\section{The Frey representation of a solution}\label{sec:frey}

Parts~II and~III use the local properties of the Frey representation of a solution and its Frobenius data at the primes above $7$ (\cref{prop:local}), together with its irreducibility (\cref{sec:irreducibility}).

\subsection{Inputs from Kraus}
Kraus \cite[\S6]{Kraus98} writes a solution with the even variable first and normalizes its sign.  His results for $p\ge17$ assume that $E_{a,b}$ is modular, which is now known \cite{BCDT01}; his note added in proof observes that it follows from the work of Conrad, Diamond and Taylor.  We only need the following consequences, which do not depend on the normalization.

\begin{proposition}\label{prop:kraus}
Let $p\ge17$ be prime and let $(a,b,c)$ be a solution of $x^3+y^3=z^p$ in coprime nonzero integers.  Then
\begin{enumerate}[(a)]
\item $3\mid c$;
\item $p\nmid c$;
\item $7\nmid abc$;
\item exactly one of $a,b$ is even.
\end{enumerate}
\end{proposition}

\begin{proof}
Replacing $(a,b,c)$ by $(b,a,c)$ or by $(-a,-b,-c)$ we may assume that we are in Kraus's normalization.  Then (a) and (d) are part of \cite[Théorème~6.1]{Kraus98} and (b) is \cite[Lemme~6.4]{Kraus98}.  For (c), \cite[Proposition~6.3~(ii)]{Kraus98} gives $\rhob_{E_{a,b},p}\isom\rhob_{W,p}$, where $W\colon Y^2=X^3+6X-7$ has good reduction at $7$ with $a_7(W)=0$ (it reduces to $Y^2=X^3-X$, and $7\equiv3\pmod 4$).  The discriminant of $E_{a,b}$ is $-432c^{2p}$ and $c_4(E_{a,b})=-144ab$, where $ab$ is prime to $c$.  If $7\mid c$, then $E_{a,b}$ has multiplicative reduction at $7$ and the congruence forces $a_7(W)\equiv\pm(7+1)\pmod p$, that is $p\mid 8$.  If $7\mid ab$, then $7\nmid c$, $E_{a,b}$ has good reduction at $7$, and modulo $7$ it is $Y^2=X^3\pm(\text{a unit})^3$, a quadratic twist of $y^2=x^3+1$, which has $a_7=-4$; the congruence then gives $0\equiv\pm4\pmod p$.  Both are impossible for $p\ge17$.
\end{proof}

\subsection{The parameter}
Let $p\ge 17$ and let $(a,b,c)$ be a solution in coprime nonzero integers.  As $a,b$ are coprime, $p\nmid a$ or $p\nmid b$.  Choose $(u,w)\in\{(a,b),(b,a)\}$ with $p\nmid u$ (when both choices are possible, nothing depends on which is made; see \cref{rem:normalisation}), and put
\[
s_0=\frac{c^p}{u^3}=1+\Big(\frac wu\Big)^3\in\Q\smallsetminus\{0,1\}.
\]

\needspace{4\baselineskip}\begin{lemma}\label{lem:s0}
We have $v_3(s_0)=p\,v_3(c)\ge p$, and $s_0\equiv2\pmod 7$ (in $\Z_{(7)}$).
\end{lemma}

\begin{proof}
By \cref{prop:kraus}~(a), $3\mid c$, and $3\nmid u$ by coprimality.  By \cref{prop:kraus}~(c), $u,w,c$ are prime to $7$; the cube of a unit modulo $7$ is $\pm1$, and $(w/u)^3\equiv-1$ would give $7\mid c$; so $(w/u)^3\equiv1$ and $s_0\equiv2\pmod 7$.
\end{proof}

\subsection{Tame specialization}

\begin{lemma}\label{lem:tame}
Let $\q\nmid3\ell$ be a prime of $K$ and $s_0\in K\smallsetminus\{0,1\}$.
\begin{enumerate}[(a)]
\item If $v_\q(s_0)=0$ and $v_\q(s_0-1)=0$, then $\VV_{s_0}$ is unramified at $\q$.
\item If $v_\q(s_0)>0$, then there is a $G_K$-stable lattice in $\VV_{s_0}$ on which $I_\q$ acts through $\sigma\mapsto\gamma_0^{v_\q(s_0)t_\ell(\sigma)}$, where $t_\ell\colon I_\q\to\Z_\ell(1)$ is the $\ell$-adic tame character (\cref{sec:conventions}) and $\gamma_0$ is a unipotent automorphism of the lattice with $(\gamma_0-1)^3=0$.
\item If $v_\q(s_0-1)>0$ and $3\mid v_\q(s_0-1)$, then $I_\q$ acts trivially on $\VV_{s_0}$.
\item If $s_0=z_0/u^3$ with $z_0,u\in\OO$, $\q\mid u$ and $\q\nmid z_0$, then $I_\q$ acts on $\VV_{s_0}$ by the scalar character $\kappa_u^{\ninf}|_{I_\q}$.
\end{enumerate}
\end{lemma}

\begin{proof}
$\VV$ is lisse on $\SSS$ over $\OO[1/3\ell]$ and tamely ramified along $0,1,\infty$ (\cref{prop:geometry,prop:compactification}); fix a lattice $\VV^\circ$.  In (a) the point $s_0$ extends to a point of $\SSS$ over the local ring at $\q$, so $\VV_{s_0}$ is unramified.  Otherwise $s_0$ reduces modulo $\q$ to one of the points $0,1,\infty$; let $z$ be the local coordinate $s$, $s-1$ or $1/s$ there.  By Abhyankar's lemma \cite[Exposé~XIII, 5.3]{SGA1}, étale locally near that point a finite étale cover of $\SSS$ trivializing $\VV^\circ/\ell^n$ is a disjoint union of Kummer covers, each obtained by extracting an $i$th root of $z$ times a unit, with $i$ prime to the residue characteristic.  Pulling back along $s_0$ shows that $I_\q$ acts on $\VV^\circ_{s_0}$ through the composite of the tame Kummer map $\sigma\mapsto(\sigma(z(s_0)^{1/i})/z(s_0)^{1/i})_i\in\hat\Z'(1)$ with the local monodromy of $\VV^\circ$ at that point, since the value of the unit at $s_0$ contributes nothing on inertia.  Now use \cref{prop:geometry}~(c).  In (b) the point is $0$ and the monodromy is the unipotent $\gamma_0$ on $\VV^\circ$, with $(\gamma_0-1)^3=0$ (one Jordan block of length three).  In (c) the point is $1$ and the monodromy factors through $\mu_3$; the image of $\sigma\in I_\q$ in $\mu_3$ is $\sigma((s_0-1)^{1/3})/(s_0-1)^{1/3}$, which is $1$ because $s_0-1$ is a unit times a cube in $K_\q$ and $\q\nmid3$.  In (d) the point is $\infty$ and the monodromy factors through $\mu_9$; the image of $\sigma\in I_\q$ is $\sigma((1/s_0)^{1/9})/(1/s_0)^{1/9}=\sigma(u^{1/3})/u^{1/3}\cdot\sigma(z_0^{-1/9})/z_0^{-1/9}$, and the second factor is $1$ on $I_\q$ since $\q\nmid9z_0$.  So the image is $\kappa_u(\sigma)\in\mu_3$, which acts by the scalar $\iota_\ell(\kappa_u(\sigma))^{\ninf}$ (\cref{prop:geometry}).
\end{proof}

