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?



Free subquotient of Galois representations coming from Hida theory
$\newcommand\T{\mathbf{T}\_{\mathfrak{m}}}$
$\newcommand\Q{\mathbf{Q}}$
$\newcommand\m{\mathfrak{m}}$
$\newcommand\F{\mathbf{F}}$
$\newcommand\Frob{\mathrm{Frob}}$
$\newcommand\rhobar{\overline{\rho}}$
$\newcommand\eps{\epsilon}$


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.


