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\begin{document}

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\section*{algebraically closed field}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{algebra}{}\paragraph*{{Algebra}}\label{algebra}

[[!include higher algebra - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{classical_invariants}{Classical invariants}\dotfill \pageref*{classical_invariants} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A [[field]] $k$ is \textbf{algebraically closed} if every non-constant [[polynomial]] (with one variable and coefficients from $k$) has a root in $k$. It follows that every polynomial of degree $n$ can be factored uniquely (up to [[permutation]] of the factors) as

\begin{displaymath}
p = c \prod_{i = 1}^n (\mathrm{x} - a_i) ,
\end{displaymath}
where $c$ and the $a_i$ are elements of $k$.

An \textbf{algebraic closure} of an arbitrary field $k$ is an algebraically closed field $\bar{k}$ equipped with a field homomorphism (necessarily an [[injection]]) $i: k \to \bar{k}$ such that $\bar{k}$ is an [[algebraic extension]] of $k$ (which means that every element of $\bar{k}$ is the root of some non-zero polynomial with coefficients only from $k$). For example, $\mathbb{C}$ is an algebraic closure of $\mathbb{R}$. An algebraic closure of $k$ can also be described as a maximal algebraic extension of $k$. The [[axiom of choice]] proves the existence of $\bar{k}$ for any field $k$, as well as its uniqueness up to [[isomorphism]] over $k$. (See [[splitting field]] for a more refined result.) However, note that $\bar{k}$ need not be unique up to \emph{unique} isomorphism, so it's not really appropriate to speak of [[the]] algebraic closure of $k$. For example, complex conjugation is a nontrivial [[automorphism]] of $\mathbb{C}$ over $\mathbb{R}$.

Without choice, the existence and uniqueness of algebraic closures may fail; see

\begin{itemize}%
\item \emph{\href{http://cs.nyu.edu/pipermail/fom/2006-May/010531.html}{Algebraic closure of Q}}, a thread on FOM started by [[Timothy Chow]];

\end{itemize}
be sure to check for improperly replied posts with the same subject in that and the next two months

\begin{itemize}%
\item possibly \emph{\href{http://doi.wiley.com/10.1002/malq.19920380136}{Algebraic closure without choice}}, a paper by somebody that I can't read


\item \emph{\href{http://math.fau.edu/richman/html/docs.htm}{The fundamental theorem of algebra: a constructive development without choice}} by [[Fred Richman]]



\end{itemize}
Even with choice, algebraic closure is not [[functor|functorial]] in any reasonable sense. For example, it is very easy to demonstrate that there is no algebraic closure functor $F \mapsto \widebar{F}$ that renders the inclusion $i: F \to \widebar{F}$ natural:

\begin{example}
\label{}\hypertarget{}{}
Supposing there were such an algebraic closure functor $F \mapsto \widebar{F}$, consider its application to the (equalizer) diagram

\begin{displaymath}
\mathbb{R} \to \mathbb{C} \underoverset{conj}{id}{\rightrightarrows} \mathbb{C}.
\end{displaymath}
We would have a commutative naturality diagram (meaning serially commutative on the right)

\begin{displaymath}
\itexarray{
\mathbb{R} & \stackrel{i}{\to} & \mathbb{C} & \underoverset{conj}{id}{\rightrightarrows} & \mathbb{C} \\ 
\mathllap{i} \downarrow & & \mathllap{id} \downarrow & & \downarrow \mathrlap{id} \\ 
\widebar{\mathbb{R}} & \stackrel{\widebar{i}}{\to} & \widebar{\mathbb{C}} & \underoverset{\widebar{conj}}{\widebar{id}}{\rightrightarrows} & \widebar{\mathbb{C}} 
}
\end{displaymath}
where serial commutativity of the right square(s) forces $\widebar{id} \neq \widebar{conj}$, but functoriality applied to the equation $id \circ i = conj \circ i$ on the top forces $\widebar{id} = \widebar{conj}$ (no matter which isomorphism $\widebar{i}$ is taken to be, $id$ or $conj$).

\end{example}
Thus, any two algebraic closures are isomorphic, but [[unnatural isomorphism|not naturally]] so.

