Prove automorphism is trivial I would like to prove the following:

Let $L\subset L'$, where $L'$ is a quadratic extension of $L$, and $\rho\in\text{Aut}(L'/L)$, the automorphism group of $L'$ which fixes $L$. Also, let $\mathfrak{p}$ be a prime ideal of $L$ which ramifies in $L'$. Then $\rho$ acts trivially on the residue class field.

This is what I've tried:
Let $\mathfrak{p'}^2=\mathfrak{p}R_{L'}$, then an element in $L'/\mathfrak{p'}$ can be written as $a+\mathfrak{p'}$ where $a\in L'$. Then, $a$ can be written as $x+y\lambda,\;x,y\in L,\;\lambda$ the basis element extension of $L$ to $L'$. So $\rho$ sends $\lambda$ to its conjugate. Then we have
\begin{align*}
\rho(x+y\lambda+\mathfrak{p'})&=x+y\overline{\lambda}+\rho(\mathfrak{p'})\\
&=x+y\lambda+\rho(\mathfrak{p'})+y\overline{\lambda}-y\lambda
\end{align*}
So I need to show that $\rho(\mathfrak{p'})+y\overline{\lambda}-y\lambda\in\mathfrak{p'}$, but I'm not too sure how. 
Thanks for the help in advance.
 A: We have $\mathfrak p'^2=\mathfrak p$, so $\rho(\mathfrak p')^2=\mathfrak p$, hence $\rho(\mathfrak p')=\mathfrak p'$. So it remains to check that $y(\lambda-\overline\lambda)\in\mathfrak p' \quad\forall y\in L$.
Now $L'/\mathfrak p'\equiv L/\mathfrak p$, because the inertia degree is $1$. So $\rho\mid_{L'/\mathfrak p'}=\iota$, where $\iota$ is the identity on the residue field. And $\rho\mid_{L'/\mathfrak p'}$ is the induced map on the residue field.
That is, $\rho(\lambda)$ and $\lambda$ lie in the same coset of $L'$ modulo $\mathfrak p'$, hence our claim.
Hope this helps.
Edit
Sorry. I used the conclusion to show your statement. But our conclusion is easy to prove, and you need not make it so complex in fact. I am referring to the fact that inertia degree being $1$ implies the triviality of $\rho$ acting on the residue field. This is what I meant, rather that $\rho$ restricted is the inclusion. Sorry again for the mis-understanding.  

Suppose $L'/\mathfrak p'\equiv (L/\mathfrak p)^f$, then $L'/\mathfrak p'\oplus \mathfrak p'/\mathfrak p'^2\equiv L'/\mathfrak p$ is of dimension $2f$ over $L/\mathfrak p$, but that dimension is also $[L'/\mathfrak p:L/\mathfrak p]=2$. Hence $f=1$, and the extension is trivial, thereby proving your claim.  

