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

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