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In Shafarevich's Basic Algebraic Geometry, he gives an example of why we use algebraically closed fields. He considers a circle $C$ and a point $P$ outside of it. He constructs the polar of $P$ as the line joining the points of contact of the two tangents to the circle that pass through $P$. If $P$ is on the circle, the tangent and the polar are the same. But these constructions can be expressed by algebraic relations between the coordinates of $P$ and the equation of $C$, and so are valid when $P$ is inside the circle.

But now, he says, the points of tangency of the tangents have complex coordinates and can't be seen in the figure in his book. But given that the data was real, we can realize certain facts: that the points of tangency must be conjugates and the polar is the line through them, and that the polar can be seen as the locus of points whose polar line passes through $P$.

Now, is there a picture for this situation in $\mathbb{C}$ where there is none in $\mathbb{R}^2$? I feel like I'm missing the point with no geometric intuition present to "refer" to. What are these algebraic relations that define this construction?

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What do you mean by "a picture for this situation in $\mathbb{C}$"? –  Qiaochu Yuan Sep 4 '12 at 5:09
If you intersect it with the cross-section of $\Bbb{C}^2$ where one of your coordinates is real and the other one is imaginary, $C$ becomes a hyperbola. Can you envision the analogous geometric construction for a hyperbola? –  Micah Sep 4 '12 at 5:24
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The picture would have to be in $\mathbb C^2\cong\mathbb R^4$, thus not easily to visualize. Note that a variety in $\mathbb C$ is just a finite set of points (zeroes of a complex polynomial) - no circles, no lines.

If you do the math with $C$ the unit circle, you'll find that $P=(x_P, y_P)$ leads to the line given by $x x_P+y y_P = 1$. This may be derived from the tangent construction for $x_P^2+y_P^2>1$, and it produces the tangent through $P$ in the limiting case $x_P^2+y_P^2=1$. As a pure algebraic transformation, it remains valid for $P$ inside the circle, but as he says, you don't have real tangent points any more.

Mind you: The $x$ and $y$ we are talking about here are not to be seen as real and complex part of a complex number $z=x + i y$, but rather as to complex numbers themselves. That's why we'd have to visualize (real) four-dimensional space toappreciate the situation. I fwe wite the circle equation with compley numbers, and substitute real and imaginary parts for thes, we obtain (now with all variables taking only real values) $(x+i y)^2+(u+iv)^2=1$, i.e. the simultanuous (real) equations $x^2-y^2+u^2-v^2=1$ and $xy+uv=0$. I don't know if there is an easy way to visualize even this "circle" in four real dimensions - one could try to use one variable ($v$ say) as time and then the whole thing is a 3D movie. It just won't loook as circular as you might expect.

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A perfectly good picture to which your geometric intuition can refer is your original picture of a point outside of a circle. From the perspective of algebraic geometry, circles don't have insides, and circles, ellipses, hyperbolas, and parabolas all have the same overall shape.

The fact that the picture you want to draw isn't helpful to you when the point is inside the circle just means you should draw the point outside the circle when you're trying to borrow from your intuition about Euclidean plane geometry.

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