# rational points of an algebraic variety

a $$K$$-rational point is a point on an algebraic variety where each coordinate of the point >belongs to the field $$K$$. This means that, if the variety is given by a set of equations

$$f_i(x_1, ..., x_n)=0,\; j=1, \cdots, m$$

then the $$K$$-rational points are solutions $$(x_1, ..., x_n)\in K^n$$ of the equations

My understanding is that the algebraic variety is the set of solutions $$(x_1, ..., x_n)\in K^n$$ of the equations so by definition all points of the algebraic variety are $$K$$-rational points.. I'm I wrong? so what is the meaning of defining rational points if all points of ther alg variety are rational?

The notion of a $K$-rational point depends very much on the field $K$. For example, consider the algebraic variety given as the zero set of the polynomial $x^2+y^2+1\in \mathbb{C}[x,y]$. Note that this variety has no $\mathbb{R}$-rational points.

Exercise 1: Prove that if we consider the algebraic variety given as the zero set of the polynomial $x^2+y^2-1\in \mathbb{R}[x,y]$, then this variety has infinitely many $\mathbb{Q}$-rational points.

In light of Exercise 1, note that the set of $\mathbb{Q}$-rational points of this variety is a proper subset of the set of $\mathbb{R}$-rational points of this variety.

In general, if $X$ is a scheme over a field $K$, then we can speak of the set of all $K$-rational points of $X$. Note that the $K$-rational points of $X$ are precisely those points of $X$ at which the residue field of the local ring at that point is isomorphic to $K$.

Definition: Let $\pi:X\to S$ be an $S$-scheme. A $\textit{section}$ of $X$ is a morphism of $S$-schemes $\sigma:S\to X$. This amounts to saying that $\pi\circ\sigma=\text{Id}_S$. The set of sections of $X$ is denoted by $X(S)$ (and also by $X(A)$ if $S=\text{Spec}(A))$).

Exercise 2: Prove that if $X$ is a scheme over a field $K$, then we can identify $X(K)$ with the set of all $K$-rational points of the scheme $X$.

Let $K$ be a field and let $X=\text{Spec }K[T_1,\dots,T_n]/I$ be an affine scheme over $K$. Let $Z$ be the zero set of some polynomial $P\in I$. It is natural to ask whether our notion of a "$K$-rational point of $X$" in the scheme-theoretic sense agrees with our naive notion of a $K$-rational point of the algebraic set defined by $P$.

Exercise 3: Prove that we have a canonical bijection $Z\to X(K)$ where $X(K)$ denotes the set of all $K$-rational points of the scheme $X$.

• if my understanding is right, every point in an algebraic variety $A$ over $K$, is $K-$rational by definition and also $K'-$rational for every field $K'\supset K$ and then the problem is to know about $K''-$rational points for subfields $K''$ of $K$. is this correct? Jun 23, 2011 at 10:27
• @palio: Dear palio, Typically, if a variety is defined over $K$, we consider points defined over the algebraic closure of $K$ (or some other algebraically closed field containing $K$, such as $\mathbb C \supset \mathbb Q$). The $K$-points alone may not determine the variety (i.e. they may not be Zariski dense), e.g. they could be empty, such as for the variety $x^2+y^2 =1$ over $\mathbb R$. Regards, Jun 23, 2011 at 16:52

An algebraic variety (over a non-algebraically-closed field) is "more than just a set of points". If X is an algebraic variety over $K$, you can make sense of the set $X(L)$ of $L$-points for any field extension $L / K$.

For instance, consider the empty subvariety of $\mathbb{Q}^1$ (defined, if you like, by the equation $1 = 0$) and the subvariety defined by $X^2 + 1 = 0$. Both of these have no $\mathbb{Q}$-points, but their $\mathbb{C}$-points are very different!

If you want to think of a variety as a set of points, it's best to think of it as a set of points in $\overline{K}^n$, where $\overline{K}$ is the algebraic closure of $K$. Then you can meaningfully ask which ones of these are contained in $K^n$.

• In fact you can think of a variety over $K$ as a set of points in $\bar{K}^n$ equipped with an action of $\text{Gal}(\bar{K}/K)$. The fixed points of this action are the $K$-rational points. (Otherwise you get the wrong notion of morphism; you want morphisms to respect the Galois action.) Jun 23, 2011 at 12:31

For example Hartshorne defines affine variety as the set of solutions in $$K^n$$ only in the case that $$K$$ is algebraically closed. I rather think that this is the norm. Without this assumption a lot of the basic stuff goes out of the window. Nullstellensatz for example. As an extreme example would you like to treat the origin, a single point, as the solution set of the equation $$x^2+y^2=0$$ over the reals, when we know' that the set of (affine) solutions really is the union of the two lines $$x=iy$$ and $$x=-iy$$.

You can study geometric objects associated with polynomial equations (with coefficients in $$K$$) over a non-algebraically closed field $$K$$, but then you need to use the language of schemes or a suitable substitute. Then we get "non-rational points" or points of degree $$>1$$. This way of bookkeeping (keeping track of the degree of the field extension $$K(x_1,\ldots,x_n)/K$$, where the coordinates of a solution are $$(x_1,\ldots,x_n)$$ ) gives us uniform answers to several questions. For example, consider the curve $$x^2+y^2=1$$ over the field of rational numbers. Superficially it would look like there are 2 points on that curve such that $$x=3/5$$, namely $$(x,y)=(3/5,4/5)$$ and $$(3/5,-4/5)$$, but no points with $$x$$-coordinate equal to $$1/2$$. With this bookkeeping in place we would say that there is a single degree two point with $$x=1/2$$, namely that with $$y=\pm\sqrt{3}/2$$. Even though this may look like two points also, it is treated as a single degree two point. This is because we cannot distinguish between the two points in the sense that if $$P(u,v)$$ is polynomial with rational coefficents, then $$P(1/2,\sqrt3/2)$$ and $$P(1/2,-\sqrt3/2)$$ are both zero or neither is. Branch points (=ramification) cause another change to the way of counting, but in the end we get two points' with a given $$x$$-coordinate with appropriate bookkeeping.

Basically the point is that some fields are too small to tell us the whole story of the set of solutions of a system of polynomial equations. I may not have listed the most natural reasons. Expect many answers to this one.