I'm looking for a brief proof of the following statement:
Let $K$ be a number field and $p$ a prime number. Then there exists a field extension $L\supset K$ of degree $p$ with $\mu(L)=\mu(K)$.
My current approach is to first show that there exist infinitely many degree $p$ field extensions of $K$ that are pairwise nonisomorphic (over $K$), and then that there are only finitely many degree $p$ field extensions $L\supset K$ with $\mu(L)\neq\mu(K)$, up to isomorphism. For this I invoke quite a few small results from algebraic number theory, and the result is that there are infinitely many such field extensions, though I only need one. So my question is;
Is there a simple, uninvolved proof of the statement above?
Choose a prime number $q \equiv 1 \bmod 4p$ not dividing the discriminant of $K$, and let $F$ be the subfield of degree $p$ inside the field of $q$-th roots of unity. Observe that $F$ is real. The extension $KF/K$ is ramified exactly at $q$. The field $L = KF$ does not contain the $q$-th roots of unity since $L \cap {\mathbb Q}(\zeta_q) = F$. If $L$ contains some $n$-th root of unity with $n$ coprime to $q$, then $L$ must contain $K(\zeta_n)$, which is unramified at $q$. But $L/K$ is completely ramified at the primes above $q$: contradiction.