# Ring isomorphisms maintain noetherian property?

If a noetherian ring R is isomorphic to a ring S is S noetherian too?

I am pretty sure that it isn't but can't find a counterexample

Thanks!

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Hm, well if that's your first instinct, then your instincts need some re-calibration! When two rings are isomorphic, they share all of their ring-theoretic properties. The isomorphism says that one ring's elements are just a relabeling of the other ring's elements. All of the addition and multiplication lines up, and the ideals line up too. –  rschwieb Mar 19 '14 at 18:08
In fact you can say even more: if the ring homomorphism is from $R$ onto $S$ (not necessarily 1-1) and $R$ is Noetherian, then so is $S$. The same thing with onto and 1-1 transposed is not true, though. –  rschwieb Mar 19 '14 at 18:10

## 1 Answer

It is indeed. Suppose $\phi:R\to S$ is an isomorphism. If $P_1\subset P_2\subset\cdots$ is an ascending chain of ideals in $S$, then $\phi^{-1}(P_1)\subset\phi^{-1}(P_2)\subset\cdots$ is an ascending chain of ideals in $R$, and $P_n=P_{n+1}$ if and only if $\phi^{-1}(P_n)=\phi^{-1}(P_{n+1})$.

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