Let $k$ be a field and $G$ a linear algebraic group over $k$. What is the group of $k$-rational points of $G$? By definition, $G$ is an algebraic variety. Suppose that $G$ is defined by polynomials $f_1, \ldots, f_n$. Then the set of $k$-rational points of $G$ is $A=\{a=(a_1, \ldots, a_n)\in k^n: f_1(a)=\ldots = f_m(a)=0\}$. Is $A = G$?

In $GL_n(k)$ case, what are the group of $k$-rational points of $GL_n(k)$? Thank you very much.

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    $\begingroup$ Yes, $A$ is the set of $k$-rational points of $G$. If $k$ is not algebraically closed, then $A$ and $G$ are fundamentally different (well at least in some cases). For instance if $A=\{ (0,0) \}$ and $k=\mathbb{R}$, then you can't tell if $G$ is defined by $x^2+y^2=0$ or by $(x=0,y=0)$ [other than the fact that only one has an algebraic group structure]. Similarly, if $k$ is finite, then $A$ doesn't tell you much about $G$. If $k$ is algebraically closed, then $A=G$ loses no information. $\endgroup$ – Jack Schmidt Jan 21 '14 at 18:05

A linear algebraic group $G$ is more than just a collection of points with an additional structure. So I think "does $A = G$" is the wrong question: one should think of these things as fundamentally different types of objects.

I am going to attempt to explain this without using too much algebraic geometry, and toward the end I am hiding some details. Please let me know if you want them filled in.

It is very common to think of a linear algebraic group $G$ defined over $k$ not as some definite object, but rather as a functor. This functor is from the category of fields over $k$ to the category of sets:

$$F \rightarrow G(F) \text{ (the F-rational points of } G).$$

So here $F$ is a field containing $k$. Actually, one allows any $k$-algebra $R$, but for this post I'm going to stick with fields. For many of the classical groups, you get what you'd expect. For example:

$$\text{GL}_n(F) = \text{collection of invertible matrices with coefficients in } F.$$

Note that $\text{GL}_{1,F}$ is defined by the equation $xy - 1$ inside $\mathbb{A}_F^2$.

However, some groups have weird behavior. Consider the circle group over $\mathbb{R}$, which cut out by the equation $x^2+y^2=1$ in $\mathbb{A}_\mathbb{R}^2$. The $\mathbb{R}$-rational points are the unit circle (whence the name), and hopefully it is clear that this is not the same as $\text{GL}_{1,\mathbb{R}}$. However when one considers $\mathbb{C}$-rational points, we have that

$$(x-iy)(x+iy) = x^2 + y^2 = 1$$

and the map sending $x \rightarrow x-iy = z$ and $y \rightarrow x+iy = w$ (with $zw = 1$) gives an isomorphism so that our group becomes $\text{GL}_{1,\mathbb{C}}$ (since we're over the complex numbers now).

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    $\begingroup$ Very nice way to summarize the difficulties that can arise without getting too technical. $\endgroup$ – Tobias Kildetoft Jan 22 '14 at 14:47
  • $\begingroup$ @rghthndsd, thank you very much. Why we called these points rational? Are these points always given by rational functions? $\endgroup$ – LJR Jan 22 '14 at 16:20
  • $\begingroup$ @LJR: I do not know the etymology of "$k$-rational", but here is what I think is a good guess. It is a generalization of the situation over the integers: A point with integral coordinates is known as an integral point, and a point with rational coordinates (i.e. in $\mathbb{Q}$) is known as a rational point. Now when one wants to consider other fields $k$, a natural name is $k$-rational. Of course, one does not say "$\mathbb{Q}$-rational" (this would be silly), we still only say "rational". $\endgroup$ – RghtHndSd Jan 22 '14 at 16:35
  • $\begingroup$ @rghthndsd, thank you very much. $\endgroup$ – LJR Jan 22 '14 at 16:37

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