Is GL(n,F) dense in Zariski topology

[Gary McGuire]

Is GL(n,F) dense in M_n(F) in the Zariski topology
over an arbitrary field F?

Here,
M_n(F) = algebra of all n-by-n matrices over F.
GL(n,F)= invertible elements of M_n(F).
Let m=n^2 and identify M_n(F) with F^m.
The Zariski topology is defined by the closed sets,
which are the sets of common zeros of finite
sets of polynomials in F[x_1,...,x_m]

I think the following is a proof for infinite fields,
but what happens for finite fields?

GL(n,F) is dense if it meets every open set.
Let U be an open set in F^m and suppose

U intersect GL(n,F) = empty set.

This is equivalent to

U' union GL(n,F)' = F^m (*)

where ' denotes set complementation.
As U' is closed it is the set of common
zeros of polynomials g_1,...,g_r.
(Assume U is not M_n(F) so r>=1.)
Also, GL(n,F)' is the set of zeros of
the single polynomial det.
Therefore, (*) says that F^m is the set of
common zeros of the products
det*g_1, ... ,det*g_r .

Now we invoke the fact that over an infinite field,
the only polynomial in F[x_1,...,x_m]
that vanishes on all of F^m
is the zero polynomial,
to obtain a contradiction.

But over finite fields, this fact is not true
(x^q-x vanishes on the field of q elements).

Does anyone know what happens here?

-Gary McGuire

[Daniel Giaimo]

"Gary McGuire" <g...@maths.may.ie> wrote in message
news:rgk85t...@forum.mathforum.com...

> Is GL(n,F) dense in M_n(F) in the Zariski topology
> over an arbitrary field F?

This is clearly false for finite fields by the following argument:
Point-sets are clearly closed in the Zariski topology as the point
(a_1, ..., a_m) is the one and only one simultaneous solution to the
equations x_1 - a_1, ..., x_m - a_m. Therefore, since the set of
matrices is finite, the topology is the discrete topology. Hence
the only dense set is the whole space. Hence GL(n,F) is not dense as
0 is not invertible.

--Daniel Giaimo

[Clarence Wilkerson]

At the risk of egg on my face, try this argument. Let
G = GL(N,F), M= etc. If the points of G are
dense in M in the Zariski topology, then each polynomial f in
the variables (x_{ij} ) which vanishes on all of G must also
vanish on all of M.

But, if F is a
finite field with p^n elements, then each element g in GL(N,F)
satifies the equation det(g)^{p^n -1} = 1, which no
non-invertible element in M satifies. In particular, it fails
for m = the zero matrix.

Hence the points of G are not dense in the points of M....
--
Clarence Wilkerson \ HomePage: http://www.math.purdue.edu/~wilker
Prof. of Math. \ Internet: wil...@NOspam.math.purdue.edu
Dept. of Mathematics \ Messages: (765) 494-1903, FAX 494-0548
Purdue University, \
W. Lafayette, IN 47907-1395 \