2

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.

  • How about this? Take $F=\Bbb R$. In the quaternions, $X^2+1$ has a lot of distinct splitting fields. – Angina Seng May 04 '19 at 11:25
  • 1
    The quaternions are not a field? (no commutativity) –  May 04 '19 at 11:35
  • But the quaternions contain a lot of commutative subfields! – Angina Seng May 04 '19 at 11:35
  • @LordSharktheUnknown You can make this an answer! Please be clear which subfields you take though. –  May 04 '19 at 11:38
  • 2
    What do you precisely mean by "not all the roots are the same"? The roots are in an algebraic closure, so there is no way to compare two sets of roots of we chose two different algebraic closures. On the other hand, given a fixed algebraic closure, there is only one splitting field that is contained in it. – asdq May 04 '19 at 11:44
  • 1
    Yes, I'm exactly asking for the situation that the roots are not comparable. They can be in two different fields that don't have a lot of elements in common. –  May 04 '19 at 11:47

2 Answers2

3

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 $X^2-2=(X-\bar{\rm e})(X+\bar{\rm e})$.

ancient mathematician
  • 14,102
  • 2
  • 16
  • 31
  • there is a interesting thing to note to complete the picture: there is a theorem that says that if $\overline{K}$ is an algebraic closure of some field $K$ and $K\subset L\subset \overline{K}$, where $L$ is a splitting field for some family of polynomials, then $L$ is unique. This means that if $L_1$ and $L_2$ are two distinct splitting fields for the same polynomials over a field $K$ then any algebraic closure of $L_1$ and $L_2$ are necessarily different (but isomorphic). Thus any algebraic closure of $K_1$ and $K_2$ will be different. – Masacroso Jul 12 '23 at 13:20
3

(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 $\pm\begin{pmatrix}0 & 1 \\ 2 & 0\end{pmatrix}$.

ancient mathematician
  • 14,102
  • 2
  • 16
  • 31