Statement about Weil representations being unramified over a certain extension

Question

Let $K = \mathbb{Q}_p$ and $\rho: W_K \to \operatorname{GL}(V)$ be a Weil representation over a finite-dimensional complex vector space $V$ (where $W_K$ denotes the Weil group over $K$).

Let $L/K$ be a finite cyclic extension of degree $n$ and ramification index $e$ such that the restriction $\rho|_L: G_L \to \operatorname{GL}(V)$ is unramified (equivalently, $\rho(I_L)=1$ where $I_L$ is the inertia group over $L$).

Let $L'/K$ be a totally ramified of degree $e$ such that $LL'/L'$ is unramified of degree $n$.

Question: Is $\rho|_{L'}$ unramified, i.e. $\rho(I_{L'})=1$?

I am certain that if this is true, the only chance to show it is by using the fact that $LL'/L'$ is unramified. But I am not sure how this should work. Could you please help me with this problem?

Answer 1

This is true, because $I_{L'} \subset I_{L}$. Hence, if $\rho(I_L) = 1$, then $\rho(I_{L'}) = 1$.

Indeed, let $L^{\mathrm{ur}}$ be the maximal unramified extension of $L$ and let $(L')^{\mathrm{ur}}$ be the maximal unramified extension of $L'$. Because $LL'/L'$ is unramified, $(L')^{\mathrm{ur}}$ contains $L$ and hence $L^{\mathrm{ur}}\subset (L')^{\mathrm{ur}}$. But $$\begin{align*}I_{L'} = \mathrm{Gal}(\overline K/(L')^{\mathrm{ur}})\subset \mathrm{Gal}(\overline K/L^{\mathrm{ur}}) = I_L.\end{align*}$$

Answer 2

I'd say use that $$\overline{K}^{\displaystyle\ker(\rho|_{L'})}(\zeta_{p^\infty-1})=\overline{K}^{\displaystyle\ker(\rho|_{L'})\cap G_{L'(\zeta_{p^\infty-1})}}=\overline{K}^{\displaystyle\ker(\rho|_{L'(\zeta_{p^\infty-1})})}$$ $$=\overline{K}^{\displaystyle\ker(\rho|_{L(\zeta_{p^\infty-1})})}=L(\zeta_{p^\infty-1})=L'(\zeta_{p^\infty-1})$$