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Jun 14 at 20:50 comment added Chris Wuthrich Sure, but if $p$ divides $v(q)$, we could have that $q$ is a $p$-th power in $K_v$ in which case $K_v=L_w$ is a trivial extension. The important part is that the extension is non-trivial of degree $p$.
Jun 14 at 19:32 comment added user267839 just to check if I understood the logic of the argumentation (even though this might be a triviality): Isn't $L_w$ by construction already $K_v\bigl(\sqrt[p]{q}\bigr)$ independently of additional condition $p >v(q)$? Or does the latter condition only serve to assure that that the extension is non trivial?
Jun 14 at 19:06 history bounty ended user267839
Jun 14 at 19:06 vote accept user267839
Jun 13 at 20:26 history answered Chris Wuthrich CC BY-SA 4.0