis admits a free sub-module $L^{+}=L\cap V^{+}$ of rank 1 stable under $G_{\mathbb{Q}_{p}}$. However, it is unclear to me whether $L$ admits a free rank 1 quotient stable under $G_{\mathbb{Q}_{p}}$. This is true if $\rho$ modulo $\mathfrak{m}$ is of the form $$\rho\sim\begin{pmatrix}\chi_{1}&*\\ 0&\chi_{2}\end{pmatrix}$$ with $\chi_{1}\neq\chi_{2}$ because then $L/L^{+}$ is generated by a single element according to Nakayama lemma. However, without this hypothesis, I don`t see an obvious proof of this fact, nor have I good reasons to believe it should be true. Does anyone know for sure?

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I went and deleted my answer because; though I had convinced myself when I wrote it, I don't believe it anymore. Sorry about that. –  Rob Harron Dec 4 '09 at 0:03
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2 Answers

up vote 7 down vote accepted

$\newcommand\T{\mathbf{T}_{\mathfrak{m}}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\m{\mathfrak{m}}$ $\newcommand\F{\mathbf{F}}$ $\newcommand\Frob{\mathrm{Frob}}$ $\newcommand\rhobar{\overline{\rho}}$ $\newcommand\eps{\epsilon}$

First, as Professor Emerton mentions, the construction of $L^{+}$ you gave is not necessarily free over $\T$. Thus, I will interpret your question as asking the following: does there exist an exact sequence:

$$0 \rightarrow L^{+} \rightarrow (\T)^2 \rightarrow L^{-} \rightarrow 0$$ of $\T[G_{\Q_p}]$-modules where $L^{+}$ and $L^{-}$ are free $\T$-modules of rank one.

( Edit Perhaps this extra remark might be useful. Suppose that $L = (\T)^2$ admits a free rank one quotient $L^{-}$. Since $L^{-}$ is free, it admits a section $L^{-} \rightarrow L$, and hence the kernel $L^{+}$ of $L \rightarrow L^{-}$ is also free. Thus the existence of a free rank one quotient asked for in the question is equivalent to the existence of the exact sequence above.)

The answer to this question, in general, is no. The following argument is implicitly contained in papers of Wiese on the failure of multiplicity one and weight one forms.

The action of $G_{\Q_p}$ on $L^{+}$ is unramified and so acts via $G_{\mathbf{F}_p}$. Thus $\Frob_p$ acts on a basis vector as multiplication by some element of $\T$. Since $\T$ is determined by its action on classical eigenforms, one may identify this element with the Hecke operator $U$. In particular, $U \in \T$ (it wasn't clear whether your $\T$ included $U$ or not).

The exact sequence remains exact after tensoring with $\T/\m$, for dimension reasons. It follows that the sequence is split as a sequence of $\T$-modules. Hence it remains exact after quotienting out by any ideal of $\T$.

Suppose that $\rhobar: G_{\Q} \rightarrow \mathrm{GL}_2(\F_p)$ is irreducible and modular (mod-$p$) of weight $1$. Suppose, moreover, that $\rhobar(\Frob_p)$ acts by a scalar $\lambda$. Associated to $\rhobar$ is a mod-$p$ weight $1$ form $f = \sum a_n q^n \in \F_p[[q]]$. If $A$ is the Hasse invariant, then then $Af$ and $f^p$ are both mod-$p$ modular forms of weight $p$. One can check that all elements of the $\F_p$-vector space $\{Af,f^p\}$ are eigenvalues for all the Hecke operators $T_l$ for $(l,p) = 1$, but the operator $T$ (and so $U$, which is the same as $T$ in weight $> 1$) satisfies $(U - \lambda)^2 = 0$ but does not act by a scalar. Since $U$ acts invertibly on this vector space, it gives rise to a surjective map: $$\T \rightarrow \F_p[\eps]/\eps^2,$$ where the image of $T_l$ lands in $\F_p$ for all $(l,p) = 1$, but $U$ does not act by a scalar. Let $I$ be the kernel.

The Galois representation on $(\T)^2/I \simeq (\F_p[\eps]/\eps^2)^2$ is equal to $\rhobar \oplus \rhobar$. This follows from a result of Boston-Lenstra-Ribet, since $T_l$ is acting by a scalar for each $(l,p) = 1$. It follows, by assumption, that the action of $G_{\Q_p}$ on $L^{+}/I L^{+} \simeq \T/I$ must also be trivial, because this is a sub-representation of $\rhobar \oplus \rhobar$. On the other hand, as we have seen, the action of Frobenius on $L^{+}$ and thus $L^{+}/I L^{+} = \T/I$ is given by $U$, which is acting non-trivially $\T/I$ by the construction of $I$. This is a contradiction.

Such representations $\rhobar$ exist (for example, with $p = 2$, and level $\Gamma_0(431)$) as mentioned in Professor Emerton's answer.

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Dear Hamburglar, It is "Wiese", not "Weise". Cheers, Matt –  Emerton Dec 19 '11 at 4:25
    
Thanks a lot for resuscitating this question and for your great answer. –  Olivier Jan 25 '12 at 13:04
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In general (i.e. without an assumption on $\bar{\rho}$ like $p$-distinguished) I don't see that $L^+$ will be free (generically free of rank 1, yes, but why free?). And I'm confident that, as you suspect, $L/L^+$ need not be free either.

I don't know an explicit counterexample off the top of my head, but one could try the non-Gorenstein examples for $p = 2$ and level $431$ found by Lloyd Kilford. (Non-Gorensteiness is a closely related property to the non-freeness you are thinking about.)

Note that one doesn't have to check on the entire Hida family. If the freeness were true over the Hida family, it would also be true over the specialization to weight 2, say, which is now a finite ${\mathbb Z}_p$-module, and is the kind of Hecke algebra considered by Kilford.

So if the freeness breaks down on the small Hecke algebra, it will break down on the big one too.

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