---
tag: NW7W4R0T85
kind: definition
name: Rational point
msc: [14G05, 11G35]
defines: SWJYG87QQH
uses: [04C5ZBG1WT, HV6YBS0YDH, 5A35TYZD02, 09XNAPTJQ7, ZT9A6JK8QF]
sources:
  - url: https://en.wikipedia.org/wiki/Rational_point
---

## Definition

Let $K$ be a [[field|04C5ZBG1WT]] and let $X$ be a [[scheme|5A35TYZD02]] over $K$, for example an [[algebraic variety|HV6YBS0YDH]] over $K$.
A *$K$-rational point* of $X$ is a morphism of $K$-schemes $\operatorname{Spec} K \to X$.
The set of $K$-rational points is written $X(K)$.

## Remarks

If $X \subseteq \mathbb{A}^n_K$ is the affine variety defined by polynomials $f_1, \dots, f_m \in K[x_1, \dots, x_n]$, then $X(K)$ is the set of $a \in K^n$ with $f_1(a) = \cdots = f_m(a) = 0$.
If $X \subseteq \mathbb{P}^n_K$ is defined by homogeneous polynomials, then $X(K)$ is the set of points $[a_0 : \cdots : a_n]$ with all $a_i \in K$, not all zero, at which the polynomials vanish.
Equivalently, when $X$ is of finite type over $K$, a $K$-rational point is a closed point of $X$ whose residue field is $K$.

Example: $(3/5, 4/5) \in X(\Q)$ for $X = V(x^2 + y^2 - 1) \subseteq \mathbb{A}^2_\Q$, since $9/25 + 16/25 = 1$.

Non-example: $(1, \sqrt{2})$ satisfies $x^2 + y^2 = 3$, but it is not a $\Q$-rational point, because $\sqrt{2} \notin \Q$.
In fact $x^2 + y^2 = 3$ has no $\Q$-rational point: clearing denominators gives $a^2 + b^2 = 3c^2$ in integers with $\gcd(a, b, c) = 1$, and reducing modulo $3$ forces $3$ to divide $a$, $b$ and then $c$.