Matrices with eingenvalue 1 in Zarisky topology

Question

Is the set of all matrices with eingenvalue 1 an irreductible set in Zarisky topology? I'm thinking that we can asscoiate the set of mtrices with the set of points $X=\{(1,0,...0), (0,1,...,0),...,(0,0,...,1)\}$ which is irreductible only if the ideal $I(X)$ is a prime ideal. It is this ideal a prime ideal? Is this the idea behind the problem? Thank you!

Answer 1

It is irreducible.

The set of all matrices with eingenvalue 1 is an hypersurface. A global equation is given by $f(X)=0$, where $f(X)=\det(X-I)\in \mathbb{F}[x_{1,1}, \dots,x_{1,n}, x_{2,1}, \dots,x_{2,n}, \dots, x_{n,1} \dots, x_{n,n}]$.

Thus, in order to prove that the set of matrices with eigenvalue one is irreducible we should prove that $f(X)$ is irreducible. Now $f(X)$ is irreducible iff $g(X)=\det (X)$ is irreducible (being the two polynomials obtained from each other by linear change of coördinates). So the proof reduces to the well known fact that the set of singular matrices is an irreducible closed set.

A nice proof of the fact that $g(X)$ is irreducible can be found here: Slick proof the determinant is an irreducible polynomial