Can a $p$-adic field admit a different valuation? Let $L/\mathbb{Q}_p$ be a finite extension. 
Question: Is it possible for $L$ to admit a henselian valuation with residue characteristic $q \not=p$? 
I would think surely not, but I can't see a quick reason why. It seems plausible that such a valuation would have to be independent of the $p$-adic valuation on $L$, giving two independent henselian valuations on $L$, which is impossible by a theorem of Schmidt. Again though I can't see a quick reason why they must be independent. 
 A: Thought about it more and came up with a proof. Posting it here in case anyone's interested (or in case I made a mistake!) Maybe there's a quicker way still though.
Let $\mathcal{O}_p$ denote the usual $p$-adic valuation ring on $L$ and suppose $\mathcal{O}_q$ is another henselian valuation ring with residue field of characteristic $q \not=p$. We know of course that the residue field wrt $\mathcal{O}_p$ is not separably closed. But in fact this must be the case also for $\mathcal{O}_q$. 
Indeed, note that $p \in \mathcal{O}_q^{\times}$ (otherwise $p$ would become zero in the residue field which has characteristic $q$). Consider, for any positive integer $n$, the polynomial $x^{p^n}-p$. Then if the residue field of $\mathcal{O}_q$ were separably closed, it would contain a root to this polynomial, which could then be lifted by Hensel's lemma (the root is non-zero since $p$ is a unit, and hence is simple since $p^n \not=0$ in the residue field). That is, $L$ contains $p^{1/p^n}$ for every $n$, contradicting the fact that $L$ is a finite extension.
But now one knows that henselian valuations with non-separably closed residue field are always comparable (cf. eg. Engler-Prestel page 105). However, this cannot be, since $1/p \in \mathcal{O}_q$ and $1/q \in \mathcal{O}_p$, so these rings are non-comparable: contradiction.
