Could you tell me what is the difference between algebraic set $X$ over $K$ and $K-$rational point $X(K)$?

Sorry to ask so much elementary question, but could you tell me what is the difference between algebraic set $X$ over $K$ and $K-$rational point $X(K)$?

Let $K$ be a field and $X$ be algebraic variety defined over $K$. Variety over $K$ is defined of a subset of certain polynomial over $K$, to me, $K-$ rational point is complete the same definition.

For example, let $E$ be an elliptic curve over $K$. Then, $E$ itsel and $E(K)$ seemingly means complete the same thing, to me.

Is there some difference?

THank you in advance.

Solution 1:

In "modern" algebraic geometry, $X$ is a scheme, which can be thought of as a machinery that comprises $X(A)$ for every commutative $K$-algebra $A$. In this case $X(K)$ is just a small part of that data (we can say that $X$ remembers the equations that are used to define it, not just the solutions over a particular algebra).

But in more old-school algebraic geometry, depending on the definitions that are used, you could say that $X$ and $X(K)$ are the same thing. Especially if $K$ is algebraically closed. When $K$ is not algebraically closed, it was common to take $X$ to be $X(\bar{K})$ (plus something to mark the fact that $X$ can be defined over $K$). So, in short, it depends.

Solution 2:

The term algebraic set has a long-ish history and has different definitions, depending on the foundations/framework for algebraic geometry you choose to work with. One thing they all share in common (as far as I know) is that they are defined with reference to an ambient affine or projective space, so perhaps it would be more proper to call them algebraic subsets. Anyway, an algebraic set defined over $K$ is any set $X$ of points (of the ambient space) for which there exist a set $\{ F_1, \ldots, F_n \}$ of polynomials with coefficients in $K$ such that a point $P$ is in $X$ if and only if $F_1 (P) = \cdots = F_n (P) = 0$.

... but what, in the first place, is a point? That is where the difference between the various foundations lies. Traditionally, one fixes an algebraically closed field $\mathbb{K}$ containing $K$ and says that a point of affine $N$-space (resp. projective $N$-space) is an $N$-tuple of elements of $\mathbb{K}$ (resp. $(N + 1)$-tuple of elements of $\mathbb{K}$ other than $(0, \ldots, 0)$, modulo multiplication by a common factor). For example, for the purposes of number theory, it can be convenient to take $\mathbb{K}$ to be the field of complex numbers even when $K$ is just some number field. Some people do not insist that $\mathbb{K}$ be algebraically closed and instead set $\mathbb{K} = K$, but this leads to complications if $K$ is not algebraically closed. People who use schemes instead say that a point of affine (resp. projective) $N$-space over $K$ is a prime ideal of the polynomial ring $K [X_1, \ldots, X_N]$ (resp. homogeneous prime ideal other than the irrelevant ideal of the polynomial ring $K [X_0, \ldots, X_N]$).

Regardless, there is agreement about what a rational point is. (This is perhaps why there are people who avoid talking about points and only talk about rational points.) If $X$ is an algebraic set defined over $K$, a $K$-rational point is basically a point of $X$ that has coordinates in $K$. This can be taken literally if you use the traditional definition of algebraic set. More generally, if you allow points to have coordinates in $\mathbb{K}$ and you have $K \subseteq L \subseteq \mathbb{K}$, you can define the $L$-rational points of $X$ to be the points whose coordinates are in $L$. We write $X (L)$ for the set of $L$-rational points of $X$. As sets of points in the traditional setting, it is true that $X$ and $X (\mathbb{K})$ are the same, but if $K \ne \mathbb{K}$, then in general $X (K) \subsetneqq X (\mathbb{K})$. Anyway, $X$ should not be considered as merely a set of points, so it is not advisable to confuse $X$ and $X (\mathbb{K})$ (let alone $X (K)$).

(One deficiency of the traditional definition of algebraic set is the implicit dependence on $\mathbb{K}$. In some sense it doesn't matter what $\mathbb{K}$ you choose, as long as it contains $K$, and if you care about $L$-rational points for some $L$, $\mathbb{K}$ needs to also contain $L$, and really $\mathbb{K}$ should be algebraically closed and preferably of infinite transcendence degree over any of the subfields that you actually care about.)