Valuations on number fields I'm trying to explicitly compute modular representations of some finite groups -- the easiest example to discuss is the cyclic group $C_3$ when $p=3$. The three ordinary irreducible modules for $C_3$, which can be realised over the Eisenstein rationals $\mathbb{Q}(\omega)$, are all one-dimensional and have characters $1,\omega$ and $\omega^2$ respectively. 
I want to know how to "reduce" these three representations modulo 3. I know from general theory that there is only one simple module for $C_3$ in characteristic three (the trivial representation), and I also know that each of the three irreducibles above should reduce to the trivial representation. 
The general theory of Brauer etc says to choose a field $K$ which is complete with respect to some valuation, whose valuation ring $\mathcal{O}$ has residue field $k$ with characteristic 3. We then perform a reduction procedure to turn representations over $K$ into representations over $k$. 

I understand we need to take $K=\mathbb{Q}(\omega)$. 


*

*Do we need to take our valuation $\nu$ to be the 3-adic valuation?

*In that case, what is the 3-adic valuation of an arbitrary element of $K$?

*What is $\mathcal{O}$? (This should be the set of elements $x$ with $\nu(x)\geq 0$?)

*What is the maximal ideal $\mathfrak{m}$ of $\mathcal{O}$? (This should be the set of elements $x$ with $\nu(x)>0$?)

*What is $k$? 

*How do we perform this reduction procedure to show that the three representations above all reduce to the trivial representation?

 A: Yes, we will take the $3$-adic valuation on $K$.  Actually, this doesn't quite make sense: the $3$-adic valuation is defined on $\mathbb Q$, not on $K$, and so actually we will have to choose some extension of it to $K$.  
How do we determine what the possible extensions are?  Well, we first consider the ring of integers $\mathcal O_K$ in $K$; this is the ring $\mathbb Z[\omega]$.  Now we consider the prime ideal $3\mathcal O_K$, and determine how it factors.  Each prime factor gives an extension of the $3$-adic valuation on $\mathbb Q$ to a valuation on $K$.
In our particular case, $3\mathcal O_K$ is a square: $(1-\omega)^2 = 1 +\omega^2 - 2\omega = - 3\omega,$ so $\bigl((1-\omega)\mathcal O_K\bigr)^2 = 3\mathcal O_K$.  
So there is a unique prime ideal dividing $3\mathcal O_K$, namely $(1-\omega)\mathcal O_K$, and so this determines the unique extension of the $3$-adic valuation on $\mathbb Q$ to a valuation on $K$.  How is this valuation defined, explicitly?  Well, given $x \in K\setminus \{0\}$, write $x = a/b$
with $a, b \in \mathcal O_K$ (as we may, since $K$ is the fraction field
of $\mathcal O_K$).  The valuation $v(x)$ will be the difference $v(a) - v(b)$.
So we are reduced to computing the valuation $v(a)$ for $a \in \mathcal O_K\setminus \{0\}$.  This is defined by setting $v(a)$ to be the largest natural number $n$ such that $a \in (1-\omega)^n\mathcal O_K$, or, if you prefer, the largest $n$ such that $(1 - \omega)^n$ divides $a$.
Now we could complete $K$ with respect to this valuation, but that is not really necessary.  If we were to do that, we would get the field $\mathbb Q_3[\omega]$ (where $\mathbb Q_3$ is the field of $3$-adic numbers, i.e. the completion of $\mathbb Q$ w.r.t. the $3$-adic valuation).  However, I am going to omit this 
step, since it isn't necessary, and adds an extra layer of baggage to the discussion.
So I am going to consider $K$ itself as the valued field. It's easy to see, using the above description of $v$, that the valuation ring $\mathcal O$ is obtained by starting with $\mathcal O_K$,
and adjoining $1/a$ for every element $a \in \mathcal O_K$ which does not lie in $(1-\omega)\mathcal O_K$.  (In commutative algebra terms, it is the localization  of  $\mathcal O_K$ at the prime ideal $(1-\omega)\mathcal O_K$.)
The maximal ideal is precisely the ideal $(1-\omega)\mathcal O$.
Finally, we come to the key point: when we reduce modulo the maximal ideal of  $\mathcal O$, we set $\omega \equiv 1$ (since the maximal ideal is generated by $1 - \omega$), and so the non-trivial characters become congruent to the trivial character.
If we replaced $3$ by $p$ (so $C_p$ rather than $C_3$ etc.) the same computation would go through: the maximal ideal in $\mathcal O$ would be generated by $1-\zeta_p$, and hence all the characters of $C_p$ reduce to the trivial character.
