Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A question about a proof of Cayley's Hamilton theorem using Zariski topology.

"The set $C$ of all matrices of size $n \times n$ (over an algebraically closed field $k$) with distinct eigenvalues is dense in the Zariski topology".

Can we argue as follows?

Since non-empty open sets in the Zariski topology are dense in $k^{n}$ then we are done if we can show the complement of $C$, i.e the set of all matrices of size $n \times n$ with repeated eigenvalues is open in the Zariski topology.

Now to each matrix $B$ of size $n \times n$ compute its characteristic polynomial $p_{B}$ and associate to this polynomial its discriminant $D(p_{B})$. Now define a map:

$f: \mathbb{A}^{n^{2}} \rightarrow \mathbb{A}^{1}$ given by $f(B)=D(p_{B})$ where where $\mathbb{A}^{n}$ denotes the $n$-affine space. I'm identifying here $\mathbb{A}^{n^{2}}$ with the set of all matrices $n \times n$ over an algebraically closed field $k$.

Here is my question. How do we know the map $f$ is continuous with respect the Zariski topology? If we can show it is continuous aren't we done? because we can take $\{0\}$ this is closed in $\mathbb{A}^{1}$ because it is finite, so by continuity of $f$, $f^{-1}(\{0\})$ is closed in $\mathbb{A}^{n^{2}}$ but this preimage is exactly the set of all matrices with repeated eigenvalues.

share|improve this question

2 Answers 2

up vote 5 down vote accepted

This map is not only continuous. It is a actually a morphism of algebraic varieties. To see this note that the characteristic polynomial of a matrix is a polynomial whose coefficients are polynomials in the entries of the matrix. This is because it is equal to $\det(tI-A)$, and the determinant of a matrix is a polynomial in its entries. Also, the discriminant of a polynomial is again, a polynomial in its coefficients, so we see that this map is a polynomial in the entries of $A$.

share|improve this answer
    
Thanks. so basically it is because the map $f$ is a regular function (being a polynomial essentially) and regular maps are continuous in the Zariski topology? –  user6495 Mar 21 '11 at 20:53
    
@user6495: yes. –  the L Mar 21 '11 at 21:07

$f$ is a polynomial hence it's continuous wrt. Zariski topology. You are almost done:you need to show that the complement of $f^{-1}(0)$ is non-empty, and this you know as there is matrix with different eigenvalues (eg. a suitable diagonal matrix)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.