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algebraically closed field

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 as

p=c i=1 n(xa i),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 . Arguably, this theorem is not entirely algebraic; the algebraic portion is that R[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 result the real numbers form a real closed field is constructively valid with the proper definitions.

An algebraic closure of an arbitrary field k is an algebraically closed field k¯ equipped with a field homomorphism (necessarily an injection) i:kk¯ such that k¯ is an algebraic extension? of k (which means that every element of k¯ is the root of some non-zero polynomial with coefficients only from k). For example, is an algebraic closure of . The axiom of choice proves the existence of k¯ for any field k, as well as its uniqueness up to isomorphism over k. However, note that 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 over .

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

Revised on September 25, 2012 18:28:25 by Todd Trimble (68.198.214.232)