Exercise 9, § 3

The colength of a simple A-module S is the dimension of the countermodule of S (over the field End$_A(S)$). We say that an A-module M of finite length has finite colength if it admits a Jordan–Hölder series whose successive quotients are of finite colength; the sum of these colengths is called the colength of M and is denoted by $\lambda (M)$. If N is a submodule of M, then we have $\lambda (M) =\lambda (N) +\lambda (M/N)$.

Let A be an autoinjective ring (Exercise 8) such that the modules $A_s$ and $A_d$ are of finite colength. We say that A is a Frobenius ring if for every simple A-module S, we have $\lambda (S) =\lambda (S^*)$.

a) Let M be an A-module of finite length. If A is a Frobenius ring, then we have $\lambda (M) =\lambda (M^*)$. If, moreover, A is an algebra of finite rank over a commutative field K, then we have dim$_K(M) =$ dim$_K(M^*)$ (use Exercise 7).

b) Let K be a commutative field, and let A be the subalgebra of $\mathbf{M}_6(K)$ with basis the elements $E_{11}+ E_{55}, E_{12}+ E_{56}, E_{21}+ E_{65}, E_{22}+ E_{66}, E_{33}+ E_{44}, E_{13}, E_{23}$, $E_{45}$, and $E_{46}$ (where the $E_{ij}$ are the matrix units of $\mathbf{M}_6(K)$). Prove that A is an autoinjective algebra but not a Frobenius algebra (observe that the minimal left ideals of A are KE$_{13}+$ KE$_{23}$, KE$_{45}$, and KE$_{46}$, and the minimal right ideals are KE$_{13}$, KE$_{23}$, and KE$_{45}+$ KE$_{46}$).