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
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.