---
tag: 02SJG1HC32
kind: definition
name: Algebraically closed field
msc: [12F05]
defines: 0PF6TM8TWT
uses: [0QA8CHH8MG, 00EHX6FWC0, 0SHMC00RJV, 00X2A9G9W3]
sources:
  - url: https://en.wikipedia.org/wiki/Algebraically_closed_field
    locator: Equivalent properties
---

## Definition

A [[field|0QA8CHH8MG]] $K$ is algebraically closed if every polynomial $f \in K[x]$ with $\deg f \ge 1$ has a [[root|00EHX6FWC0]] in $K$.

## Remarks

Equivalently, every nonzero polynomial over $K$ [[splits|0SHMC00RJV]] over $K$. Also equivalently, the [[irreducible polynomials|00X2A9G9W3]] over $K$ are exactly the polynomials of degree one.
Another equivalent condition: $K$ has no [[algebraic extension|09EYMKMSB9]] other than $K$ itself.
The field $\mathbb{C}$ is algebraically closed, by the fundamental theorem of algebra. The field $\mathbb{R}$ is not, because $x^2 + 1$ has no real root.
No finite field is algebraically closed: for a finite field $K$, the polynomial $1 + \prod_{a \in K} (x - a)$ has no root in $K$.

## Explanations

### L1 human

In an algebraically closed field, every polynomial equation of degree at least one has a solution.
You never need a bigger field to find roots.