\section{Irreducibility}\label{sec:irreducibility}

Suppose that the residual Frey representation had a one-dimensional constituent.  Such a constituent would be the reduction of one of five algebraic Hecke characters of conductor dividing $\p^2$, and the Frobenius data at the primes above $7$ exclude all five (\cref{thm:irreducible}).

\needspace{5\baselineskip}\subsection{Cubic characters at \texorpdfstring{$\p$}{p}}

\begin{lemma}\label{lem:cubic-local}
Let $u\in\Z$ with $3\nmid u$, and let $\Art_\p\colon K_\p^\times\to G_{K_\p}^{\mathrm{ab}}$ be the local Artin map.
\begin{enumerate}[(a)]
\item $\kappa_u\circ\Art_\p$ is trivial on $\Q_3^\times$.
\item If $u\equiv\pm1\pmod 9$, then $\kappa_u$ is trivial on $G_{K_\p}$.
\item If $u\not\equiv\pm1\pmod 9$, then $\kappa_u|_{G_{K_\p}}$ has conductor $\p^2$ and $\kappa_u(\Art_\p(\omega))\ne1$.
\end{enumerate}
In all cases $\kappa_u\circ\Art_\p$ is trivial on $1+\p^2$.
\end{lemma}

\begin{proof}
(a) $\kappa_u(\Art_\p(x))$ is the cubic Hilbert symbol $(u,x)_\p$ (up to inversion, depending on normalizations).  Let $u,x\in\Q_3^\times$, and let $g$ be the nontrivial automorphism of $K_\p/\Q_3$.  By Galois equivariance $g\big((u,x)_\p\big)=(gu,gx)_\p=(u,x)_\p$, and $g$ inverts $\mu_3$; so the symbol is $1$.

(b), (c) The group $\Z_3^\times/(\Z_3^\times)^3$ has order $3$: $-1$ is a cube and $1+9\Z_3=(1+3\Z_3)^3$, while $2$ is not a cube in $\Q_3$.  So $u$ is a cube in $\Q_3$ if and only if $u\equiv\pm1\pmod 9$, and otherwise $\kappa_u=\kappa_2^{\pm1}$ on $G_{K_\p}$ (a cube in $K_\p$ that lies in $\Q_3$ is a cube in $\Q_3$, as $[K_\p:\Q_3]=2$).  It remains to treat $u=2$.  The discriminant of $\Q(\sqrt[3]2)$ is $-108$, so $\Q_3(\sqrt[3]2)/\Q_3$ is totally ramified of degree $3$ with discriminant exponent $3$; the quadratic extension $K_\p(\sqrt[3]2)/\Q_3(\sqrt[3]2)$ is tamely ramified, so $K_\p(\sqrt[3]2)/\Q_3$ has discriminant exponent $2\cdot3+1=7$.  Computed through the tower $K_\p(\sqrt[3]2)\supset K_\p\supset\Q_3$, the same exponent is $3\cdot1+2f$, where $f$ is the conductor exponent of $\kappa_2$ at $\p$ (both nontrivial characters of $\Gal(K_\p(\sqrt[3]2)/K_\p)$ have conductor $\p^f$).  So $f=2$.  A character of conductor $\p^2$ factors through $(\OO/\p^2)^\times\isom\mu_6$ nontrivially, and, being cubic, is nontrivial on $\omega$.  The final assertion of the lemma follows from (b) and $f=2$.
\end{proof}

\subsection{The theorem}

\begin{theorem}\label{thm:irreducible}
Let $p\ge17$ be a prime with $p\equiv1\pmod3$, and let $\rhob$ be the residual Frey representation of a solution of $x^3+y^3=z^p$ in coprime nonzero integers (\cref{def:frey}).  Then $\rhob$ is absolutely irreducible.
\end{theorem}

\begin{proof}
Suppose not.  Since $\dim\rhob=3$, $\rhob$ has a one-dimensional constituent $\theta\colon G_K\to\Fbar_p^\times$.

\emph{Step 1: ramification.}  By \cref{prop:local}~(b), $\theta$ is unramified outside $\{\p,v,\bar v\}$.  By \cref{prop:local}~(d), $\theta|_{I_\p}=\bar\kappa_u^{-\ninf}|_{I_\p}$, which by \cref{lem:cubic-local} has order dividing $3$ and conductor dividing $\p^2$.  By \cref{prop:local}~(f) and Fontaine--Laffaille theory \cite{FontaineLaffaille} ($p\ge5$, weights in $[0,2]$), $\rhob|_{G_{K_v}}$ is the reduction of a crystalline representation with Hodge--Tate weights $\{0,1,2\}$, and every one-dimensional subquotient restricts to inertia as $\epsb^{-k}$ with $k\in\{0,1,2\}$.  So $\theta|_{I_v}=\epsb^{-k_v}|_{I_v}$ and $\theta|_{I_{\bar v}}=\epsb^{-k_{\bar v}}|_{I_{\bar v}}$ with $k_v,k_{\bar v}\in\{0,1,2\}$.

