It is well-known and easily proved that whenever $R$ is a commutative ring with unity and $S$ is a multiplicative subset of $R$, each ideal of the localization ring $R_S$ is an extended ideal (with respect to the ring homomorphism taking any $r\in R$ to $r/1$).

I have three questions:

1) Is any nilpotent (nil) ideal of $R_S$ the extension of a nilpotent (resp. nil) ideal of $R$?

2) Is any idempotent ideal of $R_S$ the extension of an idempotent ideal of $R$?

3) May it be possible that extension of two distinct ideals of $R$ in $R_S$ (with respect to the homomorphism above) be equal?

It is not difficult to see that any nilpotent (idempotent) ideal extends to a nilpotent (idempotent) ideal. I think the answer to the second question is yes when $R$ is a domain.

Any point of suggestion is appreciated!


1) Let $a/s \in R_S$ be nilpotent. Then $a/1$ is nilpotent. Hence, there is some natural number $n$ and some $s \in S$ such that $s a^n = 0$. This implies $(s a)^n=0$, i.e. that $s a$ is nilpotent. We may write $a/s = (s a)/s^2$. The claim follows. (Both for nil and for nilpotent.)

3) Sure, $R$ and $Rs$ for $s \in S$.

  • $\begingroup$ I guess that the answer to 2) is No. But I haven't found a counterexample yet. $\endgroup$ – Martin Brandenburg Jan 14 '16 at 9:18

2) Let $R=K[X,Y]/(XY)$. This ring has no non-trivial idempotents (why?). On the other side, its total ring of fractions is isomorphic to a direct product of two fields (why?), so it has non-trivial idempotent ideals.

  • $\begingroup$ Note: The idempotent ideals in a commutative noetherian ring are principal and generated by some idempotent element (see SE/74900 for instance). So to show that $R$ has only trivial idempotent ideals, it suffices to show that it has only trivial idempotent elements. $\endgroup$ – Martin Brandenburg Jan 14 '16 at 10:37

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