Intersection of ramified Galois extension and maximal unramified abelian extension

Question

Let $L/K$ be a finite Galois extension where a prime $p$ is completely ramified, and $A$ the maximal unramified abelian extension of $K$.

It should be $L \cap A =K$

By I don't' quite see why.

Clearly $K\subseteq L\cap A$, and also $L\not\subseteq A$ and $A\not\subseteq L$.

But I don't see why can't be some "unramified part" in $L$, that would belong to $A$, making the intersection bigger than $K$.

I'm sure this is just a simple exercise in field theory, but I just don't see it.

Thanks in advance for any hint or suggestion.

Answer 1

$p$ is totally ramified, which means that $e_p(L|K) = [L:K]$. But ramification degrees are multiplicative as are field extension degrees. If there is a proper extension $K\subseteq F\subseteq L$ such that $e_p(F|K) = 1$, then we get the contradiction

$$e_p(L|K) = e_p(L|F)e_p(F|K) \le [L:F] < [L:F][F:K] = [L:K] = e_p(L|K).$$