# Does $S^{-1}I \subset S^{-1}J$ imply $I \subset J$?

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...

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 $$I \subseteq J$$.

• I think $S$ contains all the powers of $2$. – user26857 May 4 at 22:17
• @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 $S$. – Clive Newstead May 5 at 12:00

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 $$a \in J$$.