Is there a bijection between the ideals of a ring and the ideals of its ring of quotients?

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 $A$... – Qiaochu Yuan Dec 13 '11 at 17:44

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 $S^{-1}A$? – Manos Dec 13 '11 at 17:56
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 $1$? – user5137 Dec 13 '11 at 18:00
Yes, because $\frac{2}{2} \in S^{-1}A$ and $\frac{2}{2}=\frac{1}{1}$. So $\frac{1}{1} \in S^{-1}A$. – Manos Dec 13 '11 at 18:03
Right. Now, generalize. – user5137 Dec 13 '11 at 18:04
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? – Manos Dec 13 '11 at 18:09

It isn't true when $A=\mathbb Z$ and $S=\mathbb Z\setminus\{0\}$.

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A counterexample? – Manos Dec 13 '11 at 17:44
@Manos: Yes. Qiaochu's comment generalizes this counterexample, and Jack Maney's first sentence generalizes it even further. – Jonas Meyer Dec 13 '11 at 17:46

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? – Manos Dec 13 '11 at 19:42

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... – Manos Dec 13 '11 at 23:35
@Manos Could you be more specific? Where? – Dylan Moreland Dec 13 '11 at 23:35
Maybe it's a problem with my browser. Some symbols seem to be missing. – Manos Dec 14 '11 at 2:56
@Manos Hm. Folks in the chat say it's fine. Apparently pressing shift-reload or clearing the cache can help with MathJax issues. – Dylan Moreland Dec 14 '11 at 4:55
I'll study your answer tomorrow. Thanks. – Manos Dec 14 '11 at 5:20