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?
  • 1
    $\begingroup$ Dear Clinton Boys, Do you know any algebraic number theory? (If you do, it will be easier to explain what is going on; but it can still be explained even if you don't.) Regards, $\endgroup$ – Matt E Nov 16 '11 at 0:01
  • $\begingroup$ I know very little number theory; algebraic or not. $\endgroup$ – Clinton Boys Nov 16 '11 at 2:55

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.

  • $\begingroup$ Why do the prime factors of that particular ideal determine extensions of the 3-adic valuation? $\endgroup$ – Clinton Boys Nov 17 '11 at 5:01
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    $\begingroup$ @Clinton: Dear Clinton Boys, The $3$-adic valuation has to do with how divisible a number is by $3$, or (more ideal-theoretically) how large of a power of the prime ideal $3\mathbb Z$ is lies in. If you want to extend this valuation to $K$, you have to choose a prime ideal in $\mathcal O_K$ lying over $3\mathbb Z$. (Primeness of the ideal is important in order for this recipe to actually give a valuation.) More details can be found in any (or at least many) algebraic number theory books. (Or you could ask another question!) Regards, $\endgroup$ – Matt E Nov 17 '11 at 7:02

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