TITLE: Proving a morphism is étale
QUESTION [7 upvotes]: I'd like some help to prove that a morphism of schemes $f:X\to Y$ if étale. 
Here are the characters: $X=\textrm{Spec}\,k[x,x^{-1}]$, $Y=\textrm{Spec}\,k[t]$ and $f$ is induced by $t\mapsto x^2$. [We may assume $k=\overline k$]
I tried to show $f$ is étale in two different manners.

By showing that it is flat and unramified. It is certainly flat, because it is dominant over a nonsingular curve (and it is dominant because $f^\sharp$ is injective).
I am not able to show it is unramified, because I cannot figure this morphism concretely.
What I can see is just that under $f^\sharp$ a polynomial $p(t)=\sum_{i\geq 0}a_it^i$ goes to $a_0+a_1x^2+a_2t^4+\dots$. Can you please help to me to actually see in the most concrete way what is the image of a point $x\in X$ under $f$?
By showing that $f$ is smooth of relative dimension $0$. It is certainly flat of relative dimension $0$, but I still need to show that the geometric fibers of $f$ are nonsingular and pure of dimension $0$. So for every closed point $y\in Y$ we have $X_y=\textrm{Spec}\,(k[x,x^{-1}]\otimes_{k[t]}k)=X_{\overline y}$ and over the generic fiber
$$
X_{\overline{\eta}}=\textrm{Spec}\,k[x,x^{-1}]\otimes_{k(t)}\overline{k(t)}.
$$
Can you help me understand what "is" this $X_{\overline y}$?

Thank you all.

REPLY [6 votes]: 0  You must assume $char.k\neq 2$, else the morphism $f$ is not étale at all.
1 The image of the generic point $(0)\in \textrm{Spec}\,k[x,x^{-1}]$ of $X$ is the generic point $(0)\in \textrm{Spec}\,k[t]$ of $Y$.
The other points of $X$ are closed and the image of such a closed point  point $(x-a) \in \textrm{Spec}\,k[x,x^{-1}] \:(a\neq 0)$ of $X$ is the closed  point $(t-a^2)\in \textrm{Spec}\,k[t]$ of $Y$.
2 The generic fiber of $f$ is the fiber of the generic point $\eta=(0)=\textrm{Spec}\,k[t]$.
That fiber  consists of the generic point $\xi=(0)\in\ \textrm{Spec}\,k[x,x^{-1}]$.
At the generic point $\eta$  of $Y$ the dual morphism of local rings is the field extension $f^{\star}_\eta:\mathcal O_{Y,\eta}=k(t)\hookrightarrow k(x)=\mathcal O_{X,\xi}:t\mapsto x^2$.
This field extension is (as it should) étale of degree 2 or, in a more  traditional terminology , separable of degree 2.