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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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1 Answer 1

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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

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