# Uniqueness of splitting field for linear representations of finite groups

If $F$ is any field, $\bar{F}$ its algebraic closure, then it is well-known that all irreducible (indecomposable) $\bar{F}$-representations of a finite group $G$ are realizable over some finite extension $E$ of $F$, and we call $E$, a splitting field for $G$. Any extension of $E$ is also a splitting field of $G$. By finiteness of $[E\colon F]$, we can have a minimal extension of $F$ which is splitting field for $G$.

I puzzled by the following question:

Is such minimal extension unique?

(For example, consider a finite group $G$, $\mathbb{F}_p$ a field of order $p$ with $(|G|,p)=1$. Then there is $n\in \mathbb{N}$ such that $\mathbb{F}_q$ ($q=p^n$) is a splitting field field of $G$.

Is it possible that there are subfields of $\mathbb{F}_q$ of order $p^{m_1}$ and $p^{m_2}$, $m_1\nmid m_2$, and $m_2\nmid m_1$ such that they are splitting fields of $G$, but their subfields are not splitting fields?)

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Are you fixing the representation of $G$ or do you want an extension $E/F$ which works for any representation ? –  François Brunault Oct 11 '11 at 13:33
I am considering representations only on splitting fields, and looking whether there is smallest subfield (whether exist) in the algebraic closure $\bar{F}$. Not fixing any representation, since every irreducible representation over a splitting field remains irreducible even after extension of that splitting field, so not considering only one representation, but all irreducible representations. Is this the thing you want to know? –  joseph Oct 11 '11 at 13:38
I'm having trouble with your example. The subfield lattice of $\mathbb{F}_{p^n}$ is a chain (corresponding under the Galois correspondence to the subgroup lattice of $\mathbb{Z}/p^n\mathbb{Z}$). So in this case, uniqueness is obvious. I expect uniqueness also holds in the general case. –  David Hill Oct 11 '11 at 15:09
@David: The $\mathbb F_{p^n}$'s are not linearly ordered, containment is given by divisibility of the exponent $n$. –  Torsten Ekedahl Oct 11 '11 at 16:45
@joseph : OK (I think there was a little ambiguity in the first sentence of your question in the sense that the extension $E$ could depend on the irreducible representation). –  François Brunault Oct 11 '11 at 17:15

## 1 Answer

You need two conditions for a field to be a splitting field for a specific irreducible representation (in characteristic zero to begin with): It must contain the character values of the representation. For this there is of course a minimal field, the field generated by those values. However, a splitting field must also split a division algebra and for that there is no unique minimal field.

As a specific example we can take the quaternion group of order $8$ over the rational numbers. The component of the group algebra corresponding to the only faithful representation is a quaternion algebra over $\mathbb Q$ ramified at $\infty$ and $2$ and hence is split by any quadratic imaginary field for which $2$ is either ramified or non-split.

To be more concrete about the quaternion representation, the ordinary real quaternion algebra makes sense over the rationals, denoted $\mathbb H_{\mathbb Q}$. It has $\mathbb Q$-basis 1,i,j,k and the usual multiplication table. Then $\pm\{1,i,j,k\}$ is a multiplicative group that is a copy of the quaternion group $Q$ and therefore we have an algebra map $\mathbb Q[Q]\to\mathbb H_{\mathbb Q}$. It is evidently surjective so that $\mathbb H_{\mathbb Q}$ is one of the factors in the group algebra (more precisely $\mathbb Q[Q]=\mathbb Q^4\times\mathbb H_{\mathbb Q}$, where the first four factors correspond to the four one-dimensional representations). Now, for a field $K$ of characteristic zero $\mathbb H_K$ has a two-dimensional irreducible representation exactly when it is split, i.e., when it is isomorphic to the algebra of $2\times2$-matrices. That is thus exactly the condition for a two-dimensional irreducible representation to exist over $K$. Now, it is well-known that the algebra is split precisely when the reduced norm form ($N(\alpha)=\alpha\overline{\alpha}$) restricted to the purely imaginary quaternions has a non-trivial zero. As $N(ai+bj+ck)=a^2+b^2+c^2$ we get the condition that David mentions in the comments.

The situation in positive characteristic is different. As the group algebra can be defined over the prime field and the Brauer group of finite fields is trivial, one only needs for (the mod $p$) character values to be in the field so in that case there is a minimal field.

Addendum: The question of a minimal field for all irreducible representations has essentially the same answer. First the field has to contain all character values, then there are still division algebras to split (in characteristic $0$). The example of the quaternion group still illustrates the problem, all characters are rational-valued and one must still kill the quaternion algebra for which there is no unique minimal field.

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Let me make the quaternion example very explicit. If $K$ is any field in which there are $a$ and $b$ with $a^2+b^2=-1$, then the quaternion $8$ group has a two dimensional representation given by the matrices $\left( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} a & b \\ b & -a \end{smallmatrix} \right)$. There is no unique minimal field in which $a^2+b^2=-1$ has solutions. –  David Speyer Oct 11 '11 at 18:00
@Torsten: Can you explain second paragraph in answer, please? (I know that Q8 has four 1-dim. irrep, and one 4-dim. irrep over Q; and over C (or Q(i)), it splits as sum of two irrep. of dim. 2 (i.e. twice the irrep. of dim 2 over $\mathbb{C}$)) –  joseph Oct 12 '11 at 4:08
Thanks! Nice Answer!! –  joseph Oct 12 '11 at 8:24