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Jun 8, 2023 at 8:44 comment added Geoff Robinson However, a beautiful theorem of Benard and Schacher ( Journal of Algebra, 1972 or so), with an alternate proof by G. Janusz ( Proc AMS, 1972) ) shows that potential odd prime divisors of Schur indices of irreducible characters can be excluded by examination of the character table: if the Schur index of a complex irreducible character $\chi$ of a finite group $G$ is divisible by an integer $m$, then the extension of $\mathbb{Q}$ generated by values of $\chi$ must contain a primitive $m$-th root of unity.
Jun 6, 2023 at 12:04 history edited Mare CC BY-SA 4.0
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Jun 6, 2023 at 10:54 comment added Dave Benson To add insult to injury, although the quaternion group has a two dimensional representation over $\mathbb{Q}[\sqrt{-35}]$, it cannot be written as two by two matrices over the ring of integers in this field.
Jun 6, 2023 at 10:35 comment added Dave Benson @Mare If you want a cautionary example, look at the quaternion group of order eight. All the entries in the character table are integers. But a field $K$ of characteristic zero is a splitting field if and only if $x^2+y^2=-1$ has a solution in $K$. The "generic" one is the field of fractions of $\mathbb{Q}[x,y]/(x^2+y^2+1)$ but there are many smaller ones such as $\mathbb{Q}[\sqrt{-35}]$.
Jun 6, 2023 at 10:14 comment added Benjamin Steinberg I don’t have access to James’s book but I’ll try to get my hands on it.
Jun 6, 2023 at 9:32 comment added Mare @DaveBenson Thanks. I try to think about 1. again then.
Jun 6, 2023 at 9:32 history edited Mare CC BY-SA 4.0
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Jun 6, 2023 at 9:31 comment added Dave Benson Character tables does not determine splitting fields, because of Schur indices.
Jun 6, 2023 at 8:10 history answered Mare CC BY-SA 4.0