Residue field

In abstract algebra, the splitting field of a polynomial is constructed using residue fields. Residue fields also applied in algebraic geometry, where to every point ${\displaystyle x}$ of a scheme ${\displaystyle X}$ one associates its residue field ${\displaystyle k(x)}$. One can say a little loosely that the residue field of a point of an abstract algebraic variety is the natural domain for the coordinates of the point.

Definition

Suppose that ${\displaystyle R}$ is a commutative local ring, with maximal ideal ${\displaystyle \mathfrak{m}}$. Then the residue field is the quotient ring ${\displaystyle R/\mathfrak{m}}$.

Now suppose that ${\displaystyle X}$ is a scheme and ${\displaystyle x}$ is a point of ${\displaystyle X}$. By the definition of a scheme, we may find an affine neighbourhood ${\displaystyle \mathcal{U}=\text{Spec}(A)}$ of ${\displaystyle x}$, with some commutative ring ${\displaystyle A}$. Considered in the neighbourhood ${\displaystyle \mathcal{U}}$, the point ${\displaystyle x}$ corresponds to a prime ideal ${\displaystyle \mathfrak{p}\subseteq A}$ (see Zariski topology). The local ring of ${\displaystyle X}$ at ${\displaystyle x}$ is by definition the localization ${\displaystyle A_{\mathfrak{p}}}$ of ${\displaystyle A}$ by ${\displaystyle A\setminus \mathfrak{p}}$, and ${\displaystyle A_{\mathfrak{p}}}$ has maximal ideal ${\displaystyle \mathfrak{m}=\mathfrak{p}{A}_{\mathfrak{p}}}$. Applying the construction above, we obtain the residue field of the point ${\displaystyle x}$:

${\displaystyle k(x):=A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}}$.

Since localization is exact, ${\displaystyle k(x)}$ is the field of fractions of ${\displaystyle A/\mathfrak{p}}$ (which is an integral domain as ${\displaystyle \mathfrak{p}}$ is a prime ideal). One can prove that this definition does not depend on the choice of the affine neighbourhood ${\displaystyle \mathcal{U}}$.

A point is called ${\displaystyle k}$-rational for a certain field ${\displaystyle k}$, if ${\displaystyle k(x)=k}$.