Why is the multiplicative group of a field an algebraic group? Reading from here: https://en.wikipedia.org/wiki/Algebraic_torus
In particular, the paragraph under the heading "Multiplicative group of a field."
So, in my mind, a multiplicative group of a field is denoted $F^\times$ and is just the group $(F \backslash \{0\}, \times)$ where $\times$ is the multiplicative operation in $F$. So, this multiplicative group is an algebraic group? That means it is also an affine variety: a set of solutions of a system of polynomial equations. What is this system of polynomial equations, is it $\{f-x : f \in F^\times \}$?
Also, the next sentence says that the multiplicate group of $F$ is such that for any field extension $E \backslash F$ the $E$-points are isomorphic to the group $E^\times$.
Are $E$-points those points that are in $E \backslash F$? What does it mean that the $E$-points are isomorphic to $E^\times$?
I'm realizing I asked a few different questions here... If anyone could answer anything it would be greatly appreciated!! Thanks, I don't know what I'd do without this website!
 A: The system of polynomial equations is $xy = 1$ in $\mathbb{A}^2$. The system you wrote down has no solutions as soon as $F^{\times}$ has at least two elements.
There are several different ways of thinking about algebraic varieties, at different levels of abstraction and requiring different levels of background, and the differences between them matter when the ground field is not algebraically closed. The simplest way to say it is that if you have a variety defined by some equations over a field $F$, then you can consider the solutions to those equations over any field extension $E$ of $F$; these are called the $E$-points of the variety. We need to do this sort of thing because a variety over a non-algebraically closed field $F$ need not have any $F$-points; consider, for example, the variety $x^2 + y^2 + 1 = 0$ over $\mathbb{R}$, which has no $\mathbb{R}$-points but does have $\mathbb{C}$-points. This is also a very natural thing to do in number theory, taking $F = \mathbb{Q}$ and $E$ some number field for example.
If the variety is an algebraic group $G$ then the solutions to the defining equations over any field extension $E$ form a group denoted $G(E)$. If $G$ is the multiplicative group, then $G(E) \cong E^{\times}$, since the set of solutions to $xy = 1$ in $\mathbb{A}^2(E)$ is the set of pairs of points $(x, x^{-1})$ where $x \in E^{\times}$ (and the multiplication is pointwise so it's the same as the ordinary multiplication in $E^{\times}$).
There's a more abstract and more powerful way to say this using the language of group schemes and functors of points also, which has the benefit of not requiring a specific choice of equations and generalizing from fields to arbitrary commutative rings.
