Algebraic geometry -- at least for the purposes of this article -- is the geometric study of the solution sets of polynomial equations. Specifically, it speaks of a certain duality between $(n+1)$-variable polynomial equations and their $n$-dimensional solution sets, called **algebraic varieties**. We are interested in the precise nature of this duality, i.e.

- How "
**injective**" is it, i.e. when do two polynomials have the same solution space? - How "
**surjective**" is it, i.e. what sort of sets can be algebraic varieties?

In more general terms:

- for a set of polynomials $S$ we write their null set (the intersection of all their null sets) as $V(S)$
- for some set of points $U$ we write the ideal of polynomials whose null set contains it as $I(U)$.

Then we can formulate our previous two questions as:

- when is $S=I(V(S))$? (this is the
**Nullstellensatz problem**) - when is $U=V(I(U))$? (this has to do with the
**Zariski topology**)

We generally prefer to work over the complex numbers (and in fact something slightly more interesting than the complex numbers) in algebraic geometry, as the fundamental theorem of algebra makes things easier. However, we are often interested in looking at real or even rational/integer solutions (the latter marks the connection between algebraic geometry and

**number theory**) to equations.**Definition (rational points):**Let $V\subseteq K$ be an algebraic variety and $K_0\le K$ be a subfield. Then $V\cap K_0$ is the set of

**$K_0$-rational points**of $V$.

**Exercise:**Find all the $\mathbb{Q}$-rational points of the equation ${x_1}^2+{x_2}^2=1$ [e.g. (3/5, 4/5, 1)] -- these correspond to the

*Pythagorean triples*, which are integers.

We can parameterize this variety (the unit circle) with rational functions using the parameter $t=\tan(\theta/2)$:

$$x(t)=\frac{1-t^2}{1+t^2};\,\,\,y(t)=\frac{2t}{1+t^2}$$

Since $t$ is the slope of the line connecting $(x,y)$ and (-1, 0), the rational points have a rational $t$, and since $(x,y)$ can be written as rational functions of $t$, rational $t$ give rise to rational points. Writing $t=u/v$, this leads to

$$x(t)=\frac{v^2-u^2}{v^2+u^2};\,\,\,y(t)=\frac{2uv}{u^2+v^2}$$

(Analogously, Fermat's last theorem is that the solution set of ${x_1}^n+{x_2}^n=1$ does

*not*have rational points.)
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