# When is a ring considered as a module over a Noetherian subring itself Noetherian?

If $S$ is a ring and $R$ is a Noetherian subring of $S$, and we know $_RS$ (i.e. $S$ viewed as a left $R$-module) is finitely generated (hence Noetherian), is $_SS$ necessarily Noetherian?

I can't figure it out... it feels like it should be true since $_SS$ contains $_RS$ and we have only finitely more things to worry about.

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«${}_SS$ contains ${}_RS$» is slightly confusing: the two things are the same set. –  Mariano Suárez-Alvarez Nov 13 '12 at 4:07

Every left ideal in $S$ is an $R$-submodule, so every ascending chain of left ideals in $S$ is also an ascending chain of let $R$-submodules of $S$. Since $S$ is a noetherian left $R$-module, such a chain must stabilize.
Alternatively, every left ideal in $S$ is finitely generated as a left $R$-module, so in particular it is finitely generated as a left ideal.