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Let $S$ be a multiplicative subset of a commutative ring with identity, and consider the ring of fractions $S^{-1}R$. Ideals in $S^{-1}R$ of are of the form $S^{-1}I$, where $I$ is an ideal in $R$.

Suppose $S^{-1}I$ and $S^{-1}J$ are two ideals in $S^{-1}R$, where $I$ and $J$ are ideals in $R$, and further suppose that $S^{-1}I \subset S^{-1}J$. Does it then follow that $I \subset J$?

This is really bugging me. If $\varphi^{-1}(S^{-1}I)$ is the contraction of $S^{-1}I$ in $R$, then we have that $I \subset \varphi^{-1}(S^{-1}I)$ (and similarly for $J$); but this doesn't get me what I want, obviously. So, let's give this a shot: Let $a \in I$. Then, $a/s \in S^{-1}I$; and so $a/s \in S^{-1}J$. But this (apparently) does not imply that $a \in J$, and I can't seem to figure out why...

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Consider $R = I = \mathbb{Z}$, $J=(2)$ and $S = \{ 2 \}$. Then $S^{-1}I = S^{-1}J = \{ \frac{a}{2^n} \mid a \in \mathbb{Z},~ n \in \mathbb{N} \}$, but we certainly don't have $I \subseteq J$.

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  • $\begingroup$ I think $S$ contains all the powers of $2$. $\endgroup$ – user26857 May 4 at 22:17
  • $\begingroup$ @user26857: That would have the same effect: if $S$ isn't multiplicatively closed, then $S^{-1}R$ is taken to mean $\bar S^{-1} R$, where $\bar S$ is the multiplicative closure of $S$. $\endgroup$ – Clive Newstead May 5 at 12:00
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Hint:

Suppose that $J \cap S \not= \emptyset$.

Then for any ideal $I$, $$S^{-1} I\subseteq S^{-1}J = S^{-1}R$$

You can use this observation to construct counterexamples with ease.

However, it's worth noting that the claim does hold in the important case that $J$ is a prime ideal of $R$ disjoint from $S$, since in that case $a \in I$ and the assumption that $S^{-1}I \subseteq S^{-1}J$ implies $sa \in J$ for some $s \in S$. Then $S \cap J = \emptyset$ implies $s \notin J$, and primeness of $J$ implies $a \in J$.

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