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We know that a subring of a noetherian ring is not necessarily noetherian.

Indeed, if we take $A:=k[X_1,\ldots,X_n,\ldots]$ where $k$ is a field, then $A$ is a non-noetherian subring of $K:=k(X_1,\ldots,X_n,\ldots)$ (and $K$ is a noetherian ring because it's a field).

So the question is:

Does there exists a subring of $\mathbb Q$ which is not noetherian?

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I'm assuming every subring contains the multiplicative identity. Every subring $R$ of $\mathbb{Q}$ is localization of $\mathbb{Z}$ at some multiplicatively closed subset. (Take $S$ to be the set of all denominators in the fully reduced representation of elements in $R$, then it is easy to see that $S^{-1}\mathbb{Z} = R$)

The localization of a Noetherian Ring is Noetherian, thus every subring of $\mathbb{Q}$ is Noetherian.

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  • $\begingroup$ Nice answer. +1 $\endgroup$ – Xam Mar 17 '17 at 2:58

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