# nLab field extension

Contents

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Idea

If a subset $k$ of a field $K$ is a subfield, then we call the larger field $K$ an extension of the smaller field $k$.

More generally, if $k \to K$ is any ring homomorphism between fields, then it must be an injection, so we may treat it as a field extension.

## Definition

Let $S$ be a set, and let $k$ be a field. Then a field extension of $k$ is a commutative $k$-algebra $K$ with injections $inj,inv:S\to\K$ such that $inj(s) \cdot inv(s) = 1$ for all $s:S$, ensuring that $K$ is a field.

## Factorization system

Every field extension can be factorized as a purely transcendental extension? followed by an algebraic extension. Indeed, by Zorn's lemma, we may construct a transcendence basis (i.e. maximal algebraically independent set) $B$, and the purely transcendental part is the subfield generated by $B$.

Unfortunately, this does not yield an orthogonal factorization system: given a field $K$, we may form the field $K (x)$ of rational functions over $K$, which is a purely transcendental extension of $K$, and we may form the algebraic closure $\overline{K (x)}$, which is an algebraic extension of $K (x)$; but we have the following commutative diagram,

$\array{ K & \to & K (x^2) \\ \downarrow & & \downarrow \\ K (x) & \to & \overline{K (x)} }$

where $K (x^2)$ is the subfield of $K (x)$ generated by $x^2$, and $K (x^2) \to \overline{K (x)}$ is algebraic, yet there is no homomorphism $K (x) \to K (x^2)$ making both evident triangles commute.

## Examples

Last revised on May 8, 2021 at 03:53:57. See the history of this page for a list of all contributions to it.