We know in a semisimple ring R, for every R-module, Noetherian is equivalent to Artinian, my question is:
If for every R-module M Noetherian is equivalent to Artinian, can we prove R is a semisimple ring?
MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.Sign up to join this community
Let $R= k[\epsilon]/\epsilon^2$. Then module is Artinian if and only if it is Noetherian if and only if it is finite-dimensional. This is clear, since if a module is finite-dimensional there can't be an infinite ascending or descending sequence, and if it's infinite dimensional than $M/\epsilon$ is an infinite-dimensional vector space, so it has infinite ascending and descending chains.
But $R$ is obviously not semisimple.