Referring to the post what is the simplest example of an etale cover which is not Galois?
where the morphism $$f:Y=\mathbb A_{\mathbb Q(\sqrt [3]2)}^1 \to X=\mathbb A_{\mathbb Q}^1$$ corresponding to the inclusion $\mathbb Q[T] \hookrightarrow \mathbb Q(\sqrt [3]2)[T]$ is etale but not Galois.
$\textbf{Edit:}$ Thanks to @reuns comment -- it is clear why the cover is not Galois ( the extension $\mathbb Q(\sqrt[3]{2})$ over $\mathbb Q $ is not Galois).
My question now is about the fibers of the constant field extension
$$g:Z=\mathbb A_{\mathbb Q(\sqrt2)}^1 \to X=\mathbb A_{\mathbb Q}^1$$ which is known to be Etale and Galois.
Here we have the preimage of $t^2-2$ is $t-\sqrt 2$ and $t +\sqrt2$ (two points) the degree of the image is two and the degree of the preimage is also two.
But what about the preimage of $t-1$? It is again $t-1$ and now we have only one point and the degrees of both are one.
If I understand correctly, despite the fact that cardinality of the preimages varies (in this case it is one or two) the extension is still etale? Also, what is the degree of this extension? (For the extension $\mathbb Q(\sqrt{t} )$ over $\mathbb Q(t) $ I can see the degree is two). And if the degree is two aren't we supposed to have two points in the preimage?