Shouldn't $t^n : \mathbb{A}^1 \rightarrow \mathbb{A}^1$ ramifies at $0$? Yo, this is probably the stupidest question ever that I've asked here. 
Let $$\varphi: \mathbb{A}^1 \rightarrow \mathbb{A}^1$$ be the map of schemes (over a field $k$) such that $\varphi (x) = x^n$. In other words, $$\varphi^{\#}: k[t] \rightarrow k[t]$$ satisfies $\varphi^{\#} (t) = t^n$. Intuitively this should be the most stupid map that is unramified everywhere except at $0$. Let $(t)$ be the point zero. Localization at $(t)$ inverts every polynomial with an invertible constant term and quotienting by $k[t]_{(t)}(t)$ kills every higher term, remaining just the constant term. So at the level of residue fields $$\varphi^{\#}_{(t)}: k \rightarrow k$$ is the identity, which is certainly separable.
More generally, apparently, if $\varphi: X \rightarrow X$ is a map of $S$-schemes such that it fixes a point $x \in X$, then pointwise $\varphi_x : \kappa (x) \rightarrow \kappa (x)$ is always separable.
Certainly I'm committing a terribly stupid mistake. What's wrong?
Thanks in advance.
 A: Let's run through this situation according to the map on local rings and we'll see where things differ from your argument.
Name the first $\mathbb{A}^1$ $X$ and the second one $Y$, so that our map is $\varphi:X\to Y$. The corresponding map on local rings is $\varphi^\#: \mathcal{O}_{Y,\varphi(x)}\to \mathcal{O}_{X,x}$. Let $\mathfrak{m}\subset \mathcal{O}_{Y,\varphi(x)}$ be the maximal ideal, and let $\mathfrak{n}=\varphi^\#(\mathfrak{m})\mathcal{O}_{X,x}$, the ideal generated by the image of of $\mathfrak{m}$ in $\mathcal{O}_{X,x}$.
The map is unramified if it's locally of finite type and if for all $y\in Y$, $\mathfrak{n}$ is the maximal ideal of $\mathcal{O}_{X,x}$ and the obvious map $\mathcal{O}_{Y,\varphi(x)}/\mathfrak{m} \to \mathcal{O}_{X,x}/\mathfrak{n}$ is a finite separable field extension.
The map is clearly locally of finite type. 1/3 so far.
Let's check whether $\mathfrak{n}$ is the maximal ideal: $\mathfrak{m}$ is principal and generated by $t$, so $\varphi^\#(\mathfrak{m})\mathcal{O}_{X,x}=\varphi^\#((t))k[t]_{(t)}=\varphi^\#(t)k[t]_{(t)}=(t^n)k[t]_{(t)}=(t^n)$ which is not maximal. Failure!
Your argument would be fine if you had some sort of guarantee that $\mathfrak{n}$ was maximal, but you do not. Your computation only holds for the case that $\mathfrak{n}$ is the maximal ideal of $\mathcal{O}_{X,x}$.
A: Let $f:X\rightarrow Y$ be a morphism locally of finite type. Then we say that $f$ is unramified at $x$ iff $$\mathcal{O}_{X,x}/\mathfrak{m}_{f(x)}\mathcal{O}_{X,x}$$
is a finite separable extension of $k(f(x))$. 
So your argument is wrong, due to incorrect definition, what was already pointed out in comments. Assuming your "definition" of unramified morphism you can for example "prove" that any quasi-finite morphism of schemes which are locally of finite type over a field of characteristic zero is unramified, because for such morphisms:
$$k(f(x)))\rightarrow k(x)$$ 
is always finite and separable(characteristic zero). 
Consider your map:
$$\phi:k[t]\rightarrow k[x]$$
given by $$\phi(t)=x^n$$
Then morphism induced on local rings:
$$k[t]_{(t)}\rightarrow k[x]_{(x)}$$
sends:
$$\frac{t}{1}\mapsto \frac{x^n}{1}$$
So in this case you derive:
$$k[x]_{(x)}/x^nk[x]_{(x)}\cong k[x]/(x^n)$$
which is not even a field.