\emph{Step 2: the type.}  By class field theory $\theta$ is a character of the idele group of $K$, trivial on $K^\times$ and on the connected component at infinity, and unramified outside $\p,v,\bar v$.  Evaluate it at the global unit $-1$: through the local Artin maps its components at $v$ and $\bar v$ contribute $(-1)^{k_v}$ and $(-1)^{k_{\bar v}}$, its component at $\p$ contributes the value at $-1$ of the cubic character $\bar\kappa_u^{-\ninf}$, which is $1$, and the other components contribute $1$; so $(-1)^{k_v+k_{\bar v}}=1$.  So the \emph{type} $(k_v,k_{\bar v})$ lies in $\mathcal T=\{(0,0),(1,1),(2,2),(0,2),(2,0)\}$.

\emph{Step 3: $\theta$ is the reduction of an algebraic Hecke character.}  For $x\in K^\times$ prime to $\p$ let $\zeta_x\in\mu_6$ be the unit with $x\equiv\zeta_x\pmod{\p^2}$.  The algebraic Hecke characters
\[
\psi_{(k,k)}((x))=(x\bar x)^k\ (k=0,1,2),\qquad \psi'((x))=(\zeta_x\bar x)^2,\qquad \psi''((x))=(\bar\zeta_x x)^2
\]
are well defined on ideals prime to $\p$ and have conductor dividing $\p^2$.  Their $p$-adic realizations (through $\iota_p$, with values $\psi(\q)$ on $\Frob_\q$) restrict to $I_v\times I_{\bar v}$ as $(\varepsilon_p^{-k},\varepsilon_p^{-k})$, and, for $\psi'$ and $\psi''$, as $(\varepsilon_p^{-2},1)$ and $(1,\varepsilon_p^{-2})$ in some order.  So there is $\psi\in\{\psi_{(0,0)},\psi_{(1,1)},\psi_{(2,2)},\psi',\psi''\}$ whose reduction $\bar\psi$ has the same restrictions to $I_v$ and $I_{\bar v}$ as $\theta$.  Then $\theta\bar\psi^{-1}$ is unramified outside $\p$ and its restriction to $I_\p$ has conductor dividing $\p^2$: it is a character of the ray class group $\Cl_{\p^2}(K)$, which is trivial because $\OO$ is principal and $\mu_6\to(\OO/\p^2)^\times$ is onto.  Hence $\theta=\bar\psi$.

\emph{Step 4: the primes above $7$.}  By \cref{prop:local}~(e), $\theta(\Frob_\q)$ is a root of the reduction of $P_\q(\xi_\q^{-1}T)$ for $\q\in\{\lp,\lpb\}$, with $\xi_\q=\kappa_u(\Frob_\q)^{-\ninf}\in\mu_3$.  Let $\mathfrak P$ be the prime of $\OO$ above $p$ defined by $\iota_p$, so that $\mathfrak P$ is $v$ or $\bar v$ (the coefficients of the characteristic polynomials lie in $\OO$).  Write $\varpi=2-\omega$, so $\lp=(\varpi)$ and $\lpb=(\bar\varpi)$.  The value $\psi(\lp)$ is $\varpi^k\bar\varpi^{k'}$ times an element of $\mu_3$, and $\psi(\lpb)$ is $\bar\varpi^k\varpi^{k'}$ times an element of $\mu_3$, where $(k,k')\in\mathcal T$ is the infinity type of $\psi$ (for $\psi'$ and $\psi''$ the units $\zeta_\varpi^2,\bar\zeta_\varpi^2$ lie in $\mu_3$).  Absorbing these roots of unity and $\xi_\q$ into $\zeta_1,\zeta_2\in\mu_3$, we find $(k,k')\in\mathcal T$ and $\zeta_1,\zeta_2\in\mu_3$ with
\begin{equation}\label{eq:test7}
P_\lp(\zeta_1\varpi^k\bar\varpi^{k'})\in\mathfrak P\quad\text{and}\quad P_\lpb(\zeta_2\bar\varpi^k\varpi^{k'})\in\mathfrak P .
\end{equation}
This test does not depend on the normalization of \cref{sec:conventions}: with arithmetic Frobenius elements, the roots of $P_\q$ and the values $\psi(\q)$ are both replaced by their inverses.

\emph{Step 5: computation.}  By \cref{prop:frob7}~(a), $(P_\lp,P_\lpb)$ is $(P^+,P^-)$ or $(P^-,P^+)$.  For both assignments, each $(k,k')\in\mathcal T$ and each $(\zeta_1,\zeta_2)\in\mu_3^2$, the two elements above are nonzero, and the ideal of $\OO$ they generate has all its prime factors above $2$, $3$, $7$ or $13$ (\cref{app:computations}, item~C2).  Since $\mathfrak P$ lies above $p\ge19$, this is a contradiction.
\end{proof}

\begin{remark}
Requiring the \emph{same} prime $\mathfrak P$ at $\lp$ and $\lpb$ matters: without that requirement, the rational primes dividing the norms of both elements include $67$, $103$ and $523$ for the types $(0,0)$ and $(2,2)$, and these primes divide the two norms through different primes of $K$.  The computation of Step~5 also allows the values of $\kappa_u$ at $\lp$ and $\lpb$ to vary independently in $\mu_3$, although they are complex conjugate.  So it covers every solution, both sheaves $\VV_{\chi^{\pm1}}$ and both choices of $u$.
\end{remark}