Why is $\mathbb{A}^{1}_{\mathbb{C}} \to \mathbb{A}^{1}_{\mathbb{R}}$ etale? Consider the morphism $\mathbb{A}^{1}_{\mathbb{C}} \to \mathbb{A}^{1}_{\mathbb{R}}$ induced by the inclusion $\mathbb{R}[T] \to \mathbb{C}[T]$. I'm having trouble proving that this is etale.
The definition of etale that I'm working with is flat and unramified. Really, I would like input on why for each $x \in \mathbb{A}^{1}_{\mathbb{C}}$, we have
$$m_{f(x)}\mathcal{O}_{\mathbb{A}^{1}_{\mathbb{C}}, x} = m_{x}.$$
Every element in $\mathbb{A}^{1}_{\mathbb{C}}$ is either the zero ideal (and then $\mathcal{O}_{\mathbb{A}^{1}_{\mathbb{R}}, (0)} = \mathbb{R}(T)$, in which case the desired conclusion is obvious) or it's of the form $(T-z)$. So the stalks at these points are the localizations $\mathcal{O}_{\mathbb{A}^{1}_{\mathbb{C}}, (T-z)} = \mathbb{C}[T]_{(T-z)}$, and the maximal ideals are $(T-z)\mathcal{O}_{\mathbb{A}^{1}_{\mathbb{C}}, (T-z)}$, but is there a nice description of this localization (or do we even need one)? And how do we obtain the above relation? In particular, what is the form of f((T-z))?
Finally, we know what the elements (and thus of the maximal ideals in the stalks) look like in $\mathbb{A}^{1}_{\mathbb{R}}$. They are either 0, of the form $(T-a)$ of of the form $(aT^2 + cT + d)$ for some irreducible degree 2 polynomial. Obviously, for this to be true, $f((T-z))$ must be degree 1, but I don't see how to show this.
 A: This is easy once you wrote down what $f((T-z))$ is.  By definition, it is the inverse image of $(T-z)$ under the inclusion map $\mathbb{R}[T]\to\mathbb{C}[T]$, also known as the intersection of $(T-z)$ and $\mathbb{R}[T]$.  That is, it is the set of polynomials with real coefficients that have $z$ as a root.  If $z$ is real, this is just the ideal of $\mathbb{R}[T]$ generated by $T-z$.  If $z$ is not real, then any real polynomial with $z$ as a root also has $\bar{z}$ as a root, so $(T-z)\cap\mathbb{R}[T]$ is the ideal of $\mathbb{R}[T]$ generated by $(T-z)(T-\bar{z})$, which has real coefficients and is irreducible in $\mathbb{R}[T]$.
I encourage you to now try to verify $m_{f(x)}\mathcal{O}_{\mathbb{A}^1,x}=m_x$ in each case on your own.  A proof is hidden below.

 In the first case, $m_{f(x)}\mathcal{O}_{\mathbb{A}^1,x}=m_x$ is obvious since both sides are the ideal in $\mathcal{O}_{\mathbb{A}^1,x}$ generated by $T-z$.  In the second case, the left side is the ideal generated by $(T-z)(T-\bar{z})$ and the right side is the ideal generated by $T-z$.  However, these ideals are the same, since $T-\bar{z}$ is a unit in $\mathcal{O}_{\mathbb{A}^1,x}$ (it is not in $(T-z)$ so it is inverted when we localize).