\subsection{The Frey representation}
Fix a prime $p\ge17$ with $p\equiv1\pmod3$ and a solution, with $u,w,s_0$ as above.  Take $\ell=p$.

\begin{definition}\label{def:frey}
The \emph{Frey representation} of the solution is
\[
\rho=\VV_{s_0}\otimes\kappa_u^{-\ninf}\colon G_K\to\GL_3(\overline\Q_p),
\]
where $\ninf\in\{1,2\}$ is as in \cref{prop:geometry}, and $\rhob\colon G_K\to\GL_3(\Fbar_p)$ is its reduction (\cref{sec:conventions}).
\end{definition}

\begin{proposition}\label{prop:local}
The representation $\rho$ has the following properties.
\begin{enumerate}[(a)]
\item $\rho$ is unramified at every prime $\q\nmid3pc$.  For every prime $\q\nmid3pcuw$ the characteristic polynomial of $\rho(\Frob_\q)$ has coefficients in $\OO$, so $\rho$ has coefficients in $K$.
\item $\rhob$ is unramified at every prime $\q\nmid3p$.
\item $\rho^\cc\isom\rho^\vee\otimes\varepsilon_p^{-2}$, and hence $\rhob^\cc\isom\rhob^\vee\otimes\epsb^{-2}$.
\item $I_\p$ acts on $\rho$ through $\kappa_u^{-\ninf}|_{I_\p}$ times a unipotent representation, so every constituent of $\rhob|_{I_\p}$ is $\bar\kappa_u^{-\ninf}|_{I_\p}$.
\item $\rho$ is unramified at $\lp$ and $\lpb$, and the characteristic polynomial of $\rho(\Frob_\q)$, $\q\in\{\lp,\lpb\}$, is $P_\q(\xi_\q^{-1}T)$ with $\xi_\q=\kappa_u(\Frob_\q)^{-\ninf}\in\mu_3$ and $P_\q$ as in \cref{prop:frob7}.
\item For each prime $v\mid p$ of $K$, $\rho|_{G_{K_v}}$ is crystalline with Hodge--Tate weights $\{0,1,2\}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(a) If $\q\nmid 3pcuw$, then $s_0$ and $s_0-1=(w/u)^3$ are $\q$-units, so $\VV_{s_0}$ is unramified at $\q$ by \cref{lem:tame}~(a); if $\q\mid w$, it is unramified at $\q$ by \cref{lem:tame}~(c) ($v_\q(s_0-1)=3v_\q(w)$); if $\q\mid u$, then $I_\q$ acts on $\VV_{s_0}$ by $\kappa_u^{\ninf}$ by \cref{lem:tame}~(d) (with $z_0=c^p$), which the twist cancels.  The character $\kappa_u$ is unramified outside $3u$.  For $\q\nmid3pcuw$ the coefficients lie in $\OO$ by \cref{lem:traces-in-K}, which applies because $s_0$ reduces to a point of $\SSS$ modulo $\q$, and because $\kappa_u$ takes values in $\mu_3\subset\OO^\times$.

(b) By (a) it remains to treat $\q\mid c$, $\q\nmid 3$ (note $p\nmid c$).  By \cref{lem:tame}~(b), on a $G_K$-stable lattice $I_\q$ acts through powers of $\gamma_0^{p\,v_\q(c)}$, since $v_\q(s_0)=p\,v_\q(c)$ and $\kappa_u$ is unramified at $\q$.  As $(\gamma_0-1)^3=0$ and $p\ge3$, $\gamma_0^p=1+p(\gamma_0-1)+\binom p2(\gamma_0-1)^2\equiv1\pmod p$; so $I_\q$ acts trivially on the reduction of this lattice, hence on $\rhob$.

(c) By \cref{lem:duality} (as $s_0\in\Q$), $\VV_{s_0}^\cc\isom\VV_{s_0}^\vee\otimes\varepsilon_p^{-2}$, and $(\kappa_u^{-\ninf})^\cc=\kappa_u^{\ninf}$ because $u\in\Z$.

(d) By \cref{lem:s0}, $v_3(s_0)\ge17>15/2$, so $I_\p$ acts unipotently on $\VV_{s_0}$ by \cref{thm:inertia3}.

(e) By \cref{lem:s0}, $s_0$ reduces modulo $\lp$ and $\lpb$ to $2\in\SSS(\F_7)$, so $\VV_{s_0}$ is unramified there and $\Frob_\q$ has characteristic polynomial $P_\q$ on it; $\kappa_u$ is unramified at $\q$ since $7\nmid 3u$.

(f) This is \cref{prop:crys} below.
\end{proof}

\needspace{5\baselineskip}\subsection{Crystallinity at \texorpdfstring{$p$}{p}}\label{sec:crys}

\begin{proposition}\label{prop:crys}
Let $p\ge5$, $p\equiv1\pmod 3$, let $v\mid p$ be a prime of $K$ (so $K_v=\Q_p$), and let $s_0\in\Q\smallsetminus\{0,1\}$ with $v_p(s_0)=0$.  Suppose that either $s_0\not\equiv1\pmod p$, or $s_0-1=t_1^3$ with $t_1\in p\Z_{(p)}$.  Then $\VV_{s_0}|_{G_{K_v}}$ (with $\ell=p$) is crystalline with Hodge--Tate weights $\{0,1,2\}$.
\end{proposition}

For the Frey representation, $v_p(s_0)=0$ by \cref{prop:kraus}~(b) and $p\nmid u$; if $p\nmid w$ then $s_0\not\equiv1\pmod p$, and if $p\mid w$ then $s_0-1=(w/u)^3$ with $w/u\in p\Z_{(p)}$.  Since $\kappa_u$ is unramified at $v$ and of finite order, \cref{prop:local}~(f) follows.

