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

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 Weise 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.

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

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.

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

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.

The answer to this question, in general, is no. The following argument is implicitly contained in papers of Weise 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.

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

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 Weise 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.