2005/11/08 | 伽罗瓦扩张域[Galois Extension Field]
类别(∑〖数学〗) | 评论(0) | 阅读(191) | 发表于 21:27
The following are equivalent definitions for a Galois extension field (also simply known as a Galois extension) of .
1. is the splitting field for a collection of separable polynomials. When is a finite extension, then only one separable polynomial is necessary.
2. The field automorphisms of that fix do not fix any intermediate fields , i.e., .
3. Every irreducible polynomial over which has a root in factors into linear factors in . Also, must be a separable extension.
4. A field automorphism of the algebraic closure of for which must fix . That is to say that must be a field automorphism of fixing . Also, must be a separable extension.

A Galois extension has all of the above properties. For example, consider , the rationals adjoined by the imaginary number , over , which is a Galois extension. Note that contains all of the roots of , and is generated by them, so it is the splitting field of . Of course, there are two distinct roots in so it is separable. The only nontrivial automorphism fixing is given by complex conjugation
[center]
(1)

whose fixed field is . The only irreducible polynomials with rational coefficients with roots of the form with are () and . Both split into linear factors over . Finally, the algebraic closure is the set of algebraic numbers in . Given an automorphism of the algebraic numbers that sends to itself, it must fix for trivial reasons. In general, it is not so simple to verify all of these properties, which makes their equivalence useful.

There are a couple of ways for an extension not to be a Galois extension. One is for it to not be a normal extension. For instance, is not normal, and hence not Galois. It is missing the complex roots of . It's only nontrivial automorphism is defined by , which not only fixes but also the subfield .

Another possibility for a non-Galois extension is for it to be not separable. The field characteristic of such a field must be finite since all polynomials are separable in characteristic zero. Moreover, all finite fields are perfect, i.e., all algebraic extensions are separable. Consider the field of rational functions with coefficients in , infinite in size and characteristic 2 ().
[center]
(2)

and the extension
[center]
(3)

For instance, and . Then is the splitting field of , as in characteristic 2, and so it is a normal extension. However, is not separable because has repeated roots in its splitting field,.
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