For a finite extension $L$ of $\Q_p$ let $L_0=W(k_L)[1/p]$ be its maximal subfield unramified over $\Q_p$, and for a
$p$-adic representation $V'$ of $G_L$, with coefficients in a finite extension $E$ of $\Q_p$, let $D_{\mathrm{st},L}(V')=(B_{\mathrm{st}}\otimes_{\Q_p}V')^{G_L}$, a module over
$L_0\otimes_{\Q_p}E$ with a Frobenius $\varphi$ and a monodromy operator $N$.  Since $B_{\mathrm{st}}^{G_L}=L_0$ and
$B_{\mathrm{st}}^{N=0}=B_{\mathrm{cris}}$ \cite[\S2.1, (4)$_{\mathrm{st}}$ and (6)$_{\mathrm{st}}$]{Tsuji02}, $V'$ is
semistable when $\dim_{L_0}D_{\mathrm{st},L}(V')=\dim_{\Q_p}V'$, and crystalline when moreover $N=0$ on
$D_{\mathrm{st},L}(V')$ \cite[Definition~2.2.7]{Tsuji02}.

\begin{lemma}\label{lem:dst-descent}
Let $L/F$ be a finite Galois extension of finite extensions of $\Q_p$, with inertia subgroup $I(L/F)$, and let $V'$ be a
$p$-adic representation of $G_F$ whose restriction to $G_L$ is semistable.  Then $\Gal(L/F)$ acts on
$D_{\mathrm{st},L}(V')$, through $g\otimes g$ for $g\in G_F$, semilinearly with respect to its action on $L_0$ and
commuting with $N$.  If $N=0$ on $D_{\mathrm{st},L}(V')$ and $I(L/F)$ acts trivially on it, then $V'$ is crystalline.
\end{lemma}

\begin{proof}
Since $G_L$ is normal in $G_F$, the action $g\otimes g$ preserves $D=D_{\mathrm{st},L}(V')$ and factors through
$\Gal(L/F)$; it commutes with $N$ because $N$ commutes with $G_F$ on $B_{\mathrm{st}}$.  The group $\Gal(L/F)$ acts
on $L_0$ through $\Gal(L/F)/I(L/F)\isom\Gal(L_0F/F)\isom\Gal(L_0/F_0)$.  If $I(L/F)$ acts trivially on $D$, then $D$ is
an $L_0$-vector space with a semilinear action of $\Gal(L_0/F_0)$, so $D=L_0\otimes_{F_0}D^{\Gal(L/F)}$ by Galois
descent for vector spaces (Hilbert's Theorem~90).  As $(B_{\mathrm{st}}\otimes V')^{G_F}=D^{\Gal(L/F)}$, this gives
$\dim_{F_0}D_{\mathrm{st},F}(V')=\dim_{L_0}D=\dim_{\Q_p}V'$, so $V'$ is semistable; and $N=0$ on
$D_{\mathrm{st},F}(V')\subset D$, so $V'$ is crystalline.
\end{proof}

For the rest of this subsection let $p\ge5$, let $v$ be a prime of $K$ above $p$ (so $K_v$ is unramified over $\Q_p$
and contains $\mu_3$; $K_v=\Q_p$ if $p\equiv1\pmod 3$), let $\bar F=\Kbar_v$, and let $F=K_v(\zeta_9)\subset\bar F$, where $\zeta_9$ is the image of $\tau_0(\zeta_9)$
under an embedding $\Kbar\hookrightarrow\Kbar_v$ over $K$ that defines the decomposition group $G_{K_v}\subset G_K$ (\cref{sec:conventions}); $F$ is an unramified extension of $K_v$.  Over $F$ the twisted forms of \cref{sec:surfaces} split: $X_s\otimes_KF=\{\prod_j(1-y_j^9)=s\}$,
$H=\mu_9^3$ acting coordinatewise, and $G_F$ acts trivially on $H(\Kbar)$; we use the model $\tilde X^\flat\otimes\OO_F$
of \cref{prop:compactification}~(a) in this split form.  Let $t_1\in\mathfrak m_F\smallsetminus\{0\}$, $s_0=1+t_1^3$, let $t_0\in\bar F$
with $t_0^3=t_1$, and $K'=F(t_0)$; as $\mu_3\subset F$, the extension $K'/F$ is Galois of degree $1$ or $3$ and tamely
ramified.  Let $k'$ be the residue field of $K'$, $\pi$ a uniformizer, and $e=v_{K'}(t_0)\ge1$.  For
$\lambda\in\mu_9$ let $\delta(\lambda)=(\lambda,\lambda,\lambda)\in\mu_9^3$.

The case $s_0\equiv1\pmod p$ of \cref{prop:crys}, where $s_0$ reduces to the point $1$, over which the model of \cref{prop:compactification} is not smooth, needs the following semistable model.  It is used through parts~(c) and~(d): the diagonal $\mu_9$ fixes the double curves pointwise and $\chi^m$ is nontrivial on it, so the $\chi^m$-part of the cohomology of the special fiber comes only from the components, which forces $N=0$; and inertia acts on each component through the diagonal $\mu_3$, on which $\chi^m$ is trivial, so it acts trivially on that part (proof of \cref{prop:crys}).

