Minimal splitting field need not be contained in a cyclotomic extension of rationals

Statement

It is possible to have a finite group $G$ and a minimal splitting field $K$ of $G$ such that $K$ has characteristic zero, but it is not contained in any cyclotomic extension of the rationals.

Note that, by the Kronecker-Weber theorem from Galois theory, being contained in a cyclotomic extension of the rationals is equivalent to being an abelian extension of the rationals, so this is equivalent to saying that it is possible for a finite group to have a minimal splitting field that is not an abelian extension of the rationals.

Related facts

Similar facts

• Minimal splitting field need not be unique
• Splitting not implies sufficiently large
• Field generated by character values is splitting field implies it is the unique minimal splitting field
• Minimal splitting field need not be cyclotomic

Opposite facts

• Sufficiently large implies splitting: This in particular shows that there exists at least one minimal splitting field that is contained in a cyclotomic extension of rationals.
• Field generated by character values is contained in a cyclotomic extension of rationals: Combined with field generated by character values is splitting field implies it is the unique minimal splitting field (a situation that occurs when all irreducible representations have Schur index one), it shows that to find an example of a minimal splitting field that is not contained in a cyclotomic extension of the rationals, we need to look for a finite group whose field generated by character values is not a splitting field.

Proof

Algebraic example

Further information: linear representation theory of quaternion group, faithful irreducible representation of quaternion group

Consider the quaternion group. Apart from the two-dimensional faithful irreducible representation of quaternion group, all other representations can be realized over the rationals, so a field in characteristic zero is a splitting field iff the two-dimensional faithful irreducible representation can be realized over it.

A sufficient condition for being a splitting field is that the field contain elements $\alpha,\beta$ such that $\alpha^2 + \beta^2 = -1$. We construct one such field and then argue that no proper subfield of it is a splitting field.

Field description

Consider the field where $\mathbb{Q}(\alpha,\beta)$ where $\alpha$ is the real cube root of 2 and $\beta$ is a square root of $-1 - \alpha^2$.

Proof that this field is a minimal splitting field for the quaternion group

We have $\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\alpha,\beta)$. The extension degree $\mathbb{Q}(\alpha)/\mathbb{Q}$ is 3 and the extension degree $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}(\alpha)$ is 2 (note that the latter extension is strict because $\mathbb{Q}(\alpha)$ can be viewed as a subfield of $\R$ but $\mathbb{Q}(\alpha,\beta)$ cannot). Overall, the extension $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$ has degree $3 \times 2 = 6$.

As noted above, the field is a splitting field for the quaternion group.

Thus, it suffices to show that it contains no proper subfield that is also a splitting field for the quaternion group. For this, note that any proper subfield which is degree three over the rationals cannot work as it is formally real, and there are no proper subfields that are degree two over the rationals.

Proof that the field is not contained in a cyclotomic extension of the rationals

Any cyclotomic extension of the rationals has an abelian Galois group, so any subfield of it must also be a Galois extension. However, $\mathbb{Q}(\alpha) = \mathbb{Q}(2^{1/3})$ is not Galois, because the automorphism mapping $2^{1/3}$ to a complex cube root of 2 does not preserve the subfield. Therefore $\mathbb{Q}(\alpha, \beta)$ is also not contained in a cyclotomic extension of the rationals.

For this part of the proof, see also this Math StackExchange question.