A linear Algebra question

Question

Let V be a finite-dimensional vector space over an algebraically closed field F.

Suppose A is a commutative subalgebra of L(V, V ), i.e. if S, T ∈ A, and c ∈ F then

• S + T ∈ A,

• ST ∈ A,

• cT ∈ A, and

• ST = T S

Prove that if A has no nonzero nilpotent elements, then dim A ≤ dim(V ).

Answer 1

This is a question regarding the Lie algebra gl_n(F) for an algebraically closed field F, where n denotes dim(V). gl_n(F) is the direct sum of its center (the 1-dimensional vector space spanned by the identity linear transformation) with the simple Lie algebra sl_n(F). The center commutes with anything, so for the purpose of this problem, we may assume that it already lies in the commutative subalgebra A.

Take any element B in sl_n(F) that lies in your commutative subalgebra A. There is a canonical decomposition B=B_s+B_n into its semi-simple part and nilpotent part. (taking a basis of V, thinking in terms of the Jordan canonical form of the matrix of B, B_s is the diagonal part of the Jordan canonical form) It is well known from a standard textbook that B_n can be expressed as a polynomial of B with no constant term. So B_n also lies in the subalgebra A. By the assumption we then have B_n=0. Thus, B itself is semi-simple. So the intersection of A with sl_n(F) is a commutative subalgebra consisting of semisimple elements, and therefore it is contained in a Cartan subalgebra of sl_n(F), which is of dimension n-1. Possibly adding the center of gl_n(F) into A, we therefore deduce that dim A is less than or equal to n, i.e. dim (V).