\begin{lemma}\label{lem:ss-model}
There is a regular scheme $\mathcal X$, projective and flat over $\OO_{K'}$, with the following properties.
\begin{enumerate}[(a)]
\item Its generic fiber is $\tilde X_{s_0}\otimes_FK'$.
\item Its special fiber $Y$ is a reduced divisor with strict normal crossings and without triple points.  Its
components are $Z'$, $\Xi$ and, if $e\ge2$, $E_1,\dots,E_{e-1}$, where $\Xi$ is the Fermat surface
$y_0^9+y_1^9+y_2^9+t^9=0$ in $\PP^3_{k'}$, $Z'$ is the blow-up of $\tilde X^\flat_1\otimes k'$ at its singular point
$y=0$, and each $E_r$ is a $\PP^1$-bundle over the Fermat curve $C\colon y_0^9+y_1^9+y_2^9=0$ in $\PP^2_{k'}$.  The
components form a chain $Z',E_1,\dots,E_{e-1},\Xi$, and each double curve maps isomorphically to $C$.
\item $H=\mu_9^3$ acts on $\mathcal X$ by $\OO_{K'}$-automorphisms, extending its action on the generic fiber and
mapping every component of $Y$ to itself.  The diagonal $\mu_9\subset H$ fixes every point of every double curve.
\item For each $\gamma\in\Gal(K'/F)$ there is an automorphism $\gamma_{\mathcal X}$ of the scheme $\mathcal X$ over
$\mathrm{Spec}(\gamma)$ which extends $1\times\mathrm{Spec}(\gamma)$ on $\tilde X_{s_0}\times_FK'$ and commutes with $H$.
If $\gamma$ lies in the inertia subgroup, then $\gamma_{\mathcal X}$ maps every component $Y_a$ of $Y$ to itself, and
its restriction to $Y_a$ is the restriction of an element $\delta_a$ of the diagonal $\mu_3\subset H$.
\end{enumerate}
\end{lemma}

\begin{proof}
\emph{The family over the $t$-line.}  Let $B=\mathrm{Spec}\,\OO_F[t,(1+t^9)^{-1}]$, mapped to $\bar\SSS$ by
$s=1+t^9$, and $\mathcal Y=\tilde X^\flat\times_{\bar\SSS}B$, projective over $B$, with $H$ acting over $B$.  By
\cref{prop:compactification}~(a), $\mathcal Y\to B$ is smooth outside the closed subset
$\Sigma=\{y=0,\ t=0\}$, a section of $\mathcal Y$ over $\OO_F$; so $\mathcal Y$ is smooth over $\OO_F$ outside
$\Sigma$.  Near $\Sigma$, $\mathcal Y$ is the hypersurface $\Phi=0$ in $\A^4_{\OO_F}$, with coordinates $y_0,y_1,y_2,t$
and
\[
\Phi=\textstyle\prod_j(1-y_j^9)-1-t^9=-(y_0^9+y_1^9+y_2^9+t^9)+\sum_{i<j}y_i^9y_j^9-y_0^9y_1^9y_2^9 .
\]
For $\zeta\in\mu_3$, $\phi_\zeta(x,t)=(x,\zeta t)$ is an automorphism of $\mathcal Y$ over the automorphism $t\mapsto\zeta t$ of $B$ (which preserves
$s=1+t^9$); it commutes with $H$ and fixes the fiber $t=0$ pointwise.

\emph{One blow-up.}  Let $\mathcal Y'$ be the blow-up of $\mathcal Y$ along $\Sigma$.  Near $\Sigma$ it is the strict
transform of $\mathcal Y$ in the blow-up of $\A^4_{\OO_F}$ along the zero section, and $H$ and $\phi_\zeta$ act on
it.  In the chart $y_1=y_0\eta_1$, $y_2=y_0\eta_2$, $t=y_0\tau$ we have $\Phi=y_0^9G$ with
\[
G=-(1+\eta_1^9+\eta_2^9+\tau^9)+y_0^9(\eta_1^9+\eta_2^9+\eta_1^9\eta_2^9)-y_0^{18}\eta_1^9\eta_2^9 ,
\]
so $\mathcal Y'=\{G=0\}$ there, and $\partial G/\partial\tau=-9\tau^8$,
$\partial G/\partial\eta_1=-9\eta_1^8(1-y_0^9)(1-y_0^9\eta_2^9)$,
$\partial G/\partial\eta_2=-9\eta_2^8(1-y_0^9)(1-y_0^9\eta_1^9)$.  On the exceptional divisor $y_0=0$ these do not
vanish simultaneously on $G=0$ (otherwise $G=-1$), and off it $\mathcal Y'=\mathcal Y\smallsetminus\Sigma$.  The
charts of $y_1$ and $y_2$ are symmetric, and in the chart $y_j=t\eta_j$ ($j=0,1,2$) we have $\Phi=t^9G_t$ with
$G_t\equiv-(1+\eta_0^9+\eta_1^9+\eta_2^9)\pmod t$, whose partial derivatives $-9\eta_j^8$ at $t=0$ do not all vanish
on $G_t=0$.  Hence $\mathcal Y'$ is smooth over $\OO_F$; its exceptional
divisor $\Xi$ is the Fermat surface $y_0^9+y_1^9+y_2^9+t^9=0$ in $\PP^3_{\OO_F}$, smooth over $\OO_F$ as $p\ne3$; and
the divisor of the function $t=y_0\tau$ near $t=0$ is $Z'+\Xi$, where $Z'$, the strict transform of the fiber
$\tilde X^\flat_1\otimes\OO_F$ of $\mathcal Y$ over $t=0$, is $\{\tau=0\}$ in the chart above.  Its fibers are
irreducible: over an algebraically closed field of characteristic $\ne3$ the surface $\prod_j(1-y_j^9)=1$ is a
hypersurface whose open part $y_0y_1y_2\ne0$ is an étale $\mu_9^3$-cover, $y_j^9=1-x_j$, of an open part of the torus
$x_0x_1x_2=1$, connected because the functions $1-x_j$ have the distinct prime divisors $\{x_j=1\}$ on the torus, so that they are
independent modulo ninth powers; and the open part is dense, the complement having dimension~$1$.  The intersection
$C=Z'\cap\Xi=\{y_0=\tau=0,\ 1+\eta_1^9+\eta_2^9=0\}$ is the Fermat curve $y_0^9+y_1^9+y_2^9=0$ in the plane $t=0$
of $\PP^3$, and along $C$ one of $\partial G/\partial\eta_1$, $\partial G/\partial\eta_2$ is a unit.  So near each
point of $C$ there are étale coordinates $(\xi,\tau,\eta)$ of $\mathcal Y'$ over $\OO_F$ (in the chart of a
coordinate $y_j$ that does not vanish at the point; we write it for $j=0$), with $\xi=y_0$ and
$\eta\in\{\eta_1,\eta_2\}$, in which $Z'=\{\tau=0\}$, $\Xi=\{\xi=0\}$ and $t=\xi\tau$; and $Z'$, $\Xi$ and $C$ are
smooth over $\OO_F$.  Off $C$, near $t=0$, the morphism $\mathcal Y'\to B$ is smooth (on $Z'\smallsetminus C$ because $\mathcal Y\to B$ is
smooth off $\Sigma$; on $\Xi\smallsetminus C$ because $\tau$ is a unit there, so that $t$ is a coordinate).  In these coordinates $\delta(\lambda)$ acts by
$(\xi,\tau,\eta)\mapsto(\lambda\xi,\lambda^{-1}\tau,\eta)$ and $\phi_\zeta$ by
$(\xi,\tau,\eta)\mapsto(\xi,\zeta\tau,\eta)$.  Globally, $\phi_\zeta$ is the identity on $Z'$ and acts on
$\Xi\subset\PP^3$ by $[y:t]\mapsto[y:\zeta t]=[\zeta^{-1}y:t]$, as $\delta(\zeta^{-1})$ does; and $\delta(\lambda)$
fixes $C$ pointwise, since it acts on the plane $t=0$ by $[y:0]\mapsto[\lambda y:0]$.

