Assume the standard context for extension of valuations.
An extension $w$ of a valuation $v$ induced from a prime ideal $\mathfrak{p}$ comes from a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$?
Is tris true? I strongly believe so, but I have no idea how to prove it.
I'll assume we're working with number fields. Let $L$ be a finite extension of $K$, and $w$ an absolute value on $L$ which, when restricted to $K$, is equivalent to an absolute value $||_P$ obtained from some prime $P$ of $K$. Let $A, B$ be the rings of integers of $K, L$. We want to find a prime $Q$ of $L$ which lies over $K$ which induces an absolute value equivalent to $w$.
Let $O = \{ x \in L : |x|_w = |x| \leq 1\}$ have unique maximal ideal $\mathfrak M = \{ x \in L : |x| < 1 \}$. We want to find a prime $Q$ of $L$ for which $O = B_Q$ and $\mathfrak M = Q B_Q$ (then we're pretty much done). Now $$O \cap K = \{ x \in K : |x| = |x|_P \leq 1 \} = A_P$$ $$ \mathfrak M \cap K = \{ x \in K : |x|_P < 1 \} = PA_P$$ Now the elements of $B$, which are integral over $A$, can also be seen to have absolute value $\leq 1$. Therefore $B \subseteq O$. We let $Q = \mathfrak M \cap B$. It follows that $Q$ lies over $P$, since $Q \cap A = PA_P \cap A = P$. Now you want to argue that $$O = B_Q$$ and $$\mathfrak M = QB_Q$$ but this isn't too hard. Let me know if you run into any difficulties.