algebraically closed field



A field kk is algebraically closed if every non-constant polynomial (with one variable and coefficients from kk) has a root in kk. It follows that every polynomial of degree nn can be factored uniquely (up to permutation of the factors) as

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

where cc and the a ia_i are elements of kk.

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

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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 FF¯F \mapsto \widebar{F} that renders the inclusion i:FF¯i: F \to \widebar{F} natural:


Supposing there were such an algebraic closure functor FF¯F \mapsto \widebar{F}, consider its application to the (equalizer) diagram

conjid.\mathbb{R} \to \mathbb{C} \underoverset{conj}{id}{\rightrightarrows} \mathbb{C}.

We would have a commutative naturality diagram (meaning serially commutative on the right)

i conjid i id id ¯ i¯ ¯ conj¯id¯ ¯\array{ \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}} }

where serial commutativity of the right square(s) forces id¯conj¯\widebar{id} \neq \widebar{conj}, but functoriality applied to the equation idi=conjiid \circ i = conj \circ i on the top forces id¯=conj¯\widebar{id} = \widebar{conj} (no matter which isomorphism i¯\widebar{i} is taken to be, idid or conjconj).

Thus, any two algebraic closures are isomorphic, but not naturally so.

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:


Two algebraically closed fields K,KK, K' are isomorphic iff they have the same characteristic pp (the nonnegative generator of the kernel of the unique ring map K\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, KK contains a subring isomorphic to /(p)[S]\mathbb{Z}/(p)[S] where SS is a transcendence basis, and similarly KK' contains a subring isomorphic to /(p)[S]\mathbb{Z}/(p)[S']. By hypothesis, there is a bijection f:SSf: S \to S', which extends uniquely to an isomorphism of integral domains /(p)[S]/(p)[S]\mathbb{Z}/(p)[S] \to \mathbb{Z}/(p)[S'], which extends uniquely to an isomorphism of their fields of fractions 𝔽(S)𝔽(S)\mathbb{F}(S) \to \mathbb{F}(S'). Then K,KK, K' are algebraic closures of these fields, and one applies a theorem that an isomorphism of fields 𝔽(S)𝔽(S)\mathbb{F}(S) \to \mathbb{F}(S') can be extended to an isomorphism KKK \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 /(p)\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 KK be an algebraically closed field, and let cl:P(K)P(K)cl: P(K) \to P(K) be the operator that takes a subset SKS \subseteq K to the smallest algebraically closed subfield that contains SS. Then clcl 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 algebraic closure of a field FF is the splitting field of the set of all monic polynomials over FF. Thus for relevant material, see

Last revised on July 12, 2017 at 16:09:07. See the history of this page for a list of all contributions to it.