\emph{Base change and resolution.}  Since $t_0\in\mathfrak m_{K'}$ and $1+t_0^9=s_0$ is a unit, $t\mapsto t_0$ is an
$\OO_{K'}$-point of $B$.  Let $\mathcal X_0=\mathcal Y'\times_B\mathrm{Spec}\,\OO_{K'}$, projective over $\OO_{K'}$,
with generic fiber $\tilde X^\flat_{s_0}\otimes_FK'$ (the center $\Sigma$ lies over $t=0$).  Its special fiber is
$(Z'+\Xi)\otimes k'$.  Off $C$, $\mathcal X_0$ is smooth over $\OO_{K'}$; near a point of $C$ it is étale over
$\mathrm{Spec}\,\OO_{K'}[\xi,\tau,\eta]/(\xi\tau-\pi^eu)$ with $u\in\OO_{K'}^\times$, which is flat over $\OO_{K'}$.  So
$\mathcal X_0$ is flat.  If $e=1$ it is regular with reduced normal-crossings special fiber, and we put
$\mathcal X=\mathcal X_0$.  If $e\ge2$, blow up $e-1$ times the ideal of the strict transform of $\Xi\otimes k'$
(locally $(\xi,\pi)$ at the first step).  For the ring $\OO_{K'}[\xi,\tau]/(\xi\tau-\pi^eu)$ the blow-up of $(\xi,\pi)$
has two charts: $\xi=\pi\xi_1$, with equation $\xi_1\tau=\pi^{e-1}u$; and $\pi=\xi\pi_1$, with $\tau=\xi^{e-1}\pi_1^eu$,
which is regular with special fiber $\xi\pi_1=0$, reduced with normal crossings.  In the first chart the special fiber
$\xi_1\tau=0$ consists of the strict transform $\{\xi_1=0\}$ of $\Xi\otimes k'$ and of a new component $E_1=\{\tau=0\}$,
mapping to $C$ with fiber coordinate $\xi_1$; the strict transform of $Z'\otimes k'$ lies in the second chart.  By
induction the result $\mathcal X$ is regular, and its special fiber is $Y=Z'\cup E_1\cup\dots\cup E_{e-1}\cup\Xi$, where
$E_r$ is the $\PP^1$-bundle over $C$ that in the chart after $r$ blow-ups is $\{\tau=\pi=0\}$, with fiber coordinate
$\xi_r=\xi/\pi^r$; consecutive components meet transversally along sections isomorphic to $C$, and there are no triple
points.  Off $C$ the ideal blown up is invertible, so nothing changes there.  This proves (a) and (b).

\emph{The actions.}  $H$ acts on $\mathcal Y'$ over $B$, hence on $\mathcal X_0$ over $\OO_{K'}$; it preserves
$\Xi\otimes k'$, so it acts on $\mathcal X$, and it maps every component to itself.  Since $\delta(\lambda)$ fixes $C$
pointwise and commutes with the projections of the double curves to $C$, which are isomorphisms, the diagonal $\mu_9$
fixes every double curve pointwise.  This proves (c).  Let $\gamma\in\Gal(K'/F)$ and $\zeta=\gamma(t_0)/t_0\in\mu_3$.
The morphisms $\phi_\zeta\circ\mathrm{pr}_1\colon\mathcal X_0\to\mathcal Y'$ and
$\mathrm{Spec}(\gamma)\circ\mathrm{pr}_2\colon\mathcal X_0\to\mathrm{Spec}\,\OO_{K'}$ induce the same morphism to $B$
(both pull $t$ back to $\zeta t_0=\gamma(t_0)$), so they define an automorphism $\gamma_{\mathcal X_0}$ of
$\mathcal X_0$ over $\mathrm{Spec}(\gamma)$.  On the generic fiber it is $1\times\mathrm{Spec}(\gamma)$ on
$\tilde X^\flat_{s_0}\times_FK'$, because $\phi_\zeta$ does not move the point of $\tilde X^\flat$; it commutes with
$H$; and it preserves $\Xi\otimes k'$, so it lifts to an automorphism $\gamma_{\mathcal X}$ of $\mathcal X$.  Now let $\gamma$ be in the inertia subgroup.  Then $\gamma$ acts trivially on $k'$, and
$\gamma(\pi)/\pi\equiv\lambda\pmod\pi$ for some $\lambda\in\mu_3$ (the value of the tame character).  On the components
$Z'$ and $\Xi$ the claim is global: there $\gamma_{\mathcal X}$ restricts to the base change of $\phi_\zeta$ (as
$\gamma$ is trivial on $k'$), which is the identity on $Z'$ and equals $\delta(\zeta^{-1})$ on $\Xi$.  Let
$1\le r\le e-1$.  In the model obtained after $r$ blow-ups, let $U_r$ be the affine chart lying over the chart of
$y_0$ in which, at each of the $r$ steps, the ideal blown up is generated by $\pi$; so $y_0=\pi^r\xi_r$ on $U_r$, the
coordinate ring of $U_r$ is generated over $\OO_{K'}$ by $\xi_r$, $\tau$, $\eta_1$, $\eta_2$, and
$\xi_r\tau=\pi^{e-r}u$ there.  Let $U=U_r\smallsetminus\{\xi_r=0\}$.  The centers of the later blow-ups lie in
$\{\xi_r=0\}$, so $U$ is an open subscheme of $\mathcal X$; its coordinate ring is generated over $\OO_{K'}$ by
$\xi_r^{\pm1}$, $\tau$, $\eta_1$ and $\eta_2$; and $E_r\cap U=\{\tau=\pi=0\}$ is a nonempty, hence dense, open
subscheme of the irreducible scheme $E_r$, with coordinate ring generated over $k'$ by $\xi_r^{\pm1}$, $\eta_1$ and
$\eta_2$.  Both $\gamma_{\mathcal X}$ and every $\delta(\lambda')$ map $U$ to itself, since they multiply
$y_0,y_1,y_2,t,\pi$ and each $\xi_i=y_0/\pi^i$ by units.
Both pull-backs fix $\eta_1$ and $\eta_2$; pull-back by $\gamma_{\mathcal X}$ is $\xi_r\mapsto(\pi/\gamma(\pi))^r\xi_r$,
$\tau\mapsto\zeta\tau$, together with $\gamma$ on $\OO_{K'}$, and pull-back by $\delta(\lambda')$ is
$\xi_r\mapsto\lambda'\xi_r$, $\tau\mapsto\lambda'^{-1}\tau$, and the identity on $\OO_{K'}$.  On the coordinate ring of $E_r\cap U$ both are $k'$-algebra maps, and they agree when
$\lambda'=\lambda^{-r}$.  So $\gamma_{\mathcal X}$ and $\delta(\lambda^{-r})$ agree on the dense open subscheme
$E_r\cap U$ of the reduced scheme $E_r$, hence on $E_r$, as $\mathcal X$ is separated.  This proves (d).
\end{proof}

