Since each $M_n(D)$-submodule of $M_n(D)$ is also a $D$-submodule, we have $M_n(D)$ artinian as a ring.

21 Views Asked by At

$D$ is a division ring so it is artinian, hence so is $M_n(D)$ as a $D$-module. Since each $M_n(D)$-submodule of $M_n(D)$ is also a $D$-submodule, we have $M_n(D)$ artinian as a ring.

I am not sure about this "Since each $M_n(D)$-submodule of $M_n(D)$ is also a $D$-submodule, we have $M_n(D)$ artinian as a ring." What it implies? Every descending chain stops? Any help would be appreciated!