Abstract Algebra: Theory and Applications · Problem 22.4.17

Galois Theory — Thomas Judson

Mathematics — Abstract Algebra · finite-fields-separable-extensions

Problem

Let F⊂E⊂KF \subset E \subset K be fields. If KK is a separable extension of FF, show that KK is also separable extension of EE. Hint. If p(x)∈F[x]p(x) \in F[x], then p(x)∈E[x]p(x) \in E[x].
Intuitive explanation

The key insight is that separability is about minimal polynomials having no repeated roots, and this property is inherited by larger fields. When KK is separable over FF, every element of KK has a minimal polynomial over FF with distinct roots. Since any element of KK also lies in EE, its minimal polynomial over EE must divide its minimal polynomial over FF—and a polynomial with distinct roots can only have divisors with distinct roots. The hint reminds you that coefficients in FF are automatically in EE, so you're not losing information about the original minimal polynomial.

Intuitive explanation

The key insight is that separability is about minimal polynomials having no repeated roots, and this property is inherited by larger fields. When KK is separable over FF, every element of KK has a minimal polynomial over FF with distinct roots. Since any element of KK also lies in EE, its minimal polynomial over EE must divide its minimal polynomial over FF—and a polynomial with distinct roots can only have divisors with distinct roots. The hint reminds you that coefficients in FF are automatically in EE, so you're not losing information about the original minimal polynomial.

Related theory

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