nLab field extension




If a subset kk of a field KK is a subfield, then we call the larger field KK an extension of the smaller field kk.

More generally, if kKk \to K is any ring homomorphism between fields, then it must be an injection, so we may treat it as a field extension.


Let SS be a set, and let kk be a field. Then a field extension of kk is a commutative k k -algebra KK with injections inj,inv:SKinj,inv:S\to\K such that inj(s)inv(s)=1inj(s) \cdot inv(s) = 1 for all s:Ss:S, ensuring that KK is a field.


Factorization system

Every field extension can be factorized as a purely transcendental extension? followed by an algebraic extension. Indeed, by Zorn's lemma, we may construct a transcendence basis (i.e. maximal algebraically independent set) BB, and the purely transcendental part is the subfield generated by BB.

Unfortunately, this does not yield an orthogonal factorization system: given a field KK, we may form the field K(x)K (x) of rational functions over KK, which is a purely transcendental extension of KK, and we may form the algebraic closure K(x)¯\overline{K (x)}, which is an algebraic extension of K(x)K (x); but we have the following commutative diagram,

K K(x 2) K(x) K(x)¯\array{ K & \to & K (x^2) \\ \downarrow & & \downarrow \\ K (x) & \to & \overline{K (x)} }

where K(x 2)K (x^2) is the subfield of K(x)K (x) generated by x 2x^2, and K(x 2)K(x)¯K (x^2) \to \overline{K (x)} is algebraic, yet there is no homomorphism K(x)K(x 2)K (x) \to K (x^2) making both evident triangles commute.


Last revised on May 8, 2021 at 03:53:57. See the history of this page for a list of all contributions to it.