Algebraically closed field
In abstract algebra, an algebraically closed field F contains a root for every non-constant polynomial in F[x], the ring of polynomials in the variable x with coefficients in F.
As an example, the field of real numbers is not algebraically closed, because the polynomial equation x2 + 1 = 0 has no solution in real numbers, even though all its coefficients (1 and 0) are real. The same argument proves that no subfield of the real field is algebraically closed; in particular, the field of rational numbers is not algebraically closed. Also, no finite field F is algebraically closed, because if a1, a2, …, an are the elements of F, then the polynomial (x − a1)(x − a2) ··· (x − an) + 1 has no zero in F. By contrast, the fundamental theorem of algebra states that the field of complex numbers is algebraically closed. Another example of an algebraically closed field is the field of (complex) algebraic numbers.
Given a field F, the assertion "F is algebraically closed" is equivalent to other assertions:
The field F is algebraically closed if and only if the only irreducible polynomials in the polynomial ring F[x] are those of degree one.
The assertion "the polynomials of degree one are irreducible" is trivially true for any field. If F is algebraically closed and p(x) is an irreducible polynomial of F[x], then it has some root a and therefore p(x) is a multiple of x − a. Since p(x) is irreducible, this means that p(x) = k(x − a), for some k ∈ F \ {0}. On the other hand, if F is not algebraically closed, then there is some non-constant polynomial p(x) in F[x] without roots in F. Let q(x) be some irreducible factor of p(x). Since p(x) has no roots in F, q(x) also has no roots in F. Therefore, q(x) has degree greater than one, since every first degree polynomial has one root in F.
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