\hypertarget{classical_invariants}{}\subsection*{{Classical invariants}}\label{classical_invariants}

Putting aside the concerns of constructive mathematics, and freely adopting the principle of the [[excluded middle]] and the [[axiom of choice]], algebraically closed fields are characterized (up to non-unique isomorphism) by just two cardinal invariants:

\begin{theorem}
\label{}\hypertarget{}{}
Two algebraically closed fields $K, K'$ are isomorphic iff they have the same characteristic $p$ (the nonnegative generator of the [[kernel]] of the unique [[ring]] map $\mathbb{Z} \to K$) and the same [[transcendence degree]] (the [[cardinality]] of any maximal set of algebraically independent elements).

\end{theorem}
In outline, the proof is simple in structure. The ``only if'' statement is clear, provided we allow that transcendence degree is \emph{well-defined}. For the ``if'' statement, $K$ contains a subring isomorphic to $\mathbb{Z}/(p)[S]$ where $S$ is a transcendence basis, and similarly $K'$ contains a subring isomorphic to $\mathbb{Z}/(p)[S']$. By hypothesis, there is a bijection $f: S \to S'$, which extends uniquely to an isomorphism of [[integral domains]] $\mathbb{Z}/(p)[S] \to \mathbb{Z}/(p)[S']$, which extends uniquely to an isomorphism of their fields of fractions $\mathbb{F}(S) \to \mathbb{F}(S')$. Then $K, K'$ are algebraic closures of these fields, and one applies a theorem that an isomorphism of fields $\mathbb{F}(S) \to \mathbb{F}(S')$ can be extended to an isomorphism $K \to K'$ of their algebraic closures.

The full details of such a proof carry some themes important in [[model theory]]:

\begin{itemize}%
\item There is a notion of algebraic closure of a subset,


\item There are prime models (algebraic closure of prime field $\mathbb{Z}/(p)$),


\item There are notions of independence and basis, and well-defined degree or dimension,


\item There are extensions of isomorphisms of independent sets to isomorphisms of their algebraic closures.



\end{itemize}
Perhaps the most subtle in the list is the notion of independence and well-definedness of (transcendence) degree, which notably involves verification of the Steinitz exchange axiom:

\begin{lemma}
\label{}\hypertarget{}{}
Let $K$ be an algebraically closed field, and let $cl: P(K) \to P(K)$ be the operator that takes a subset $S \subseteq K$ to the smallest algebraically closed subfield that contains $S$. Then $cl$ is a [[geometric stability theory|pregeometry]].

\end{lemma}
\begin{proof}
For the moment, please consult Jacobson, Basic Algebra II, Theorem 8.34. This may be expanded upon a little later.

\end{proof}
Well-definedness of transcendence degree then follows from abstract considerations of pregeometries; see \href{/nlab/show/matroid#welldefined}{this result}.

\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

\begin{itemize}%
\item The \textbf{[[fundamental theorem of algebra]]} is, classically, the statement that the [[complex number]]s form an algebraically closed field $\mathbb{C}$.

Arguably, this theorem is not entirely algebraic; the algebraic portion is that $R[\mathrm{i}]$ is algebraically closed whenever $R$ is a [[real-closed field]]. Unusually, this algebraic portion is \emph{not} (as stated) valid in [[constructive mathematics]], while the analytic result (that the [[real numbers]] form a real closed field $\mathbb{R}$) is constructively valid with the usual definitions.


\item The algebraic closure $\overline{\mathbb{Q}}$ of the [[rational numbers]] $\mathbb{Q}$ is the [[algebraic numbers]].



\end{itemize}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[Galois group]]


\item [[separable closure]]


\item [[geometric point]]



\end{itemize}
The algebraic closure of a field $F$ is the splitting field of the set of all [[monic polynomials]] over $F$. Thus for relevant material, see

\begin{itemize}%
\item [[splitting field]]

\end{itemize}
[[!redirects algebraically closed field]] [[!redirects algebraically closed fields]]

[[!redirects algebraic closure]] [[!redirects algebraic closures]]



\end{document}