[Math] Show that if $M$ is a semisimple artinian module then $M$ is finitely generated.

abstract-algebramodulesnoetherianring-theory

The exercise is as follows:

Show that for a semisimple module $M$ over any ring, the following conditions are equivalent:

$(1)$ $M$ is finitely generated;

$(2)$ $M$ is Noetherian;

$(3)$ $M$ is Artinian;

$(4)$ $M$ is finite direct sum of simple modules.

I managed to do the following implications: $(1) \Rightarrow (4)$, $(2) \Rightarrow (4)$, $(4) \Rightarrow (2)$ and $(4) \Rightarrow (3)$.

Thus leaving $(3) \Rightarrow (1)$.

Best Answer

If $M=\oplus_{i\in I} S_i$ is a direct sum of simple modules, then it's obvious that if $M$ is Artinian, $I$ has to be finite.

If you take one nonzero element from each of these simple modules, prove those elements generate $M$.

Related Question