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 .