Re: Diagonalizable matrices

Michael Barr wrote:
> José Carlos Santos <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
>
> > Let k be an algebraically closed field and let n be a natural number.
> > Consider the space M of all square matrices with n lines, n rows and
> > entries in k. Let D be the subset of M of all diagonalizable matrices.
> > My question is: is D a dense subset of M with respect to the Zariski
> > topology? Of course, the answer is yes if k has characterist 0, because
> > D contains all matrices whose characteristic polynomial p has n distinct
> > roots and this is equivalente to (p,p') = 1. But if the characteristic
> > of k is greater than 0, this argument fails.
>
> Yes it is dense.  The reason is that any matrix with distinct
> eigenvalues is diagonalizable over an algebraically closed field.  And
> that is Zariski open because the characteristic polynomial of the
> matrix has coefficients that are polynomials in the entries of the
> matrix and the discriminant of the characteristic polynomial is also a
> polynomial in the entries and the non-vanishing of this discriminant
> is therefore Zariski open.  Since every non-empty open is dense in
> Zariski, the conclusion follows.  This gives a very elegant proof of
> the Cayley-Hamilton theorem.

Thanks. It did not occur to mo to use the descriminant. My line of
thought was:

p has n distinct roots => (p,p') = 1 => res(p,p') <> 0

but the first implication fails in characteristic greater than 0.

Best regards,

Jose Carlos Santos