Here's a quick question on noetherian rings. I know that for a ring $R$, the following are equivalent.

  • $R$ is left noetherian
  • Every finitely generated left $R$-module is noetherian
  • Every submodule of a finitely generated left $R$-module is finitely generated.

Is there a corresponding result with 'left' replaced by 'right' throughout?

  • $\begingroup$ I am guessing there is. The reason why I ask is that I'm not sure if the proof of the above result I read in Curtis and Reiner's book carries over to the right noetherian case. Regards $\endgroup$
    – user165614
    Jul 26, 2014 at 14:34
  • 1
    $\begingroup$ A right module over $R$ can be considered as a left module over the opposite ring $R^{\mathrm{op}}$; a ring is right noetherian if and only if its opposite ring is left noetherian. Just if you want to be fussy. $\endgroup$
    – egreg
    Jul 26, 2014 at 14:39

1 Answer 1


The answer is yes.

All of the proofs would go through with "left" replaced by "right" and with the arguments carried out with right modules instead of left modules. That goes for all theorems of this sort, not just the one you mention.


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