\begin{lemma}\label{lem:hk-model}
Let $\mathcal X$ and $Y$ be as in \cref{lem:ss-model}, and let
$D=H^2_{\mathrm{log\text{-}crys}}(Y/W(k'))\otimes\Q$ be the log-crystalline (Hyodo--Kato) cohomology of $Y$
(with the log structure induced by $\mathcal X$) over $W(k')$ (with the log structure $\mathbb N\to W(k')$, $1\mapsto0$).  It is a
$K'_0$-vector space with $\varphi$ and $N$.
\begin{enumerate}[(a)]
\item The restriction of $H^2(\tilde X_{s_0}\otimes\bar F,\Q_p)$ to $G_{K'}$ is semistable, and there is an
isomorphism $D_{\mathrm{st},K'}(H^2(\tilde X_{s_0}\otimes\bar F,\Q_p))\isom D$ compatible with $\varphi$, with $N$ up to a nonzero rational factor depending on the normalizations, with
the actions of $H$, and with the action of the inertia subgroup $I(K'/F)$, in which each $\gamma\in I(K'/F)$ acts on $D$ through the automorphism of the log
scheme $Y$ induced by $\gamma_{\mathcal X}$ (\cref{lem:ss-model}~(d)), lying over the Teichmüller lift to $W(k')$ of the automorphism that $\gamma$ induces on the log point $\mathrm{Spec}\,k'$ (with the log structure induced by $\OO_{K'}$).
\item There is a finite increasing filtration $P_\bullet$ of $D$, stable under $H$ and under $\gamma_{\mathcal X}$ for
$\gamma$ in the inertia subgroup $I(K'/F)$, such that $N(P_r)\subset P_{r-2}$ and $\mathrm{gr}^P_rD$ is, compatibly with
these actions, a subquotient of
\[
E_1^{-r,2+r}=\bigoplus_{j\ge\max(0,-r)}H^{2-r-2j}_{\mathrm{crys}}\big(Y^{(2j+r+1)}/W(k')\big)(-j-r)\otimes\Q ,
\]
where $Y^{(q)}$ is the disjoint union of the $q$-fold intersections of components of $Y$, and $H$ and
$\gamma_{\mathcal X}$, $\gamma\in I(K'/F)$, act on $E_1^{-r,2+r}$ by pull-back along their restrictions to the strata
$Y^{(q)}$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a)  Let $\tilde X_{\bar F}=\tilde X_{s_0}\otimes_F\bar F$, on which $G_F$ acts through the second
factor, and let $F^{\mathrm{nr}}$ be the maximal unramified extension of $F$ in $\bar F$ (Beilinson's $K_0^{\mathrm{nr}}$; his $K_0$ is our $F$).  We use, with base field $F$, the semistable comparison isomorphism constructed by Beilinson \cite{Beilinson13} (the comparison theorem is due to Tsuji \cite{Tsuji99}), which involves no choice of a
uniformizer: the base log structures are the canonical ones, generated by all nonzero elements of $\OO_{K'}$ and of $W(k')$
\cite[1.16]{Beilinson13}.  Beilinson attaches to $\tilde X_{\bar F}$ its Hyodo--Kato cohomology $H^2_{\mathrm{HK}}(\tilde X_{\bar F})$, an
$F^{\mathrm{nr}}$-vector space with $\varphi$ and $N$ on which $G_F$ acts by transport of structure
\cite[2.4, 2.5]{Beilinson13}, and constructs a canonical isomorphism
\[
\rho_{\mathrm{HK}}\colon H^2_{\mathrm{HK}}(\tilde X_{\bar F})\otimes_{F^{\mathrm{nr}}}B_{\mathrm{st}}
\xrightarrow{\ \sim\ }H^2(\tilde X_{\bar F},\Q_p)\otimes_{\Q_p}B_{\mathrm{st}},
\]
compatible with $\varphi$, $N$ and the action of $G_F$ \cite[(3.3.1)]{Beilinson13}; here $B_{\mathrm{st}}$ is Fontaine's
ring, with its $\varphi$ and $N$ \cite[1.17, last paragraph]{Beilinson13}.

