# Setting up Brauer character theory

My question relates to p. 147 of Serre's Linear Representations of Finite Groups, where he is setting up the definitions relevant to Brauer character theory.

Having fixed an algebraically closed field $K$ of characteristic $p$, we have a finite group $G$ and the set $G^p$ of $p$-regular elements of $G$, i.e. those elements whose order $\left|g\right|$ is coprime to $p$.

Assume $K$ is complete with respect to a discrete valuation $\nu:K^\times\to \mathbb{Z}$ with valuation ring $A=\{x\in K\mid \nu(x)\geq 0\}$. By general results about discrete valuation rings, $A$ has a unique maximal ideal $\mathfrak{m}=\{x\in K\mid \nu(x)\geq 1\}$, so the quotient $A/\mathfrak{m}$ is a field $k$, called the residue field of $K$.

Let $m'=\mathop{\mathrm{lcm}}_{g\in G^p}\left|g\right|$.

Q0: Is $\mathrm{char}(k)=p$?

Q1: Why must $K$ contain the group $\mu_K$ of $m'$th roots of unity? (I understand this is to do with $K$ being algebraically closed, but can't the fact that $K$ has characteristic $p$ screw things up? What if $m'>p$?)

Q2: What is the map $K\to k$ given by reduction modulo $\mathfrak{m}$ and why must this be an isomorphism $\mu_K\to \mu_k$, where $\mu_k$ is the group of $m'$th roots of unity in $k$? And again, why must $\mu_k\subseteq k$?

Q3: For $\lambda\in \mu_k$, what does he mean by $\overline{\lambda}$, the "element of $\mu_K$ whose reduction modulo $\mathfrak{m}$ is $\lambda$"?

Q4: Is there a nice example of all of this theory using a particularly simple field of characteristic $p$ with a nice valuation that I would be familiar with and could do some basic computations with?

That's quite a lot of questions, I should leave it there. I hope someone can help me out -- but it definitely just helps to write it out anyway!

Cheers,

Clinton

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I think you've misunderstood a few things. See page 115 for the definition of K (which is a field of characteristic 0 containing the |G|th roots of unity).

In the first part I'll answer as many of these as makes sense, but probably no such field K as you describe exists. In the second part I'll answer the questions with the correct definitions.

A0: Yes. The residue field in characteristic p stays characteristic p if p > 0. In general the characteristic of R/I is a divisor of the characteristic of R, and R/M does not have characteristic 1.

You may need to consider all the amazing things that are true about rings of characteristic p. They all contain Z/pZ for instance.

A1: The polynomial f=xm′−1 has derivative f′=m′xm′−1 which is relatively prime to f, and so f is a separable polynomial over the prime field Z/pZ. In particular, K, being algebraic closed, contains these roots.

A2: There is no map from K to k, but there is a map from A to k given by reduction modulo m. Every root of unity is contained in A, since it has norm 0 in any valuation. The reductions mod m of these roots of unity are still roots of unity (reduction mod m is a homomorphism), and they do not lie in the kernel (having norm 0), so the restriction is an isomorphism.

A3: An isomorphism is invertible. λ-bar is the inverse image of λ under the isomorphism.

A4: No, but the story started on the wrong path anyways.

A0: Yes, by assumption on page 115.

A1: Yes, by assumption on page 115.

A2: Again there is no such map. The map from A to k=A/m is literally reduction mod m. It is an isomorphism on the roots of unity (of order coprime to p) for the same reason.

A3: Same. Inverse image under an isomorphism.

A4: Yes.

• Take G to be the cyclic group of order 4, p=2, K to be the Gaussian field Q[i], the valuation describes how divisible by (1+i) something is, and A is the subring of Q[i] consisting of all a+bi so that a,b are rational numbers whose denominators are odd. The maximal ideal m consists of all those a+bi such that a-b is a rational number with odd denominator and whose numerator is even. The μ group is {1}, and A/m is equal to the reduction of the set {0,1} and the unique field of size 2. The character of a representation is simply its dimension, χV(g) = dim(V), but not all representations of the same dimension are isomorphic, showing that Brauer characters still just count composition factors, and that is no longer enough to decide isomorphism.

• Take G to be the cyclic group of order 4, p=3, K to be the Gaussian field Q[i], the valuation describes how divisible something is by 3, and A is the subring of Q[i] consisting of all a+bi so that a,b are rational numbers whose denominators are coprime to 3. The maximal ideal m consists of all those a+bi whose denominators are coprime to 3, but both of whose numerators are divisible by 3. The μ group is {1,−1,i,−i} and A/m is equal to the reduction of {0,1,−1,i,1+i,−1+i,−i,1−i,−1−i} and is the unique field of order 9. {1,−1,i,−i} is a subgroup of the multiplicative group isomorphic to μ and G. It turns out here that Brauer characters and ordinary characters are about the same thing. (χ(1)+χ(−1))/2 tells you how many composition factors are the trivial (central) module, (χ(1)−χ(−1))/2 tells you many composition factors are the sign module, and then you use χ(i) to find how many of the composition factors are the obvious embedding i→i, and how many are its inverse, i→−i.

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Very nice answer! @Clinton in the future, it is better to post different questions separately. Otherwise you might not find such a generous answerer as Jack who will write an essay for you. – Alex B. May 6 '11 at 3:47
Thanks a lot, that makes much more sense. So in order to think about modular representations (at least in the sense Serre talks about them in this book) over some field $k$ of characteristic $p$, we first need to find a characteristic-zero field $K$ and a valuation on $K$ such that the residue field is $k$? Is there some general procedure for doing this? – Clinton Boys May 6 '11 at 6:19
modular characters* rather – Clinton Boys May 6 '11 at 6:37
More or less, yes. However, the eigenvalues are always roots of unity, so we don't need any fancy k, we just need k a finite field of size p^n where n is the multiplicative order of p mod m′. This ensures k has the required roots of unity. K is then chosen as a cyclotomic splitting field of G (so Q adjoin an |G|'th root of unity) and then any valuation extending the p-adic valuation works. For |G|=4, the cyclotomic field is Q[i]. For p=2, the valuation was the (1+i)-adic valuation, and for p=3, it was the 3-adic valuation. – Jack Schmidt May 6 '11 at 7:21
Brauer characters have a bit of choice to them too. I chose the (1+i)-adic valuation, but one could have chosen the (1-i)-adic valuation. Since k={0,1} is so small, this didn't hurt anything, but in larger examples the values of the Brauer characters can actually change. In the literature, a convention is made that we always use Conway polynomials to define which root of unity is intended. This only matters for non-p-solvable groups, but I believe even for one of the small Mathieu groups it matters. – Jack Schmidt May 6 '11 at 7:24