notefield-theory

Defined Normal Extension

Recall Splitting Field

Theorem 1

If , normal, then such that

Proof

a basis for as a vector space over .

Let .

Claim isa. splitting field for over .

Theorem 2

If field, nonzero, is a finite normal extension of .

Note

Not all are roots of , so statement is not trivial. Why does necessarily split as a product of polynomials in ?

\usepackage{tikz-cd}
 
\begin{document}
 
\begin{tikzcd} && k &&& k \\ 
\\ 
&&&&&&& {\phi(A(\alpha))A(\beta)\ \forall A[x] \in k[x] } 
\\ && {k[\alpha]} &&& {k[\beta]} && {\phi(\alpha)=\beta} 
\\ 
\\ 
E \\ 
&& F &&& F 
\arrow["{\operatorname{id}_k}", tail reversed, from=1-3, to=1-6] 
\arrow[hook, from=1-3, to=4-3] 
\arrow[hook, from=1-6, to=4-6] 
\arrow["\sim", from=4-3, to=4-6] 
\arrow["\circlearrowleft", shift left=5, curve={height=-30pt}, draw=none, from=4-3, to=4-6] 
\arrow["\circlearrowleft"', shift right=5, curve={height=30pt}, draw=none, from=4-3, to=4-6] 
\arrow[hook, from=4-3, to=6-1] 
\arrow[hook, from=4-3, to=7-3] 
\arrow[hook, from=4-6, to=7-6] 
\arrow[hook, from=6-1, to=7-3] 
\arrow["{\exists \varphi \text{ isom.}}", from=7-3, to=7-6] 
\end{tikzcd}
 
\end{document}

Remark

We know how to construct many elements of .

Example

Compute .

Solution: . If . sends to (sends roots to conjugates). injective (automorphism), permutes .