Checking if this is an etale morphism Consider the map $\mathbb{A}(k)$ to $\mathbb{A}(k)$ where $t$ goes to $t^2$.
This induces a map $\phi$ from $k[T]$ to $k[T]$ where $T$ goes $T^2$. we know that is map is not etale at $0$. I want to check that explicitly. Now, morphism $\phi$ is a flat morphism. I want to show that is ramified at $0$ which corresponds to the ideal $T$. So, $\phi$ induces a map from $k[T]/[T]$ to $k[T]/[T]$ where $T$ goes to $T^2$. The map indiced by $phi$ on $k[T]/[T]$ seems to be the identity map. So, it is unramified at $0$, which is not the case. I'm consumed where exactly I am making the mistake. 
 A: Let's make explicit our definition of ramification. Here's all the notation I'm going to need to make the definition a 1-liner: Let $\varphi: X\to Y$ be a map. 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 $\mathcal{O}_{X,x}$ where $\varphi^\#$ is the induced map on local rings, $\varphi^\#: \mathcal{O}_{Y,\varphi(x)}\to \mathcal{O}_{X,x}$.  
Definition: A map $\varphi: X\to Y$ is unramified if it's locally of finite type and for each $y\in Y$, $\mathfrak{n}$ 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. A map is ramified if it's not unramified.
Unsurprisingly, locally of finite type is clear in your context.
The real work starts at verifying  whether $\mathfrak{n}$ is the maximal ideal at $0$: $\mathfrak{m}$ 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. So your map is ramified at 0 just like you suspected.
