# Characterization of ideals in rings of fractions

Let $R$ be a commutative unital ring. Let $S$ be a multiplicative subset.

Is there a characterisation of the ideals in the ring of fractions $S^{-1}R$ in terms of ideals $I$ in $R$ and $R$?

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I should think that the ideas of the $S^{-1}R$ should somehow depend on both $R$ and $S$. If $R$ is an integral domain and $S = R - \{0\}$, then $S^{-1}R$ is a field and has no nontrivial ideals. However if $S = \{1\}$, then the ideals of $S^{-1}R$ correspond to the ideals of $R$. – William Jun 13 '12 at 8:10
@William Good point, thank you! – Rudy the Reindeer Jun 13 '12 at 8:13

I don't believe in general you can say much about how ideals in $R$ are related to those in the localization. I will state two instances where we do have some correspondence:

Prime ideals:

The prime ideals of $S^{-1}R$ are in one to one correspondence with prime ideals of $R$ that don't meet $S.$ That is to say prime ideals $\mathfrak{p} \subset R$ such that $\mathfrak{p} \cap S = \emptyset$. The correspondence is given by $\mathfrak{p} \leftrightarrow S^{-1}\mathfrak{p}$. The proof of this fact is not hard, one direction is easy and for the other you just need to use the isomorphism $\overline{S}^{-1}(R/I) \cong S^{-1}R/S^{-1}I$. This isomorphism comes from applying $S^{-1}$ to the exact sequence

$$0 \longrightarrow I \longrightarrow R \longrightarrow R/I \longrightarrow 0$$

of $R$ - modules. By $\overline{S}$ I mean the image of $S$ in the quotient, which is still a multiplicative set.

Ideals $I\subset R$ such that $xs \in I$ implies that $x \in I$ for all $s \in S$:

It is not hard to show that for any ideal $J$ of $S^{-1}R$, we have that $(J^c)^e = J$ where $()^c$ denotes contraction, $()^e$ denotes extension. Now if we are given an ideal $I$ of $R$ instead, it is not true that $(I^e)^c = I$. What is true though (which is not hard to prove) is that

$$(I^e)^c = \{r \in R: rs \in I \hspace{2mm} \text{for some} \hspace{2mm} s \in S\}.$$

It is easy to see from here that the processes of extension and contraction define the following bijective correspondence:

$$\{ \text{ ideals I of R | xs \in I \implies x \in I for all s \in S}\} \leftrightarrow \{ \text{ideals of S^{-1}R}\}.$$

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Thank you! And can we also say something about non-prime ideals? – Rudy the Reindeer Jun 13 '12 at 8:31
@MattN. Please see my edit. – user38268 Jun 13 '12 at 8:33
Ooh, this is the stuff about contractions and extensions I'm about to read in the next chapter! I should be more patient : ) You are awesome. Thanks! – Rudy the Reindeer Jun 13 '12 at 8:34
@MattN. No problem. You have flooded math.se with a lot of commutative algebra posts recently! By the way for your tensor products question, I have posted an answer showing how to show the isomorphism using the universal property of the tensor product. – user38268 Jun 13 '12 at 8:36
So the down vote is from you as a punishment for flooding SE with CA questions? I'll have a look at it later, I'm in quite a rush to get through the book. – Rudy the Reindeer Jun 13 '12 at 8:38

Proposition: Let $R$ be a commutative ring with unity. Proper ideals of the ring of fractions $D^{-1}R$ are of the form $\displaystyle D^{-1}I = \bigg\{ \frac{i}{d} : i \in I,\ d \in D\bigg\}$ with $I$ an ideal of $R$ and $I \cap D = \emptyset$.

Proof: Let $J$ be a proper idea of $D^{-1}R$. Let $I = J \cap R$ and observe that $I$ is an ideal of $R$. Suppose to the contrary that $I \cap D \neq \emptyset$. Let $d \in I \cap D$ then $d \in I$. Observe that $\displaystyle \frac{d}{1} \in J$. Moreover, since $J$ is an ideal, it must absorb any element from $D^{-1}R$. Observe that $\displaystyle \frac1d \in D^{-1}R$. Hence it must follow that $\displaystyle \frac1d \cdot \frac{d}{1} = 1 \in J$ and thus $J$ contains a unit which implies $J = D^{-1}R$, a contradiction to $J$ being proper. Thus $I \cap D = \emptyset$.

Let $j \in J$. Observe that $\displaystyle j = \frac{i}{d} = \frac{1}{d}\frac{i}{1}$ for some $i \in R$, $d \in D$. Since $J$ is an ideal of $D^{-1}R$ it must absorb $\displaystyle \frac{d}{1}$ and thus $\displaystyle \frac{d}{1}\bigg(\frac{1}{d}\frac{i}{1}\bigg) = \frac{i}{1} \in J$. Now since $I = J \cap R$ and $\frac{i}{1} = i \in J \cap R$, it follows that $i \in I$. Hence $J \subseteq D^{-1}I$.

Let $x \in D^{-1}I$ where $\displaystyle x = \frac{i}{d}$ for some $i \in I$ and some $d \in D$. Since $i \in I = J \cap R$, then $i \in J$. Hence $x \subseteq D^{-1}I$.

Thus we can conclude that $J = D^{-1}I$.

As an example you might like to consider $R = \mathbb Z$ and $D = \{12^i : i = 0, 1, 2, \ldots\}$. You can see that $I = D^{-1}2\mathbb Z$ is not a proper ideal because you will get a unit with $r = 6$, $i = \frac{2}{12}$.

Proposition: Let $R$ be a commutative ring with unity. Prime ideals in $D^{-1}R$ are of the form $D^{-1}P$ where $P$ is a prime ideal of $R$ and $P \cap D = \emptyset$.

Proof:Suppose $Q$ is a prime ideal of $\displaystyle D^{-1}R = \bigg\{ \frac{r}{d} : r \in R,\ d \in D \bigg\}$. Set $P = Q \cap R$. Suppose to the contrary that $P \cap D \neq \emptyset$. Choose $d \in P \cap D$. Observe that $\frac{d}{1} \in Q$. Moreover $\frac1d \in D^{-1}R$. Since $Q$ is a prime ideal, it follows that $\frac1d \frac{d}{1} = 1 \in Q$ which implies that $Q = D^{-1}R$ a contradiction to $Q$ being a prime ideal, which by definition is proper. Hence $P \cap D = \emptyset$.

Let $q \in Q$. Then $q = \frac{r_1}{d_1}$ for $r_1 \in R$ and $d_1 \in D$. Observe that $\frac{d_1}{1} \in D^{-1}R$. So, py property of $Q$ being an ideal, $\frac{d_1}{1} \frac{r_1}{d_1} = \frac{r_1}{1} \in Q$ and since $P = Q \cap R$, then $r_1 \in P$. Hence we have $\frac{r_1}{1} \in D^{-1}P$. Thus $Q \subseteq D^{-1}P$.

Let $x \in D^{-1}P$. Then $x = \frac{p_1}{d_1}$ where $p_1 \in P$ and $d_1 \in D$. Since $P = Q \cap R$ and $p_1 \in P$ it follows that $p_1 \in Q$ so $x \in Q$. Hence $D^{-1}P \subseteq Q$.

Conclude that $Q = D^{-1}P$.

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Using a similar argument as I did for proper ideals, I showed the case for prime ideals which @BenjaLim showed. – Robert Cardona Feb 7 '13 at 20:11