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 $S$?
 A: 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 $R$.
A: 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.
A: It isn't true when $A=\mathbb Z$ and $S=\mathbb Z\setminus\{0\}$.
A: 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 $\rm\ S^{-1} A\ \cong A\:.$
