7
$\begingroup$

Let $G$ be a finite group and $p$ be a prime number dividing $|G|$.

Let $k$ be the algebraic closure of $\mathbb{F}_p$.

Let $K$ be another algebraically closed field of characteristic $p$ which is not isomorphic to $k$.

Is there a difference between the module categories $kG$-mod and $KG$-mod ?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ There certainly can be a difference, depending on 𝐾 and 𝐺. Here is one way to observe a difference. There exist uncountable fields of characteristic 𝑝. Suppose 𝐾 is the algebraic closure of such a field. Then π‘˜ is countably infinite, while 𝐾 is uncountable. So in 𝐾𝐺-mod, if an object has countably many endomorphisms, then that object is zero. But in π‘˜πΊ-mod, this is not so: there are nonzero objects with countably many endomorphisms. Do you want to amend your question, to put in some restrictions that rule out or somehow ignore these kinds of differences? $\endgroup$
    – user164898
    Commented Jun 28, 2022 at 5:23
  • $\begingroup$ I'm assuming you mean the vaguer "the representation theory of ..." rather than "the categories". As A.S. pointed out, the module categories are surely different. However the representation theories are essentially the same, if I'm not mistaken : I think any morphism $k\to K$ induces a bijection between simple objects over either field, and on projective objects over either field. $\endgroup$
    – Maxime Ramzi
    Commented Jun 28, 2022 at 12:36
  • $\begingroup$ @MaximeRamzi all projectives and injectives are defined over the algebraic closure of the prime field. In fact by Brauer in positive characteristic you just need to adjoin an n^th root of unity. I think the real question is whether all indecomposable module are defined over the algebraic closure of the prime field. I believe that is true too but requires more thought $\endgroup$
    – Benjamin Steinberg
    Commented Jun 28, 2022 at 14:26
  • $\begingroup$ @BenjaminSteinberg , I suspect the same as you do, when it comes to fin-dim'l reps of $G$. But for $p$-groups $G$ which are bigger than $\mathbb{Z}/p\mathbb{Z}$, the ring $kG$ will generally not be of finite rep type, so you cannot rely on Auslander-Ringel-Tachikawa to ensure that every rep is a direct sum of fin-dim'l ones. I worry that inf-dim'l $k$-linear reps which do not decompose into fin-dim'l ones could be quite different from the $K$-linear ones. Perhaps such reps could come from $G$ permuting elements of a transcendence basis for $K$ over $k$. Perhaps Karpenko-Reichstein do this. $\endgroup$
    – user164898
    Commented Jun 28, 2022 at 16:49
  • $\begingroup$ @A.S., I interpreted the lower case mod to mean finite dimensional modules but the question is not so clear. The Karpenko-Reichstein paper works with finite dimensional representations but it doesn't seem very direct. They show the functor from fields to sets taking a field K to isomorphism classes of KG-modules maps onto a functor giving the set of projective varieties of a certain sort and use algebraic geometry to show these need arbitrarily large transcendence degree to define $\endgroup$
    – Benjamin Steinberg
    Commented Jun 28, 2022 at 16:54

1 Answer 1

Reset to default
5
$\begingroup$

The question is not completely clear about what is meant by a difference. Let me give one interpretation. Every representation of a finite group $G$ over an algebraically closed field is defined over the algebraic closure of $\mathbb Q$ in that field, and so in a sense there is no difference between working over different algebraically closed fields of characteristic $0$. Of course, the module categories are not equivalent for the reasons in the comments but the basic structure doesn't change.

If $k$ is an algebraically closed field whose characteristic divides the order of $G$, then it's true that every simple and projective indecomposable $kG$-module is defined over the algebraic closure of the prime field, but this seems not to be the case for arbitrary representations (and hence for arbitrary indecomposable representations). This seems to have been proved as part of the theory of essential dimension, of which I know little about, and so there is a non-zero probability that I have misunderstood this. But from what I have gathered, Proposition 14.1 of A NUMERICAL INVARIANT FOR LINEAR REPRESENTATIONS OF FINITE GROUPS by NIKITA A. KARPENKO AND ZINOVY REICHSTEIN implies that if $G$ is a finite group containing $\mathbb Z/p\mathbb Z\times \mathbb Z/p\mathbb Z$ and if $k$ is a field of characteristic $p$, then for any $n\geq 0$, there is an extension field $K/k$ of transcendence degree at least $n$ such that there is a $KG$-module which is not defined over any intermediate subfield $k\subseteq L\subseteq K$ with $L$ having transcendence degree less than $n$ over $k$. In particular, there are indecomposable modular representations of $G$ not defined over the algebraic closure of the prime field of $G$ if I understood this correctly.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ The fact that there are indecomposable representations not defined over the algebraic closure of the prime field is much more elementary than the theory of essential dimension. It is easy to explicitly write down two-dimensional examples for the group $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$. It may be that essential dimension gives more significant differences between module categories over different fields than just individual modules that are only defined in one of the categories, but I'm no expert. $\endgroup$
    – Jeremy Rickard
    Commented Jun 29, 2022 at 7:59
  • 1
    $\begingroup$ @JeremyRickard, it might make a good answer to write these examples. The essential dimension stuff seems to say no matter what field of characteristic p you start with you will have representations that you can't get without adding arbitrarily many transcendentals. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 29, 2022 at 18:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .