Algebraic closure

Existence of an algebraic closure and splitting fields

Let ${\displaystyle S=\{f_{\lambda }\mid \lambda \in \mathrm{\Lambda }\}}$ be the set of all monic irreducible polynomials in ${\displaystyle K[x]}$. For each ${\displaystyle f_{\lambda }\in S}$, introduce new variables ${\displaystyle u_{\lambda ,1},\dots ,u_{\lambda ,d}}$ where ${\displaystyle d=\mathrm{d}\mathrm{e}\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e}(f_{\lambda })}$. Let ${\displaystyle R}$ be the polynomial ring over ${\displaystyle K}$ generated by ${\displaystyle u_{\lambda ,i}}$ for all ${\displaystyle \lambda \in \mathrm{\Lambda }}$ and all ${\displaystyle i\le \mathrm{d}\mathrm{e}\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e}(f_{\lambda }).}$ Write

${\displaystyle f_{\lambda }-\prod _{i=1}^d(x-u_{\lambda ,i})=\sum _{j=0}^{d-1}{r}_{\lambda ,j}\cdot x^j\in R[x]}$

with ${\displaystyle r_{\lambda ,j}\in R}$. Let ${\displaystyle I}$ be the ideal in ${\displaystyle R}$ generated by the ${\displaystyle r_{\lambda ,j}}$. Since ${\displaystyle I}$ is strictly smaller than ${\displaystyle R}$, Zorn's lemma implies that there exists a maximal ideal ${\displaystyle M}$ in ${\displaystyle R}$ that contains ${\displaystyle I}$. The field ${\displaystyle K_1=R/M}$ has the property that every polynomial ${\displaystyle f_{\lambda }}$ with coefficients in ${\displaystyle K}$ splits as the product of ${\displaystyle x-(u_{\lambda ,i}+M),}$ and hence has all roots in ${\displaystyle K_1}$. In the same way, an extension ${\displaystyle K_2}$ of ${\displaystyle K_1}$ can be constructed, etc. The union of all these extensions is the algebraic closure of ${\displaystyle K}$, because any polynomial with coefficients in this new field has its coefficients in some ${\displaystyle K_n}$ with sufficiently large ${\displaystyle n}$, and then its roots are in ${\displaystyle K_{n+1}}$, and hence in the union itself.

It can be shown along the same lines that for any subset ${\displaystyle S}$ of ${\displaystyle K[x]}$, there exists a splitting field of ${\displaystyle S}$ over ${\displaystyle K}$.

Separable closure

An algebraic closure ${\displaystyle K^{\text{alg}}}$of ${\displaystyle K}$ contains a unique separable extension ${\displaystyle K^{\text{sep}}}$ of K containing all (algebraic) separable extensions of ${\displaystyle K}$ within ${\displaystyle K^{\text{alg}}}$. This subextension is called a separable closure of ${\displaystyle K}$. Since a separable extension of a separable extension is again separable, there are no finite separable extensions of ${\displaystyle K^{\text{sep}}}$, of degree > 1. Saying this another way, ${\displaystyle K}$ is contained in a separably-closed algebraic extension field. It is unique (up to isomorphism).

The separable closure is the full algebraic closure if and only if ${\displaystyle K}$ is a perfect field. For example, if ${\displaystyle K}$ is a field of characteristic ${\displaystyle p}$ and if ${\displaystyle X}$ is transcendental over ${\displaystyle K}$, ${\displaystyle K(X)(\sqrt[p]{X})\supset K(X)}$ is a non-separable algebraic field extension.

In general, the absolute Galois group of ${\displaystyle K}$ is the Galois group of ${\displaystyle K^{\text{sep}}}$ over ${\displaystyle K}$.