# Contents

## Idea

A field $k$ is 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

$p = c \prod_{i = 1}^n (\mathrm{x} - a_i) ,$

where $c$ and the $a_i$ are elements of $k$.

The fundamental theorem of algebra is, classically, the statement that the complex numbers 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 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.

An 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 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

be sure to check for improperly replied posts with the same subject in that and the next two months

## 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:

###### Theorem

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).

In outline, the proof is simple in structure. The “only if” statement is clear, provided we allow that transcendence degree is 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:

• There is a notion of algebraic closure of a subset,

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

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

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

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:

###### Lemma

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 pregeometry.

###### Proof

For the moment, please consult Jacobson, Basic Algebra II, Theorem 8.34. This may be expanded upon a little later.

Well-definedness of transcendence degree then follows from abstract considerations of pregeometries; see this result.

Revised on July 5, 2015 23:24:41 by Todd Trimble (67.81.95.215)