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If every proper subring is noetherian then the ring is noetherian ?

For example every proper subring of $\mathbb{Z} $ is $m\mathbb{Z}$ and we know that $m\mathbb{Z} $ is noetherian.

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    $\begingroup$ Pardon mee, but $m\mathbf Z$ is usually considered an ideal. In commutative algebra, rings and subrings usually have a multiplicative unit. $\endgroup$ – Bernard Jan 8 '17 at 10:07
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If you require that the ring $R$ has a unit but do not require your subrings to contain the unit, then this is trivial. Indeed, if $R$ is any non-Noetherian unital ring, let $I\subset R$ be a non-finitely generated proper ideal (a non-finitely generated ideal in $R$ must be proper since $R$ has a unit). Then $I$ is a non-Noetherian subrng of $R$ (since $I$ itself is a non-finitely generated ideal of $I$).

If you do not require $R$ to have a unit, then here's a counterexample. Let $R=\mathbb{Z}[1/p]/\mathbb{Z}$ equipped with the product such that $rs=0$ for all $r,s\in R$. Then $R$ is not Noetherian, since it is not finitely generated as an ideal over itself. But any proper subrng is Noetherian, since any proper additive subgroup of $R$ is finite.

If you require $R$ to have a unit and the subrings to contain the unit, you can get a counterexample by unitizing the previous example. That is, let $S=\mathbb{Z}\oplus R$ be the unitization of the rng $R$ from the previous paragraph. Then $S$ is not Noetherian, since $R$ is an ideal of $S$ which is not finitely generated. But every (unital) subring of $S$ has the form $\mathbb{Z}\oplus A$ for some additive subgroup $A\subseteq R$, and so any proper such subring is Noetherian (since, for instance, it is finitely generated as an abelian group).

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