TITLE: If $R$ is an etale extension of $\mathbb Z$, then $R = \mathbb Z^n$?
QUESTION [10 upvotes]: Related question.
Let $A$ be a ring, and let $B$ be an $A$-algebra which is projective and finite as an $A$-module.  Then the trace $B \rightarrow A$ can be defined.  Let's say that $B$ is separable over $A$ if the trace induces an isomorphism of $A$-modules $B \rightarrow \operatorname{Hom}_A(B,A)$.  For the purposes of this question, let's say that $B$ is an etale $A$-algebra if it is finite, projective, and separable.  
Is it true that every etale $\mathbb Z$-algebra is of the form $\mathbb Z^n$?  This is stated in these notes, example 1.1.13(ii).  
It does not seem so easy to prove.  If this is true, then combined with the much easier result, "if $L/K$ is an unramified extension of number fields, then $\mathcal O_L$ is an etale $\mathcal O_K$-algebra," we would obtain an algebraic proof that there are no nontrivial unramified extensions of $\mathbb Q$.  I had thought that the only known proofs of this result relied on Minkowski's bound for the discriminant.  Perhaps any proof that etale $\mathbb Z$-algebras are of the form $\mathbb Z^n$ must rely on this fact.

REPLY [4 votes]: Let $X$ be a normal integral scheme. Then $\pi_1^\mathrm{et}(X)$ is isomorphic to $\mathrm{Gal}(K(X)^\mathrm{un}/K(X))$, where $K(X)^\mathrm{un}$ is the compositum of all finite extensions $F$ of $K(X)$ (in a fixed algebraic closure of $K(X)$) for which $X$ is unramified in $F$. Applying this to $X = \mathrm{Spec}(\mathbf{Z})$, it follows the result you are asking about is equivalent to Minkowski's theorem (that there are no nontrivial unramified extensions of $\mathbf{Q}$).