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Let $A$ be a commutative ring and $S$ a submonoid of the multiplicative structure of $A$. Then consider the quotient ring of $A$ by $S$, denoted $S^{-1}A$, see e.g. Lang p.107. Is it true that there is a bijection between the ideals of $A$ and the ideals of $S^{-1}A$? I have proved so, but i want to verify it.

Thanks.

Added(1): Consider the map that takes an ideal $\alpha$ of $A$ to the ideal of $S^{-1}A$ denoted $S^{-1} \alpha = \left\{\frac{x}{s} : x \in \alpha, s \in S\right\}$. Is this map surjective?

Added(2): Is the above mentioned map injective only if restricted on ideals that are prime and do not intersect with $S$?

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    Let $A$ be an integral domain and $S$ be the set of nonzero elements of $A$...2011-12-13

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Not quite...if $I\subseteq A$ is an ideal and if $s\in I\cap S$, then $\frac{s}{1}\in U(S^{-1}A)$ (note that I'm assuming that $U(A)\subseteq S$. This is a harmless assumption (why?)).

What is true, though, is that the mapping $P\mapsto S^{-1}P=\{\frac{p}{s}\,\vert\,p\in P,s\in S\}$ from the set of prime ideals of $A$ that have empty intersection with $S$ to the prime ideals of $S^{-1}A$ is a bijection.

In fact, a subset $S$ of $R\setminus \{0\}$ is multiplicative and saturated (ie it contains all of its divisors) precisely when $S$ is the set-theoretic complement of a union of prime ideals--ie $S=R\setminus \left(\bigcup_{P\in \mathcal{P}}P\right)$ for some family $\mathcal{P}$ of prime ideals of $R$.

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    It is a harmless assumption because elements of $S$ are invertible when injected in $S^{-1}A$?2011-12-13
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    Hmmm...not quite. Let's think of a particular example, say $A=\mathbb{Z}$ and $S=\{2^n\,\vert\,n\geq 1\}$. Note that $1\notin S$. Can you find a fraction in $S^{-1}A$ that ends up having a denominator of $1$?2011-12-13
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    Yes, because $\frac{2}{2} \in S^{-1}A$ and $\frac{2}{2}=\frac{1}{1}$. So $\frac{1}{1} \in S^{-1}A$.2011-12-13
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    Right. Now, generalize.2011-12-13
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    I am confused: i just showed that $\frac{1}{1} \in S^{-1}A$, which implies that $1 \in S$. But we assumed that $1$ is not in $S$. Contradiction?2011-12-13
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    I seem to remember that you can form $(S^{-1}I) \cap A$ and get back something larger than $I$, even if $S^{-1}I$ is not the unit ideal. Will try to think of an example.2011-12-13
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    I think i realize the point of my confusion: the fact that $\frac{2}{2}$ is equivalent with $\frac{1}{1}$ does not mean that $S$ contains all representatives of this equivalence class. So the fact that $\frac{1}{1} \in S^{-1}A$ does not mean that $1 \in S$.2011-12-13
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    Right, so in the example, let $S_1=1\cup S=\{2^n\,\vert\,n\geq 0\}$. Is $S^{-1}A$ the same as $S_1^{-1}A$?2011-12-13
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    @JackManey: The reason that we can assume that $U(A) \subset S$ is the following: we defined $\frac{a}{x}=\frac{\tilde{a}}{\tilde{s}}$ if there exists $\bar{s} \in S$ such that $\bar{s}(a \tilde{s} - \tilde{a} s)=0$. Now this relation does not change if we multiply it by a unit of $A$. What do you think?2011-12-13
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    @JackManey: Yes they are the same :-)2011-12-13
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    @JackManey: Jack, in order to have a bijection we need the ideal $I$ to have empty intersection with $S$. Do we also need it to be prime?2011-12-13
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    Here's an example of what I was talking about. Take $A = \mathbf Z$, $S = A - (2)$, $I = (6)$. Then $I \cap S = \emptyset$, and yet $S^{-1}I$ is equal to the extension of $(2)$ to $S^{-1}A = A_{(2)}$. I think. If you only consider prime ideals, then you are definitely safe.2011-12-13
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    @DylanMoreland: Interesting. What do you think about the statement Jack made regarding the bijection? Do we need the ideals to prime?2011-12-13
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    @Manos I don't think we need to restrict to primes, although that's certainly the case of most interest. I think it's enough for $S$ to not have any zero-divisors mod $I$, i.e. $s \in S$, $a \in A$ and $sa \in I$ imply $a \in I$. I could add a supplementary answer, or Jack could add to his (assuming I'm right). This is definitely in Eisenbud's book somewhere.2011-12-13
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    @DylanMoreland: But then if we do not restrict to prime ideals, our map is not injective.2011-12-13
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    @Manos Can you prove that? Restricting to prime ideals gives a bijection, but I'm not yet convinced that you can't expand the domain slightly.2011-12-13
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    @DylanMoreland: I can't. But i can't prove that just the condition of empty intersection with $S$ gives an injection.2011-12-13
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    @Manos It isn't injective in general, so that's good. I'm just saying that there's probably an intermediate restriction that we can use.2011-12-13
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    @DylanMoreland: That's what i wanted to clarify. So strictly speaking, in order to have a bijection, we need a stronger condition on the ideals than having empty intersection with $S$.2011-12-13
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    @Manos I think my example above shows this, doesn't it?2011-12-13
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    @DylanMoreland: Yes it does.2011-12-13
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    @JackManey: Jack, can you please edit your answer regarding the primality of the ideals for the bijection to exist, and i will choose it as the best answer.2011-12-13
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    Sorry, work's been busy. I'll get back to this later this afternoon, but upon further reflection, yes, I think that primality may be necessary.2011-12-13
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    @Manos - Sorry again for the delay. Going back to $S=\{2^n\,\vert\,n\in \mathbb{N}\}$ and $A=\mathbb{Z}$, if we let $I=6A$ and $J=12A$, then $I_S=J_S=3A_S$. I'll make the edit for prime ideals.2011-12-14
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It isn't true when $A=\mathbb Z$ and $S=\mathbb Z\setminus\{0\}$.

