# Extension of Noetherian rings [closed]

Let $A \subset B$ rings. If $B$ is noetherian then $A$ is noetherian. Is false or true?

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## closed as off-topic by user26857, Andres Caicedo, Davide Giraudo, RecklessReckoner, Mike MillerAug 6 '14 at 22:58

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This is false in general. Take $A$ to be an integral domain, and $Q(A) = B$ its field of fractions. $B$ is trivially Noetherian since it has only the zero ideal ; but not all integral domains are Noetherian.
The problem is essentially that the $A$-submodules of $A$ are not $B$-submodules of $B$ in general, so you can't really use the Noetherianness of $B$ to conclude anything.