Proving a morphism is étale 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.
 A: 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. 
A: Showing that it is unramified seems easier to me. Indeed, we have to show that for any $P\in X$ and $Q=f(P)$, you have that $f^\sharp(\mathfrak m_Q)\cdot\mathcal O_{X,P}=\mathfrak m_P$. Easily enough, $Y$ is just the line, so it is one-dimensional and smooth, and we know that $\mathfrak m_Q=(\pi)$ is a principal ideal, generated by a uniformizing parameter of the form $\pi=t-a$. Then, $f^\sharp(\pi)=x^2-a$. Since we are over an algebraically closed field, $f^\sharp(\pi)=(x-a_1)(x-a_2)$ for certain $a_1,a_2\in k$. Note that $a_1=a_2$ if and only if $a=0$ (this is where I assume sheepishly that the characteristic is not 2). We know for sure that  $f^\sharp(\mathfrak m_Q)\subseteq\mathfrak m_P$ and $\mathfrak m_P=(x-b)$ for some $b\ne 0$, so we have $(x-a_1)(x-a_2)=\rho\cdot(x-b)$ for some $\rho\in\mathcal O_{X,P}$. Since $\mathcal O_{X,P}$ is the localization of the UFD $k[x]$, it is again a UFD. Since $x-a_i$ is either irreducible or a unit and $x-b$ is irreducible, we may assume without loss of generality that $b=a_1$ and $\mathfrak m_P=(x-a_1)$. Note that this already implies $a\ne 0$. Since $x-a_2$ is a unit in $\mathcal O_{X,P}$, we are done.
