Exercise 8, § 3

Let A be a ring such that the A-modules $A_s$ and $A_d$ are of finite length. We identify $A_d$ with the dual of $A_s$. For every left ideal $\mathfrak{s}$ of A, the orthogonal $\mathfrak{s}^0$ of $\mathfrak{s}$ is the right annihilator of $\mathfrak{s}$ in A; likewise, for every right ideal $\mathfrak{d}$, the orthogonal $\mathfrak{d}^0$ is the left annihilator of $\mathfrak{d}$. Prove that the following properties are equivalent:

(i) Every simple (left or right) A-module is reflexive.

(ii) The dual of any (left or right) simple A-module is simple.

(iii) For every minimal left ideal $\mathfrak{s}$ and every minimal right ideal $\mathfrak{d}$, we have $\mathfrak{s}^{00}=\mathfrak{s}$ and $\mathfrak{d}^{00}=\mathfrak{d}$.

(iv) The mapping $\mathfrak{s}\mapsto \mathfrak{s}^0$ is a bijection from the set of left ideals of A to the set of right ideals.

When these properties hold, we say that the ring A is autoinjective (cf. X, §1, p. 171, exercice 26).

a) Suppose that A is autoinjective; let M be a finitely generated (left or right) A-module. Prove that the following properties hold:

(i’) The A-module M is reflexive.

(ii’) The length of $M^*$ is equal to that of M.

(iii’) For every submodule N of M, we have $N^{00}= N$.

(iv’) The mapping $N\mapsto N^0$ is a bijection from the set of submodules of M to the set of submodules of $M^*$.