Theorem - Uniqueness of Splitting Fields

theoremfield-theory

Intuition

Formal Statement

For a field $k$, and $P(X)=a_n{x}^n+\cdots +a_{1}x+a_0\in K[X]$. If ${}^E{/}_K$, ${}^F{/}_K$ are splitting fields of $P(X)$ over $k$, then there exists $\phi :{}^E{/}_k\to {}^F{/}_k$ an isomorphism of field extensions ($\phi {\mid }_k={\mathrm{id}}_k$).

Corollaries

1. Up to isomorphism, we can define the splitting field of $P(X)$. Denote it ${\mathrm{split}}_k(P(X))$ or $split({}^{P(X)}{/}_k)$.

Proof

We prove a slightly stronger statement by induction on $n$: Claim: If $\varphi :k\to K$ an isomorphism of fields and $\Phi :k[X]\to K[X]$ the induced isomorphism given by $\Phi (b_m{x}^m+\cdots +b_{1}x+b_0)=\varphi (b_m)x^m+\cdots +\varphi (b_1)x+\varphi (b_0)$ with $b_i\in k$ and $\varphi (b_i)\in K$.

If ${}^E{/}_k$ a splitting field for $P(X)$ over $k$, ${}^F{/}_K$ is a splitting field for $\Phi (P(X))$ over $K$, then

\usepackage{tikz-cd}
\begin{document}
\begin{math}
\begin{tikzcd} 
k &&& K \\
&&&&& {\varphi \circ i = j \circ \phi} \\
&&&&& {\varphi \mid_k = \phi} \\
E &&& F 
\arrow["{\phi}", from=1-1, to=1-4] 
\arrow["i"', hook, from=1-1, to=4-1] 
\arrow["j", hook, from=1-4, to=4-4] 
\arrow["{\varphi}"', dashed, from=4-1, to=4-4] 
\end{tikzcd}
\end{math}
\end{document}

Remark

$k=K$, $\phi ={\mathrm{id}}_k$ is the Theorem.

Proof of Claim by Induction on $n=\mathrm{deg}P(x)$

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Remarks

If $k\le E\le {\mathrm{split}}_k(P(X))$, then $\mathrm{spli}{\mathrm{t}}_{\mathrm{k}}(P(X))$ is also a splitting field of $P(X)$ over $E$.