Splitting fields are not unique?

Question

Let $F$ be a field and $0 \neq f \in F[X]$. I have proven that any two splitting field extensions $K_1,K_2$ are $F$-isomorphic.

Can anyone give an example of $2$ splitting field extensions of $f$ such that $f$ has two different decompositions in linear factors? With different I mean that not all the roots are the same in the two decompositions.

Answer 1

Take a couple of different transcendentals in $\mathbb{C}$, say $\pi$ and $\rm e$.

Let $F:=\mathbb{Q}$, $K_1:=\mathbb{Q}[\pi]/\left(\pi^2-2\right)$, and $K_2:=\mathbb{Q}[\rm e]/\left({\rm e}^2-2\right)$.

Then $K_1, K_2$ both split $X^2-2$ over $\mathbb{Q}$: in $K_1[X]$ we have $X^2-2=(X-\bar{\pi})(X+\bar{\pi})$, and in $K_2[X]$ we have $X^2-2=(X-\bar{\rm e})(X+\bar{\rm e})$.

Answer 2

(If you think my other answer is somehow perverse you may prefer this one.)

Let $A$ be the ring of $2\times 2$ rational matrices.

Let $$ F:=\left\{\begin{pmatrix}p & 0 \\0 & p\end{pmatrix} : p\in\mathbb{Q} \right\} $$ which is a field isomorphic to $\mathbb{Q}$.

Let $$ K_1:=\left\{\begin{pmatrix}p & 0 \\0 & p\end{pmatrix}+\begin{pmatrix}0 & 2q \\q & 0\end{pmatrix} : p,q\in\mathbb{Q} \right\} $$ which is a field splitting $X^2-2$ over $F$, the roots being $\pm\begin{pmatrix}0 & 2 \\ 1 & 0\end{pmatrix}$.

Let $$ K_2:=\left\{\begin{pmatrix}p & 0 \\0 & p\end{pmatrix}+\begin{pmatrix}0 & q \\2q & 0\end{pmatrix} : p,q\in\mathbb{Q} \right\} $$ which is a field splitting $X^2-2$ over $F$, the roots being $\pm\begin{pmatrix}0 & 1 \\ 2 & 0\end{pmatrix}$.