# Finiteness of the degree of residue fields of valuation rings

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.

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 $$[D^*:D] \leq [L:K]$$.