Posted by: Steven Sam | March 17, 2011

## Dickson invariants

Let ${q}$ be a prime power, ${V}$ a vector space of dimension ${n}$ over the finite field ${{\bf F}_q}$, and ${{\bf GL}(V)}$ the group of invertible linear transformations on ${V}$. In this post, I’ll give a proof of the following nice theorem of Dickson:

Theorem (Dickson). The ring of invariants ${{\rm Sym}(V)^{{\bf GL}(V)}}$ is a polynomial algebra on ${n}$ variables. The degrees of the generators are ${q^n - q^i}$ for ${i=0,\dots,n-1}$. In particular, they are unique up to scalars.

From the theorem, we see that these generators aren’t stable under base change. For example, if we extend scalars to some finite extension of ${{\bf F}_q}$, these are no longer invariants (by looking at degrees). This is consistent with the fact that if we replaced ${{\bf F}_q}$ with its algebraic closure, then there are no non-constant invariants.

I’ll follow the exposition from here in some sense, but I’ll add some more details and change some notation that I found confusing.

Choose a basis ${x_1, \dots, x_n}$ for ${V}$. Let ${K}$ be the field of fractions of ${{\rm Sym}(V)}$, i.e., ${K = {\bf F}_q(x_1, \dots, x_n)}$. Define a polynomial ${f_n(t) \in K[t]}$ by

$\displaystyle f_n(t) = \prod_{\lambda \in V} (t - \lambda).$

I claim that ${f_n(t)}$ can be written in the form

$\displaystyle f_n(t) = t^{q^n} + \sum_{i=0}^{n-1} c_{n,i} t^{q^i}$

where ${c_{n_i} \in {\bf F}_q[x_1, \dots, x_n]}$ has degree ${q^n - q^i}$. To see this, first consider the ${(n+1) \times (n+1)}$ matrix

$\displaystyle M_n(t) = \begin{pmatrix} x_1 & x_2 & \cdots & x_n & t \\ x_1^q & x_2^q & \cdots & x_n^q & t^q \\ \vdots \\ x_1^{q^n} & x_2^{q^n} & \cdots & x_n^{q^n} & t^{q^n} \end{pmatrix}$

and let ${\Delta_n(t) = \det M_n(t)}$. Note that ${\Delta_n(\lambda) = 0}$ whenever ${\lambda \in V}$ because ${M_n(t)}$ will have a linear dependency amongst its columns (since ${\lambda}$ is a linear combination of the ${x_i}$). Since both ${\Delta_n(t)}$ and ${f_n(t)}$ are polynomials in ${t}$ of degree ${q^n}$, and they have the same roots, we must have ${\Delta_n(t) = cf_n(t)}$ for some constant ${c \in K}$. But ${f_n(t)}$ is monic, and we can see that from the definition that the leading coefficient of ${\Delta_n(t)}$ is ${\Delta_{n-1}(x_n)}$, so ${c = \Delta_{n-1}(x_n)}$.

If ${\Delta_{n-1}(t)}$ is not identically 0, then all of its roots lie in the span of ${\{x_1, \dots, x_{n-1}\}}$, which implies that ${\Delta_n(t)}$ is not identically 0. Since ${\Delta_1(t) = x_1 f_1(t) \ne 0}$, we see that all of the ${\Delta_n(t)}$ are nonzero polynomials. Hence if we let ${C_i}$ be the determinant of the submatrix of ${M_n(t)}$ obtained by deleting the last column and ${i}$th row, we see that ${c_{n,i} = (-1)^{n-i} C_i / \Delta_{n-1}(x_n)}$.

That ${c_{n_i}}$ is a polynomial in the ${x_i}$ and has degree ${q^n - q^i}$ can be seen from the original definition of ${f_n(t)}$. Furthermore, ${c_{n,i}}$ is invariant under ${{\bf GL}(V)}$ because any change of basis can only scale ${\Delta_n(t)}$, and ${\Delta_{n-1}(x_n)}$ will also be scaled by the same amount. Also from the identity

$\displaystyle \prod_{\lambda \in V} (t-\lambda) = t^{q^n} + \sum_{i=0}^{n-1} c_{n,i} t^{q^i},$

we see that each ${\lambda \in V \subset {\rm Sym}(V)}$ satisfies a monic polynomial equation with coefficients in ${R = {\bf F}_q[c_{n,0}, \dots, c_{n,n-1}]}$, so the same is true for all of ${{\rm Sym}(V)}$ by basic properties of integral extensions of rings. Passing to their fields of fractions ${F(R)}$ and $K = F({\rm Sym}(V))$, we get an algebraic extension, and hence both of them have the same transcendence degree (namely, ${n}$) over ${{\bf F}_q}$. Since ${R}$ is generated by ${n}$ elements over ${{\bf F}_q}$, they must be algebraically independent (we hadn’t even shown they were nonzero previously!), so that ${R}$ is a polynomial ring.

All that remains to show is that we have found all of the invariants. Note that ${F({\rm Sym}(V))}$ is the splitting field over ${F(R)}$ of the polynomial ${f_n(t)}$, so it is in fact a Galois extension. Let ${W}$ be the Galois group. Since ${{\bf GL}(V)}$ leaves ${F(R)}$ pointwise fixed, we have ${{\bf GL}(V) \subseteq W}$. On the other hand, ${W}$ permutes the roots of ${f_n(t)}$, i.e., ${W}$ acts on ${V}$. Furthermore, ${W}$ acts ${{\bf F}_q}$-linearly on ${V}$ since ${W}$ acts by field automorphisms on ${F({\rm Sym}(V))}$, and since ${W}$ fixes ${{\bf F}_q}$ pointwise. Hence ${W \subseteq {\bf GL}(V)}$ and hence we get equality ${W = {\bf GL}(V)}$.

This implies in particular that ${F({\rm Sym}(V))^{{\bf GL}(V)} = F(R)}$, so that ${{\rm Sym}(V)^{{\bf GL}(V)} \subset F(R)}$. Finally, ${R}$ is integrally closed (being a polynomial ring), and we have already seen that ${{\rm Sym}(V)}$ (and hence ${{\rm Sym}(V)^{{\bf GL}(V)}}$) is integral over ${R}$, so we conclude that ${{\rm Sym}(V)^{{\bf GL}(V)} = R}$.

-Steven