Plot showing smoothness of polynomial over complex numbers I would like to get the "picture" - honestly, if it could literally be a graph, it would be great - behind an explanation in Quora on the algebraic closure of the complex numbers by Alon Amit:

A polynomial from the complex numbers to themselves is therefore a smooth map from the real plane to itself. And it's a very nice smooth map: it only has finitely many critical points, which are the points where the derivative vanishes. The derivative of a polynomial is a polynomial, and a polynomial cannot have infinitely many roots. That's a simple algebraic fact.
So we have a smooth map from the plane to itself with only finitely many points of criticality. To every point in the range of this map we can attach a natural number which counts how many preimages it has: for a given $y,$ how many $x$'s are there with $f(x)=y$? Think of this as a kind of coloring: the white points have no solutions at all, the blue points have one, the red have two solutions and so on.
Now the plane is splashed with colors, every single point. The thing is, the color puddles are actually nice-looking: they are open sets. Whenever a point has exactly 7 preimages, so do all of the points in a small circle around it. This requires a bit of technicality to show (it's easier to compactify the plane into a sphere), but intuitively it should be quite clear. Think of the graph of a smooth, real function: if a line cuts it at 7 points, so do all adjacent lines as long as you're away from local maxima/minima.
But a connected set cannot be partitioned into open sets, and the plane minus the critical values is connected. So in fact we just have one color, and that color cannot be white (it's not possible for the polynomial to miss every value), so it must be some other color, and so with finitely many exceptions (points where the derivative is 0), every value is obtained the same nonzero number of times. In particular, the equation $p(z)=0$ has a solution.

This passage makes reference to John Milnor's proof of the Fundamental Theorem of Algebra.
I have a sketchy idea of the concept of a polynomial ring over a field, but not much more. My problems start off in the first sentence: "smooth map from the real plane to itself." What does that mean geometrically? He's making reference to polynomials with complex coefficients, I presume. So how does the real plane (hyperplane, I guess?) comes back into the picture?
Evidently, the passage is very pictorial, including points of different colors, so it would be great to get a plot.
 A: It's hard to frame an answer without knowing just how much mathematics you know, but I'll try.
First, the real two dimensional plane is part of the picture because the complex numbers are often represented that way graphically: think $z = x + iy$ where $x$ and $y$ are real. Then a polynomial with complex coefficients is a function whose domain is the real plane and whose codomain is also the real plane, since the values of that polynomial are complex numbers.
Second, what about smooth? Well if you vary some particular $z$ by a little bit then $p(z)$ can only change by a little bit. That should be clear by analogy with real valued polynomials with real coefficients, where you can see a graph.
Third, those colors. The colors are attached to points in the plane viewed as the codomain of the polynomial. Well, the whole point of the argument is to show that they are not there. Or, rather, for any particular polynomial all the points (except for finitely many) will have the same color. So  the "splashed with colors" is really rather boring. It's a vivid part of the narrative that leads to the goal of the exercise: the fact that the polynomial must have a root.
Why is the codomain monochromatic? The essential idea is that smoothness means the points near a blue point must be blue and the points near a red point must be red. That means there's no way for a blue region to touch a red region (there's a little topology at work here). Since the regions fill the plane but can't touch, there can be only one region. 
