I proved the following statement, but I am very unsure that it is correct, since this proposition is not stated in my books for general ideals but only for prime ideals. Please point out where the mistake is if it is incorrect. ($I^e$ denotes the extended ideal of I which also equals $S^{-1}I$, $J^c$ denotes the contraction of $J$)

Proposition: Let R be a ring and let S be an multiplicative set. The ideals of $S^{-1}R$ are in one to one correspondence with the ideals I of R which do not meet S. The correspondence is $I \Leftrightarrow S^{-1}I$. Under this corresponcence prime ideals correspond to prime ideals.

Proof: Let $I$ be an ideal of $R$ which do not meet $S$. For any $\frac{a}{b}, \frac{c}{d} \in S^{-1}I$ and any $\frac{r}{s} \in S^{-1}R $, $\frac{a}{b}\frac{r}{s}=\frac{ar}{bs}$ is in $ S^{-1}I$ since $ar\in I$ and $cs\in S$. Futhermore, $\frac{a}{b}+ \frac{c}{d}= \frac{da+bc}{db}$ is an element in $ S^{-1}I$ by the same argument. Converserly, let $J$ be any ideal of $S^{-1}R$, if $\frac{x}{s}\in J$ then $\frac{x}{1}\in J$, hence $x \in J^c$, (with respect to the homomorphism $\Phi: R \rightarrow S^{-1}R$, given by $\Phi(x)=\frac{x}{1}$), hence $J\subseteq J^{ce}$, further it is easy to check that $J^{ce}\subseteq J$ for any homomorphism. Hence $J=J^{ce}$. Since $J^{c}=I$ for some ideal $I$ of $R$ we have that $J=S^{-1}I$. Furthermore if $I$ is prime in$R$ it is easy to see that $S^{-1}I$ is prime in $S^{-1}R$. Converserly, if $J$ is any prime ideal $J \subset S^{-1}R$ then $J=S^{-1}I$ some ideal $I$ of $R$, if $ac\in I$ then by construction $\frac{ac}{bd} \in S^{-1}I$. Since $J$ is prime this imply that $\frac{a}{b}$ or $\frac{c}{d}$ is in $J$. This imply that $a$ or $c$ is in $I$

  • $\begingroup$ I see no problem here. Your book may focus on the prime ideals because that's all it takes to show the localization is actually a local ring. $\endgroup$ Nov 6 '12 at 0:49
  • 1
    $\begingroup$ If you remove the "one-to-one" part, then the description of the ideals of $S^{-1}R$ is correct. $\endgroup$
    – user26857
    Dec 2 '12 at 11:57

Your proposition is not true in general.

($R$ is commutative) Take an ideal $I$ with $S\cap I=\emptyset$ and $S^{-1}I\ne S^{-1}R$. Let $s\in S$ not invertible in $R$. Then $sI\cap S=\emptyset$ and, in general, $sI\ne I$. But $S^{-1}(sI)=S^{-1}I$.

Concrete example: $R=\mathbb C[t]$, $S=\mathbb C[t]\setminus t\mathbb C[t]$, $I=t\mathbb C[t]$. Then $S^{-1}tI=S^{-1}I$ but $tI\ne I$.

As pointed out by YACP in the comment, every ideal of $S^{-1}R$ is of the form $S^{-1}I$ for some ideal $I$ of $R$ and your proof is correct. The mistake is on the uniqueness of $I$.

Edit If we restrict to the ideals $I$ of $R$ such that all $s\in S$ are regular in $R/I$ (i.e. for all $s\in S, x\in R$, if $sx\in I$ then $x\in I$). Then they are in 1-1 correspondence with the ideals of $S^{-1}R$.

  • $\begingroup$ Minor nitpick: $I \neq 0$ is necessary to ensure $sI \neq I$. $\endgroup$ Dec 2 '12 at 10:24
  • $\begingroup$ @NilsMatthes: yes thanks. I edited the answer. $\endgroup$
    – user18119
    Dec 2 '12 at 16:12

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