A basic result in algebraic number theory is that every proper finite extension of $\mathbb Q$ is ramified over $\mathbb Q$. This is proved by considering Minkowski's lower bound on the discriminant.
I read somewhere that there is no known proof of this result which is purely algebraic. However, can we argue like this?
Let $A$ be a ring, and let $B$ be an algebra over $A$ which is finite and projective as an $A$-module. Then the trace map $\operatorname{Tr}: B \rightarrow A$ can be defined. We say that $B$ is separable over $A$ if $b \mapsto \phi_b, \phi_b(b') = \operatorname{Tr}(bb')$ defines an isomorphism of $A$-modules $B \rightarrow \operatorname{Hom}_A(B,A)$. Let's say that $B$ is etale over $A$ if it is finite, projective, and separable over $A$.
Then the result that every proper finite extension of $\mathbb Q$ is ramified follows from these two results:
(i): Let $A$ be a Dedekind domain with quotient field $K$, and let $B$ be the integral closure of $A$ in a finite separable extension $L$ of $K$. If no prime of $A$ ramifies in $B$, then $B$ is etale as an $A$-algebra.
(ii): Every etale ring extension of $\mathbb Z$ is of the form $\mathbb Z^n$.
As far as I can tell, these two results are not hard. The first follows from the answer to my question here, and I'm currently working out the second.
Does this argument give a simple proof for why there are no unramified extensions of $\mathbb Q$? It seems a little too easy.