Proposition 5

a) Let $\mathfrak{m}$ be a regular maximal left ideal of A. Then the pseudomodule $A_s/\mathfrak{m}$ is simple, and there exists a nonzero element of $A_s/\mathfrak{m}$ with annihilator $\mathfrak{m}$.

b) Let M be a simple pseudomodule, and let $x$ be a nonzero element of M. We have $M = Ax$, the annihilator $\mathfrak{m}$ of $x$ is a regular maximal left ideal of A, and when passing to the quotient, the mapping $a\mapsto ax$ defines an isomorphism from $A_s/\mathfrak{m}$ to M.

c) Let M be a pseudomodule not reduced to 0. Then M is simple if and only if we have $M = Ax$ for every nonzero element $x$ of M.

Let $\mathfrak{m}$ be a regular maximal left ideal of A. We have seen that there exists a nonzero element $z$ of $A_s/\mathfrak{m}$ with annihilator equal to $\mathfrak{m}$ and such that $Az= A_s/\mathfrak{m}$; in particular, we have $A(A_s/\mathfrak{m})\not= 0$. Moreover, every sub-pseudomodule of $A_s/\mathfrak{m}$ is of the form $\mathfrak{n}/\mathfrak{m}$, where $\mathfrak{n}$ is a left ideal of A containing $\mathfrak{m}$. Since $\mathfrak{m}$ is maximal, the only possibilities are $\mathfrak{n}=\mathfrak{m}$ and $\mathfrak{n}= A$, so that the pseudomodule $A_s/\mathfrak{a}$ is simple. This proves a).

Under the assumptions of b), the set of elements $y$ of M such that $Ay= 0$ is a sub-pseudomodule of M different from M and therefore reduced to 0. Consequently, $Ax$ is a nonzero sub-pseudomodule of M, which implies $M = Ax$. Assertion b) then follows from the remarks before Definition 2.

Let M be a nonzero pseudomodule. Suppose that we have $Ax= M$ for every nonzero element $x$ of M. In particular, we have AM $\not= 0$. Let N be a nonzero sub-pseudomodule of M, and let $x$ be a nonzero element of N. We have $Ax= M$, and therefore N = M. Hence M is simple. Conversely, if M is simple, then we have $M = Ax$ for every $x\not= 0$ by b).