Let be a prime power, a vector space of dimension over the finite field , and the group of invertible linear transformations on . In this post, I’ll give a proof of the following nice theorem of Dickson:

**Theorem (Dickson).** The ring of invariants is a polynomial algebra on variables. The degrees of the generators are for . 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 , these are no longer invariants (by looking at degrees). This is consistent with the fact that if we replaced 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 for . Let be the field of fractions of , i.e., . Define a polynomial by

I claim that can be written in the form

where has degree . To see this, first consider the matrix

and let . Note that whenever because will have a linear dependency amongst its columns (since is a linear combination of the ). Since both and are polynomials in of degree , and they have the same roots, we must have for some constant . But is monic, and we can see that from the definition that the leading coefficient of is , so .

If is not identically 0, then all of its roots lie in the span of , which implies that is not identically 0. Since , we see that all of the are nonzero polynomials. Hence if we let be the determinant of the submatrix of obtained by deleting the last column and th row, we see that .

That is a polynomial in the and has degree can be seen from the original definition of . Furthermore, is invariant under because any change of basis can only scale , and will also be scaled by the same amount. Also from the identity

we see that each satisfies a monic polynomial equation with coefficients in , so the same is true for all of by basic properties of integral extensions of rings. Passing to their fields of fractions and , we get an algebraic extension, and hence both of them have the same transcendence degree (namely, ) over . Since is generated by elements over , they must be algebraically independent (we hadn’t even shown they were nonzero previously!), so that is a polynomial ring.

All that remains to show is that we have found all of the invariants. Note that is the splitting field over of the polynomial , so it is in fact a Galois extension. Let be the Galois group. Since leaves pointwise fixed, we have . On the other hand, permutes the roots of , i.e., acts on . Furthermore, acts -linearly on since acts by field automorphisms on , and since fixes pointwise. Hence and hence we get equality .

This implies in particular that , so that . Finally, is integrally closed (being a polynomial ring), and we have already seen that (and hence ) is integral over , so we conclude that .

-Steven

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