Definition 36. Normal field extension.
A field extension \(E/F\) is said to be a normal field extension if \(E\) is the splitting field of some family of polynomials in \(F[x]\text{.}\)
Section Normal field extension
Note 37.
Theorem 38.
The following are equivalent for a field extension \(E/F\text{:}\)
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\(E/F\) is a normal extension.
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If \(\sigma\colon E\hookrightarrow\overline{F}\) is an \(F\)-embedding of \(E\) into an algebraic closure of \(F\) containing \(E\) then, \(\sigma\) is an automorphism of \(E\text{.}\)
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Every irreducible polynomial in \(F[x]\) that has at least one root in \(E\) completely splits in \(E[x]\text{.}\)
Proof.
Suppose that \(E/F\) is a normal extension, and that \(\sigma\colon E\hookrightarrow\overline{F}\) is an \(F\)-embedding of \(E\) into an algebraic closure of \(F\text{.}\) We need to show that \(\sigma\) is an automorphism of \(E\text{.}\) Assume that \(E\) is a splitting field of \(\{f_i\}\subseteq F[x]\text{.}\) Therefore, \(\sigma(E)\) is also a splitting field of \(\{f_i\}\) over \(F\text{.}\) As \(E\) and \(\sigma(E)\) are generated over \(F\) by the same roots, we have \(\sigma(E)=E\text{.}\)
Now assume that every \(F\)-embedding of \(E\hookrightarrow\overline{F}\) is an automorphism of \(E\text{.}\) Let \(f\in F[x]\) be an irreducible polynomial with a root \(a\in E\text{.}\) For any other root \(b\in\overline{F}\) there is an isomomorphism \(\sigma_b\colon F(a)\to F(b)\) given by \(a\mapsto b\) and \(\alpha\mapsto\alpha\) for all \(\alpha\in F\text{.}\) By the extension theorem for field homomorphisms Proposition 32, \(\sigma_b\) extends to an \(F\)-embedding \(\sigma\colon E\hookrightarrow\overline{F}\text{.}\) By our assumption, \(\sigma\) is an automorphism of \(E\text{.}\) Therefore, \(b=\sigma(a)\in E\text{.}\) Thus, every root of \(f\) lies in \(E\text{.}\)
Now assume that every irreducible polynomial in \(F[x]\) that has at least one root in \(E\) completely splits in \(E[x]\text{.}\) Let \(\{f_i\}\subseteq F[x]\) be the set of all irreducible polynomials that have at least one root in \(E\text{.}\) Then, by our assumption, each \(f_i\) completely splits in \(E[x]\text{.}\) Therefore, \(E\) is the splitting field of \(\{f_i\}\text{.}\)