The scheme $\mathcal X$, with the
log structure defined by its special fiber, is log smooth over $\OO_{K'}$, its special fiber is of Cartier type over the log point \cite[\S4.3]{Tsuji02}, its reduction modulo $p$ is of Cartier type by the same computation in the charts of \cref{lem:ss-model}, and
its log structure is trivial on the generic fiber.  So \cite[Proposition in 2.6]{Beilinson13} gives a canonical
isomorphism
\[
\iota\colon H^2_{\mathrm{HK}}(Y)\otimes_{K'_0}F^{\mathrm{nr}}\xrightarrow{\ \sim\ }H^2_{\mathrm{HK}}(\tilde X_{\bar F}),
\]
where $H^2_{\mathrm{HK}}(Y)$ is the Hyodo--Kato cohomology \cite[(1.16.1)]{Beilinson13} of the special fiber with its
log structure, and $G_{K'}$ acts on the source through the second factor \cite[3.3~(ii)]{Beilinson13}.  By
\cite[Remark~(i) in 1.16 and Remark~(iv) in 1.15]{Beilinson13}, $H^2_{\mathrm{HK}}(Y)$ is the log-crystalline
cohomology of $Y$ over $W(k')$ with the log structure obtained from that of the log point $\mathrm{Spec}\,k'$ through
Teichmüller representatives; choosing the class of a uniformizer as a generator identifies it with $D$, with its
$\varphi$, and with its $N$ up to the factor $e(K'/\Q_p)$ by which the two normalizations differ \cite[Remark~(ii) in 1.16 and Theorem-construction~(ii) in 1.15]{Beilinson13}.  We use only the vanishing of $N$ and the property $N(P_r)\subset P_{r-2}$ of~(b), which do not depend on this factor.  Taking $G_{K'}$-invariants of
$\rho_{\mathrm{HK}}\circ(\iota\otimes1)$ and using $B_{\mathrm{st}}^{G_{K'}}=K'_0$ gives
$D_{\mathrm{st},K'}(H^2)\isom D$, compatible with $\varphi$, and with $N$ up to a nonzero rational factor; as $\dim_{K'_0}D=\dim_{\Q_p}H^2$
\cite[Proposition in 2.5~(ii)]{Beilinson13}, $H^2$ is semistable over $K'$.

The Hyodo--Kato complex of the model is functorial for automorphisms of the log scheme $\mathcal X$ lying over automorphisms of $\mathrm{Spec}\,\OO_{K'}$ with its canonical log structure, by the compatibility with base change of the functor from $(\varphi,N)$-modules to $F$-crystals through which that complex is defined \cite[Theorem-construction~(ii) in 1.15, and (1.16.1)]{Beilinson13}; \cite[3.3~(ii)]{Beilinson13} is the case $\gamma=1$ of the naturality used below.  All these constructions are canonical, so
they are compatible with every automorphism of their input (transport of structure, which is also how the action of
$G_F$ is defined \cite[2.4]{Beilinson13}): with $H$, which acts on $\mathcal X$ over $\OO_{K'}$ and on $\tilde X_{\bar F}$; and,
for $g\in G_F$ with image $\gamma\in I(K'/F)$, with the automorphism $\gamma_{\mathcal X}\otimes g$ of
$\mathcal X\otimes_{\OO_{K'}}\OO_{\bar F}$, which is $g$ on $\tilde X_{\bar F}$ and lies over the automorphism $\gamma$ of
$\mathrm{Spec}\,\OO_{K'}$ with its canonical log structure.  Since $\gamma$ fixes $K'_0$, the element $g$ acts on
$H^2_{\mathrm{HK}}(Y)$ through the automorphism of the log scheme $Y$ induced by $\gamma_{\mathcal X}$, which lies over
the automorphism of the log point induced by $\gamma$ (and its Teichmüller lift to $W(k')$); under the identification with
$D$ this is the action of $\gamma_{\mathcal X}$ used in~(b).

(b)  This is the $p$-adic weight spectral sequence of Mokrane \cite[3.23]{Mokrane93}, in the corrected form of
Nakkajima \cite[(9.9.4), (9.10.2) and Definition~9.11~(2)]{Nakkajima05}.  It is the spectral sequence of the
$p$-adic Steenbrink complex $W\!A_Y$, the total complex of $W\!A_Y^{ij}=W\Lambda_Y^{i+j+1}/P_jW\Lambda_Y^{i+j+1}$
\cite[3.8]{Mokrane93}, \cite[proof of Theorem~4.1]{Nakkajima05}, filtered by the weight filtration $P$
\cite[(4.1.4)]{Nakkajima05}, whose graded pieces are identified with the de Rham--Witt complexes of the strata by
Poincaré residues \cite[(9.9.3)]{Nakkajima05}.  Its $E_1$-terms are as displayed and its abutment is $D$, so $\mathrm{gr}^P_rD=E_\infty^{-r,2+r}$ is a subquotient of $E_1^{-r,2+r}$.  Under this identification,
the monodromy operator $N$ of Hyodo and Kato is induced by the operator $\nu$
\cite[Proposition~11.10 and Corollary~11.5~(2)]{Nakkajima05}, which is defined by the projections
$W\!A_Y^{ij}\to W\!A_Y^{i-1,j+1}$ \cite[Definition~11.6~(2)]{Nakkajima05}; by the definition of the filtration
\cite[(4.1.4)]{Nakkajima05} it maps $P_r$ into $P_{r-2}$.

Every step of the construction is functorial for the automorphisms used here.  The complex $W\Lambda_Y$ above is
the complex $W\widetilde\Lambda_Y$ of \cite[proof of Theorem~4.1]{Nakkajima05}, which is presented, level by level, in terms of $W_n(\mathcal O_Y)$ and of
$M_Y^{\mathrm{gp}}$ modulo the units coming from $k'$, where $M_Y$ is the log structure of $Y$ \cite[(11.0.2) and
Theorem~11.1~(1)]{Nakkajima05}; so every automorphism $g$ of the log scheme $Y$ lying over an automorphism of the log
point acts on it and preserves its filtration $P$ \cite[3.5]{Mokrane93}, which counts logarithmic factors.  The section
$\theta$ is the $d\log$ of the image in $M_Y$ of a generator of the log structure of the log point, and it does not
depend on the choice of the generator \cite[the paragraph before Theorem~11.1, and Theorem~11.1~(1)]{Nakkajima05};
so $g^*\theta=\theta$, and $g$ acts on the filtered complex $(W\!A_Y,P)$ \cite[(2.2.1;~$\star$), (4.1.3), (4.1.4)]{Nakkajima05}, on the
spectral sequence and on $\nu$.  On the abutment this is the action of $g$ on the log-crystalline cohomology of $Y$
over the canonical lift of the log point \cite[beginning of \S7, (7.12.3) and Proposition~7.13]{Nakkajima05}, whose log structure comes
from the Teichmüller section, that is, the action on $D$ of~(a): the quasi-isomorphism $\theta\wedge$
\cite[(4.1.2)]{Nakkajima05} commutes with $g$, and so does the comparison isomorphism \cite[Theorem~7.19]{Nakkajima05},
by transport of structure: it does not depend on the embedding system used to construct it (loc.\ cit., proof), $g$
carries an embedding system to another one, and the maps into the log-crystalline cohomology sheaves of $Y$ from
which it is built are compatible with $g$ \cite[Lemma~9.1~(1)]{Nakkajima05}.  If $g$ maps every component of $Y$ to itself, it acts on the
$E_1$-terms by pull-back along its restrictions to the strata, since multiplying a local equation of a component by a
unit changes a logarithmic form only by a form of lower weight, which does not affect the Poincaré residue.  The
elements of $H$ (over the identity of the log point) and the automorphisms $\gamma_{\mathcal X}$, $\gamma\in I(K'/F)$
(over the automorphism of the log point that multiplies its generator by the unit $\gamma(\pi)/\pi$) are of this kind,
by \cref{lem:ss-model}~(c),~(d); this is the compatibility used in \cite[\S8, after (8.6)]{Saito09}.
\end{proof}

