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It seems to me from a quick glance at several sources describing the complex and modular irreducible representations of $\mathrm{GL}(2,p)$ that any field $K$ containing a primitive $(p-1)$-root of unity is a splitting field for $\mathrm{GL}(2,p)$ (but maybe I need to read the details more carefully). Is this correct? And if so, what is a good source to cite for this besides folklore?

I flipped through a number of books, but they never bother to talk about field of definition for the irreducibles. But the constructions seem to always involve characters of $\mathbb F_p^\times$, which makes me suspect this is correct.

Edit. The original version of this question was incorrect, as pointed out in the comments. Here is the revised version.

  1. Is any field of characteristic $0$ containing a primitive $(p^2-1)$-root of unity a splitting field for $GL(2,p)$? Note that $GL(2,p)$ has exponent $p(p^2-1)$, so I am trying to dump the factor $p$.
  2. Is a any field of characteristic $p$ a splitting field for $\mathrm{GL}(2,p)$?
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    $\begingroup$ @DaveBenson, thanks. I don't really know the representation theory of $GL(n,q)$ so well, unfortunately. I found a nice paper doing the $p=2$ case and it seemed they constructed $p^2-p$ irreducibles over $\mathbb F_p$ using symmetric powers and twisting by the powers of the determinant and that seems to be the number of $p$-regular conjugacy classes if my back of the envelope calculation is correct. But the characteristic $0$ looked messier so I wasn't sure how the characters of the anisotropic torus come in. So it wasn't clear if I needed corresponding roots of units and $p^{th}$-roots $\endgroup$ Commented Jun 5, 2023 at 17:47
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    $\begingroup$ Dave is right about your second question. And you may be interested in knowing that $\mathbb F_q$ is even a splitting field for the semigroup $M_n(\mathbb F_q)$. $\endgroup$ Commented Jun 5, 2023 at 18:17
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    $\begingroup$ @BenjaminSteinberg This sounds very interesting. I look forward to seeing the outcome of this work. $\endgroup$ Commented Jun 5, 2023 at 22:00
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    $\begingroup$ By the way, I find your use of "any" in 1 and 2 linguistically ambiguous. It could mean "does there exist ..." or "is this true for all ...". $\endgroup$
    – Derek Holt
    Commented Jun 6, 2023 at 6:08
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    $\begingroup$ The Schur indices of the characters in question are 1 by projecteuclid.org/journals/… and so it just comes down to character values. The character table is well known, and so the result should hold for Q1. $\endgroup$ Commented Jun 7, 2023 at 16:10

2 Answers 2

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I realised that my previous version of this answer was irrelevant to the question (though what was said was accurate, as far as it went).

I think you can make the desired conclusion if you resort to modular representation theory, specifically, the theory of blocks with cyclic defect group (for this question we only need the cyclic defect theory with defect group of order $p$, which was done by R. Brauer himself). In a block with cyclic defect group, all decomposition numbers are $0$ or $1$. This means that every irreducible character occurs with multiplicity $1$ in some projective indecomposable module (over an appropriate local ring of characteristic zero).

Let $\chi$ be an irreducible character of a finite group $G$ which has order divisible by $p$ and not $p^{2}.$ Let $\theta$ be the character of a suitable projective indecomposable module as above, with $\langle \theta, \chi \rangle = 1.$

It is a theorem of W. F. Reynolds (a consequence of Brauer's characterization of characters), that $\theta$ is a $\mathbb{Z}$-linear combination of characters, each of which is induced from a Brauer elementary $p^{\prime}$-subgroups of $G$.

This means that $\chi$ occurs with multiplicity $1$ in the difference $\alpha - \beta$, where $\alpha$ and $\beta$ are characters of representations which are visibly respectively realizable over $\mathbb{K} = \mathbb{Q}[\omega]$, where $\omega$ is a primitive $m$-th root of unity and $m$ is the $p^{\prime}$-part of the exponent of $G$ ( in this case, just $\frac{e}{p},$ where $e$ is the exponent of $G$).

The Schur index $m_{\mathbb{K}}(\chi)$ divides both $\langle \alpha, \chi \rangle$ and $\langle \beta, \chi \rangle$ by the defining properties of the Schur index . Since $\langle \alpha - \beta, \chi \rangle = 1,$ we conclude that $m_{\mathbb{K}}(\chi) = 1.$

In the case of $G = {\rm GL}(2,p),$ this gives a positive answer to part $1$ of the question (every element of order divisible by $p$ has central $p^{\prime}$-part, and every element of order $p$ is conjugate to all its powers. Hence all character values on $p$-singular elements take values in the field $\mathbb{K}$ as above. and clearly all character values at $p$-regular elements take values in $\mathbb{K}$ (recall that $p$-regular means of order prime to $p$ and $p$-singular means of order divisible by $p$)).

Well, I suppose that strictly speaking, the answer as it stands does not now cover the irreducible characters of degree (divisible by) $p$,it only deals with the irreducible charcters in $p$-blocks of defect $1$. But for any irreducible character $\chi$ in a $p$-block of defect zero of any finite group $G$, the irreducible character $\chi$ is afforded by a projective $RG$-module, where $R$ is a suitable local ring of characteristic zero with residue field of characteristic $p$. Using the result of Reynolds, and arguing as above, we obtain $m_{\mathbb{K}}(\chi) = 1$, where $\mathbb{K}$ is the cyclotomic field extension of $\mathbb{Q}$ generated by a primitive $m$-th root of unity, with $m$ being the $p^{\prime}$-part of the exponent of $G$. In this case, $\chi$ vanishes on elements of order divisible by $p$, so that all character values taken by $\chi$ are in $\mathbb{K}.$

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  • $\begingroup$ I deleted what was written previously ( which was essentially irrelevant), and wrote a completely different naswer. $\endgroup$ Commented Jun 7, 2023 at 10:07
  • $\begingroup$ Thanks Geoff. This answer makes sense to me. $\endgroup$ Commented Jun 7, 2023 at 13:05
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(too long for a comment.)

  1. The character table of GL(2,p) in characteristic 0 can be found for example the book "A Journey through representation theory" by Gruson and Serganova on page 147.

  2. seems to be true, but I do not know a reference (maybe books by Gordon James?). Here is a way to test it for a given prime with MAGMA (you can input it in http://magma.maths.usyd.edu.au/calc/ ). The code also works for any other finite group to get a minimal splitting field over prime fields.

    p:=7; G:=GeneralLinearGroup(2, GF(p)); SIMS := AbsolutelyIrreducibleModules(G,GF(p)); temp_field_sizes:=[]; for i in SIMS do Append(~temp_field_sizes,#BaseRing(i)); end for; MaX := Maximum(temp_field_sizes); MaX;

when the result is p again, then the prime field is a splitting field. Magma confirms that the prime field is a splitting field for all primes <20.

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  • $\begingroup$ Character tables does not determine splitting fields, because of Schur indices. $\endgroup$ Commented Jun 6, 2023 at 9:31
  • $\begingroup$ @DaveBenson Thanks. I try to think about 1. again then. $\endgroup$
    – Mare
    Commented Jun 6, 2023 at 9:32
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    $\begingroup$ @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 $\mathbb{Q}[\sqrt{-35}]$. $\endgroup$ Commented Jun 6, 2023 at 10:35
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    $\begingroup$ 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. $\endgroup$ Commented Jun 6, 2023 at 10:54
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    $\begingroup$ 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. $\endgroup$ Commented Jun 8, 2023 at 8:44

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