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The first of them is the splitting field of ; the second has normal closure that includes the complex cubic roots of unity, and so is not a splitting field. In fact, it has no automorphism other than the identity, because it is contained in the real numbers and x 3 − 2 {\displaystyle x^{3}-2} has just one real root.
This equality implies that, if [E : F] is finite, and U is an intermediate field between F and E, then [E : F] sep = [E : U] sep ⋅[U : F] sep. [20] The separable closure F sep of a field F is the separable closure of F in an algebraic closure of F. It is the maximal Galois extension of F.
If is an extension of , which is in turn an extension of , then is said to be an intermediate field (or intermediate extension or subextension) of /. Given a field extension L / K {\displaystyle L/K} , the larger field L {\displaystyle L} is a K {\displaystyle K} - vector space .
Given a separable extension K′ of K, a Galois closure L of K′ is a type of splitting field, and also a Galois extension of K containing K′ that is minimal, in an obvious sense. Such a Galois closure should contain a splitting field for all the polynomials p over K that are minimal polynomials over K of elements of K ′.
In the following examples is a field, and ,, are the fields of complex, real, and rational numbers, respectively. The notation F ( a ) indicates the field extension obtained by adjoining an element a to the field F .
An example is the field of complex numbers. Every field has an algebraic extension which is algebraically closed (called its algebraic closure), but proving this in general requires some form of the axiom of choice. [11] An extension L/K is algebraic if and only if every sub K-algebra of L is a field.
Let / be an algebraic extension (i.e., L is an algebraic extension of K), such that ¯ (i.e., L is contained in an algebraic closure of K).Then the following conditions, any of which can be regarded as a definition of normal extension, are equivalent: [3]
The algebraic closure of K is also the smallest algebraically closed field containing K, because if M is any algebraically closed field containing K, then the elements of M that are algebraic over K form an algebraic closure of K. The algebraic closure of a field K has the same cardinality as K if K is infinite, and is countably infinite if K ...