\begin{proof}[Proof of \cref{prop:crys}]
In both cases the Hodge--Tate weights are $\{0,1,2\}$, by \cref{lem:hodge} and the de Rham comparison theorem
\cite[Theorem~A1]{Tsuji02}.  The comparison theorem is functorial \cite[(A1.1)]{Tsuji02}, hence compatible with the
action of $H$, and \cref{lem:hodge} holds with $\chi^m$ replaced by any of the six characters $\chi^{m'}$,
$m'\in(\Z/9)^\times$, so the choice of embeddings of $E$ does not matter.  By
\cref{lem:dst-descent} for the unramified extension $F/K_v$, it suffices to show that $V=\VV_{s_0}$ is crystalline as
a representation of $G_F$.  By \cref{prop:compactification},
$V=(H^2(\tilde X_{s_0}\otimes\bar F,\Q_p)\otimes_{\Q_p}E)^{\chi^m}$, where $E\subset\overline\Q_p$ is a finite
extension of $\Q_p$ containing $\iota_p(\zeta_9)$.

\emph{The case $s_0\not\equiv0,1\pmod p$.}  The pull-back of $\tilde f$ along $s_0\in\SSS(\OO_{K_v})$ is a smooth
projective model of $\tilde X_{s_0}$ over $\OO_{K_v}$, so $H^2(\tilde X_{s_0}\otimes\bar F,\Q_p)$ is crystalline
\cite[Theorem~1.1 and the remarks before it]{Tsuji02} (in the good reduction case the monodromy operator vanishes), and
so is its direct summand $V$, cut out by the idempotent attached to $\chi^m$, which commutes with $G_{K_v}$.

\emph{The case $s_0-1=t_1^3$ with $v_p(t_1)>0$.}  Let $\mathcal X$, $Y$, $D$ and $P_\bullet$ be as in
\cref{lem:ss-model,lem:hk-model}, and $D_E=D\otimes_{\Q_p}E$.  Taking $\chi^m$-parts is exact and commutes with taking
$G_{K'}$-invariants, so by \cref{lem:hk-model}~(a) the restriction of $V$ to $G_{K'}$ is semistable and
$D_{\mathrm{st},K'}(V)\isom D_E^{\chi^m}$, compatibly with $N$ up to a nonzero factor and with the inertia subgroup $I(K'/F)$.  By \cref{lem:dst-descent} it
remains to show that $N$ and the inertia subgroup $I(K'/F)$ act trivially on $D_E^{\chi^m}$.

\emph{The $\chi^m$-part has a single weight.}  Since $Y$ has no triple points, the term $E_1^{-r,2+r}$ of
\cref{lem:hk-model}~(b) vanishes for $|r|\ge2$, and for $r=1$ and $r=-1$ it is $H^1_{\mathrm{crys}}(Y^{(2)}/W(k'))(-1)\otimes\Q$
and $H^1_{\mathrm{crys}}(Y^{(2)}/W(k'))\otimes\Q$.  The diagonal $\mu_9$ fixes every point of $Y^{(2)}$
(\cref{lem:ss-model}~(c)), so it acts trivially on these terms, while $\chi^m$ is nontrivial on it
(\cref{lem:characters}).  Hence $(\mathrm{gr}^P_rD_E)^{\chi^m}=0$ for $r\ne0$, that is, $(P_{-1}D_E)^{\chi^m}=0$ and
$(P_0D_E)^{\chi^m}=D_E^{\chi^m}$.

\emph{$N=0$.}  $N(D_E^{\chi^m})=N\big((P_0D_E)^{\chi^m}\big)\subset(P_{-2}D_E)^{\chi^m}=0$.

\emph{The inertial type is trivial.}  By the previous step $D_E^{\chi^m}=(\mathrm{gr}^P_0D_E)^{\chi^m}$, which by
\cref{lem:hk-model}~(b) is, compatibly with $I(K'/F)$, a subquotient of the $\chi^m$-part of
$E_1^{0,2}\otimes E=\bigoplus_aH^2_{\mathrm{crys}}(Y_a/W(k'))\otimes_{\Q_p}E$, the sum over the components $Y_a$ of $Y$
(the other summand, from $Y^{(3)}$, is zero).  Let $\gamma\in I(K'/F)$.  By \cref{lem:ss-model}~(d) the restriction of
$\gamma_{\mathcal X}$ to $Y_a$ equals that of an element $\delta_a$ of the diagonal $\mu_3$, so $\gamma$ acts on
$(H^2_{\mathrm{crys}}(Y_a/W(k'))\otimes E)^{\chi^m}$ as $\delta_a^*$ does, that is, by $\chi^m(\delta_a)=1$
(\cref{lem:characters}).  So $\gamma$ acts trivially on the $\chi^m$-part of $E_1^{0,2}\otimes E$, hence on its
subquotient $D_E^{\chi^m}$.  Thus $V$ is crystalline over $F$, and therefore over $K_v$.
\end{proof}

\begin{remark}\label{rem:normalisation}
When $p\mid a$ the choice $(u,w)=(b,a)$ puts the variable divisible by $p$ at the point $s=1$, where \cref{prop:crys} applies.  With $(u,w)=(a,b)$ the parameter would be close to $\infty$ at $p$.  The two choices of $(u,w)$ give the same local data at $\p$: since $3\mid c$, $a\equiv-b\pmod 9$, and $-1$ and $1+9\Z_3$ are cubes in $\Q_3$, so $\kappa_a=\kappa_b$ on $G_{K_\p}$.  Apart from the requirement $p\nmid u$, nothing below depends on which of $a,b$ is called $u$, and the parity of $u$ plays no role.
\end{remark}