symmetric monoidal (∞,1)-category of spectra
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
where $c$ and the $a_i$ are elements of $k$.
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
possibly Algebraic closure without choice, a paper by somebody that I can't read
The fundamental theorem of algebra: a constructive development without choice by Fred Richman
Even with choice, algebraic closure is not 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:
Supposing there were such an algebraic closure functor $F \mapsto \widebar{F}$, consider its application to the (equalizer) diagram
We would have a commutative naturality diagram (meaning serially commutative on the right)
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 $conj$).
Thus, any two algebraic closures are isomorphic, but not naturally so.
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:
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:
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.
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.
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.
The algebraic closure $\overline{\mathbb{Q}}$ of the rational numbers $\mathbb{Q}$ is the algebraic numbers.
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
Last revised on September 25, 2018 at 11:30:42. See the history of this page for a list of all contributions to it.