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    A counterexample?2011-12-13
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    @Manos: Yes. Qiaochu's comment generalizes this counterexample, and Jack Maney's first sentence generalizes it even further.2011-12-13
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HINT $\ $ A ring hom $\rm\:h\:,\:$ injective on ideals, preserves nonunits: $\rm\: (n)\ne (1)\ \Rightarrow\ h(n)\ne (1)\:.\:$

This is contra the point of localization - to ignore certain nonunits by mapping them to units. Thus the inclusion hom from $\rm\:A\:$ into $\rm\: S^{-1}\:A\:$ is never injective on ideals, except in the trivial case when it adjoins no new inverses, i.e. when all elements of $\rm\:S\:$ are aleady units in $\rm\:A\:,\:$ so $\rm\ S^{-1} A\ \cong A\:.$

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    The monoid homomorphism that takes the ideal $\alpha$ of $A$ to the ideal $S^{-1} \alpha$ of $S^{-1}A$ is injective only if $\alpha, S$ have empty intersection and $\alpha$ is prime?2011-12-13
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Let me use the old notation for extension and contraction along the canonical homomorphism $A \to S^{-1}A$. It's true that any ideal $J$ of $S^{-1}A$ is an extended ideal, and in fact $J^{ce} = J$ (try proving this). Now the question becomes: which ideals $I$ of $A$ are contractions?

If $I = J^c$ then certainly $I^e \subset J$, and it's also clear that $I \subset I^{ec}$. So we really want to characterize $I$ such that $I = I^{ec}$, and it isn't enough to require that $I$ not meet $S$; for example, take $A = \mathbf Z$, $S = \mathbf{Z} - 2\mathbf Z$, and $\mathfrak{a} = 6\mathbf Z$. Then $2 \in I^{ec}$.

Now, an element $a \in A$ is in $\mathfrak{a}^{ec}$ if and only if $a/1 = x/s$ for some $x \in I$ and $s \in S$. Writing out the equivalence relation on fractions, this happens if and only if there exists an $s_1 \in S$ such that $s_1(sa - x) = 0$. Equivalently, there exists an $s' \in S$ such that $s'a \in I$. It follows that $I = I^{ec}$ if and only if the image of $S$ in $A/I$ consists of non-zerodivisors.

As a special case of this we see that prime ideals $P$ of $A$ which do not meet $S$ are contractions, since each $A/P$ is a domain.

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    There is some problem with the typing...2011-12-13
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    @Manos Could you be more specific? Where?2011-12-13
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    Maybe it's a problem with my browser. Some symbols seem to be missing.2011-12-14
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    @Manos Hm. Folks in the chat say it's fine. Apparently pressing shift-reload or clearing the cache can help with MathJax issues.2011-12-14
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    I'll study your answer tomorrow. Thanks.2011-12-14
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    I think i know where the problem is: for some reason by browser does not read the "\mathfrak" command. Can you use some alternative or just plain font?2011-12-14
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    @Manos I see. I will switch to $I$'s and $J$'s. This stuff is all in the relevant chapters of Atiyah-Macdonald and Eisenbud, alternatively.2011-12-14