Why is my approach for finding the number of R-submodules of E incorrect?

[lugita15]

Homework Statement

Let E be an n-dimensional vector space over a field k. Then if R is the ring of diagonal n-by-n matrices over k, E can be considered as a module over R, with the scalar multiplication diag(λ_1,...,λ_n)(a_1*e_1+...+a_n*e_n)=λ_1*a_1*e_1+...+λ_n*a_n*e_n, where e_1..._e_n form a basis for E as a vector space over k. Find the number of R-submodules of E.

Homework Equations

The Attempt at a Solution

If W is a nontrivial R-submodule of E, then it is a k-vector subspace of E (this is trivial), so it has a basis w_1...w_m which can be extended to a basis w_1...w_n of E as a vector space over k. Then let the linear transformation T from E to E be defined by T(w_m)=w_m+1 and T(w_i)=0 for i not equal to m. Then T is diagonalizable, so it has an eigenbasis v_1...v_n for E, such that a subset, say v_1...v_m is a basis for W. With respect to the basis v_1...v_n, the matrix representation A of T is diagonal and thus A is an element of R. So we have Aw_m=w_m+1, which is not an element of W, so W is not closed under scalar multiplication with respect to R and thus W is not an R-submodule. Thus the only R-submodules of E are {1} and E.

It turns out that my answer is wrong, and if you want I can provide a link to the correct solution. But where am I going wrong?

 

[lugita15]

Om, any thoughts on this?

 

[micromass]

Then T is diagonalizable

Why??

so it has an eigenbasis v_1...v_n for E, such that a subset, say v_1...v_m is a basis for W.

Why??