Throughout this post $G$ denotes $GL\_{n}(\mathbb{F})$ where $\mathbb{F}$ denotes the finite field of $q$ elements.


I'm currently reading the aforementioned book to understand how the irreducible representations of $G$ are constructed. To get started, we need to understand the conjugacy class of $G$.


In the book he explains how given a vector space $V$, for each $g \in G$ we can equip $V$ with an $\mathbb{F}[t]$ module structure by setting $t \cdot v = g(v)$ and extending it linearly. When we are regarding $V$ as an $\mathbb{F}[t]$ module for a specific $g$, we denote it as $V^{g}$.


From here I understand that $V^{g\_{1}} \cong V^{g\_{2}}$ as $\mathbb{F}[t]$ modules if and only if $g\_{1}$ and $g\_{2}$ belong to the same conjugacy class. Therefore if we only care about $V^{g}$ upto isomorphism, we can denote it as $V^{\mathcal{C}}$ where $\mathcal{C}$ is the conjugacy class of $g$.


Next he goes on to prove that for any given conjugacy class $\mathcal{C}$, we have $V^{\mathcal{C}} \cong \displaystyle \bigoplus\_{i = 1}^{s -1} \mathbb{F}[t]/(f\_{i})^{e\_{i}}$ where $f\_{i}$ are irreducible polynomials of $\mathbb{F}[t]$. The proof of this is basically using the structure theorem for finitely generated modules over a PID.


Here is where my confusion begins. He says that for each $f\_{i}$ in the decomposition of $V^{\mathcal{C}}$, we can assign a partition and if I'm understanding correctly, it looks like we can construct a partition for $n$. Let $d\_{i}$ denote the degree of $f\_{i}$. Based on the setup so far, my questions are as follows.


1. What exactly is this partition based on the above decomposition? Is the partition on $d\_{i}$ or is the partition on $n$?
2. Does the partition have anything to do with
\begin{align\*}
\displaystyle \sum\_{i = 1}^{s} d\_{i}e\_{i} = n
\end{align\*}


Any help is greatly appreciated. Thanks! :)



Chapter 4 Section 2 of Macdonald's Symmetric Functions and Hall Polynomials
The point is that the same polynomial $f\_i$ may occur many times. So for each polynomial $f$, we have a partition whose parts are the exponents $e\_i$ in all the different cases where $f\_i=f$. So this is not a partition of $n$ or of $d\_i$.


Assume for a moment that $f(u)=u-a$ for some $a$; then the resulting partition is the lengths of the Jordan blocks with $a$ on the diagonal. The other cases are similar, but Jordan decomposition is more complicated over a field that isn't algebraically closed.


