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Question: For any valuation $\lambda$, I will use $R_{\lambda}$ to denote the valuation ring, $m_{\lambda}$ to denote its unique maximal ideal, and $D_{\lambda}:=R_{\lambda}/m_{\lambda}$ the residue field. Suppose that $L$ is a finite dimensional (f.d.) extension of $K$. Let $v$ be a valuation of $K$ and $w$ an $L$-extension of $v$. Let $D:= D_v$, and $D^*:= D_w$. Must we have that $D^*$ is a f.d. extension of $D$?

My progress so far: For the record, this is true in the case that $D^*$ is separable over $D$ (the proof uses the primitive element theorem and the fact that $[L:K]$ is a uniform bound of the degrees of the f.d. subextensions of $D^*$ over $D$). Hence, the above question holds if we can just prove that the result holds in the case that $D^*$ is purely inseparable over $D$, so I will assume this below:

It can be shown that the degrees of f.d. $\textbf{primitive}$ subextensions of $D^*$ over $D$ have a uniform bound (namely $[L:K]$), but I have spent a lot of time struggling to progress this more with no luck. Any ideas would be appreciated.

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Ok well I think this works (Thanks to Wojowu for giving me the main ideas for this):

Suppose that $a_1,...,a_n\in L$ are linearly dependent over $k$, say $\sum c_ia_i = 0$ with $c_i\in k$. WLOG we can suppose that $c_1$ has minimal value among the $c_i$. Let $d_i:= c_i/c_1$ $\forall i$. So $\sum d_i a_i = 0$ $(\dagger)$, while $d_i\in R^*:= R_{v^*}$ $\forall i$ (by construction) and at least one $d_i = 1$ (and hence does not lie in $m^*:= m_{v^*}$). Reducing $(\dagger)$ modulo $m^*$ gives a dependence relation among the $a_i+m^*$. So the $a_i+m^*$ are linearly dependent over $D$ and hence we are done. In fact, we have proven that $[D^*:D] \leq [L:K]$.

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