Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 5

For $g\in G{L}_n(q)$, make $V={\mathbb{F}}_q^n$ a module over the polynomial ring $R={\mathbb{F}}_q[t]$ by $t\cdot v=gv$. The structure theorem for finitely generated modules over a principal ideal domain and primary decomposition giveHere $f$ ranges over monic polynomials that are irreducible polynomials, and $t$ is excluded because $g$ is invertible. Associate the integer partition ${\lambda }_f$ having $m_j(f)$ parts equal to $j$. It has finite support and satisfies ${\sum }_f\mathrm{deg}(f)|{\lambda }_f|=n$. Two matrices are conjugate exactly when their R-modules are isomorphic, so this gives a bijective parameterization of conjugacy classes.

The centralizer is the R-module automorphism group. Homomorphisms between distinct primary components vanish: coprimeness of their annihilating polynomials makes each such map zero. Fix one $f$ of degree $d$, put $Q=q^d$, and write $m_j=m_j(f)$. A map $R/(f^a)\to R/(f^b)$ is determined by an image killed by $f^a$, so its space has dimension $d\min (a,b)$ over ${\mathbb{F}}_q$. Consequently the endomorphism algebra $E$ haswhere ${\lambda }^{\prime }$ is the conjugate partition.

Its semisimple quotient is ${\prod }_j{M}_{m_j}({\mathbb{F}}_Q)$: reduce the maps between the equal-length summands modulo $f$ and discard the maps between unequal lengths. Composition through a different length acquires a factor of $f$ when it returns to the original length, so these reductions define an algebra homomorphism. It is onto, and its kernel is nilpotent. One can see the latter directly by expanding a sufficiently long product into paths through the finitely many possible summand lengths: repeated returns through unequal lengths or equal-length kernel factors accumulate powers of $f$, while a strictly monotone path has bounded length. Eventually every path vanishes in every summand.

An element is invertible exactly when its image in that quotient is invertible, since a nilpotent error is inverted by a finite geometric series. The fraction of units is therefore ${\prod }_j|G{L}_{m_j}(Q)|/Q^{m_j^2}={\prod }_j{\prod }_{r=1}^{m_j}(1-Q^{-r})$. Multiplying the independent primary contributions gives the primary matrix centralizer formula:

The generating function for integer partitions is ${\sum }_{\lambda }{u}^{|\lambda |}={\prod }_{i\ge 1}(1-u^i{)}^{-1}$, since independently choosing the multiplicity of each part size gives a geometric series. Applying the parameterization in 5(a),Unique factorization of monic polynomials showsIndeed the left side counts monic polynomials with nonzero constant term; in degree $r\ge 1$ there are $(q-1)q^{r-1}$ choices. Reordering the formal products is legitimate because only finitely many factors contribute to any fixed coefficient. Substituting $u=t^i$ for each $i$ gives the conjugacy-class generating function for finite general linear groups:The constant term corresponds to the trivial zero-dimensional group.

Translation $X\mapsto I+X$ is a bijection from nilpotent matrices to unipotent matrices. We can count the former without requiring a further partition identity. Let $a_n$ be their number, with $a_0=1$.

Every linear operator $T$ has a unique decomposition from the Fitting lemmaon which it is respectively nilpotent and invertible. To justify the decomposition, kernels and images stabilize by step $n$; if $T^{n}v\in \ker T^n$, then $v\in \ker T^{2n}=\ker T^n$, so $T^{n}v=0$. The dimensions then show the sum is all of $V$, and stabilization makes $T$ invertible on the image. These two summands are determined by $T$, so there is no overcounting.

For a nilpotent summand of dimension $k$, the number of ordered complementary subspaces is $|G{L}_n(q)|/(|G{L}_k(q)||G{L}_{n-k}(q)|)$, by transporting two fixed bases. The restrictions of $T$ have $a_k$ and $|G{L}_{n-k}(q)|$ choices. Summing all $q^{n^2}$ matrices givesWrite ${\varphi }_n(q^{-1})={\prod }_{i=1}^n(1-q^{-i})$, so $|G{L}_n(q)|=q^{n^2}{\varphi }_n(q^{-1})$. Subtract the same equation for $n-1$:Thus counting nilpotent matrices by the Fitting decomposition provesThe supplied product identity is consistent with the resulting mass series: taking its variables to be $s=u$, $t=q^{-1}$ gives ${\sum }_n{a}_n{u}^n/|G{L}_n(q)|={\prod }_{i\ge 1}(1-u{q}^{-i}{)}^{-1}$.

For the fixed degree-two irreducible polynomial $f$, a matrix with minimal polynomial a power of $f$ has only one primary component. Its conjugacy classes correspond to integer partitions $\lambda$ of $m$. By the primary matrix centralizer formula, its centralizer has exactly the same order $c_{\lambda }(q^2)$ as the centralizer of the corresponding unipotent matrix in $G{L}_m(q^2)$.

Sum the conjugacy class sizes:The same sum times $|G{L}_m(q^2)|$ is the number of unipotent matrices of $G{L}_m(q^2)$, which 5(c) shows is $(q^2{)}^{m^2-m}$. Hence counting matrices with a fixed primary polynomial givesIn particular the exponent $a$ of the minimal polynomial is not fixed in advance: summing all integer partitions includes every possible largest block size. The usual statement takes $m\ge 1$; dimension zero can instead be included with its identity convention.