Preservation of ideals in localized rings

Refer to localization of rings in Lang's Algebra p. 108. In particular, let $A$ be a commutative ring, $S$ a subset of $A$ that is a submonoid of the multiplicative monoid structure of $A$, containing the identity. Then $S^{-1}A$ is the set of elements $\frac{a}{s}$ with $a \in A$ and $s \in S$. We have $\frac{a_1}{s_1}=\frac{a_2}{s_2}$ if there exists $\bar{s} \in S$ such that $\bar{s}(s_2a_1 - s_1 a_2)=0$. Now consider the injection of $A$ into $S^{-1}A$ given by $x \mapsto \frac{x}{1}$. This is a ring homomorphism. Let $J$ be an ideal of $A$. Ring homomorphisms preserve ideals and so the injection of $J$ into $S^{-1}A$, call it $J^{*}$, will be an ideal. My problem is that i don't see why for any $\frac{a}{s} \in S^{-1}A$ and any $\frac{j}{1} \in J^{*}$ we will have that $\frac{aj}{s}$ will be in $J^{*}$, i.e. why there must exist an element $\kappa \in J$ such that $\frac{aj}{s}=\frac{\kappa}{1}$.

Just a note, since Chris has already cleared things up. When people talk about moving ideals from $A$ to $S^{-1}A$, or more generally from $A$ to $B$ using a homomorphism $f\colon A \to B$, they really mean: take an ideal $\mathfrak{a}$ of $A$ and form the ideal $f(\mathfrak{a})B$ of $B$, i.e. the ideal of $B$ generated by the set $f(\mathfrak{a})$. This is called the extension of the ideal and is sometimes denoted by $\mathfrak{a}^e$, or $S^{-1}\mathfrak{a}$ in the case of localization.

For localizations the story of how ideals extend (and contract) is very interesting. See Proposition 3.11 of Atiyah-Macdonald or Proposition 6.4 of Milne's notes for a discussion of this.

The claim is false: ring homomorphisms do not preserve ideals. For example, take $A=\mathbb{Z}$ and $S=\mathbb{Z} \setminus \{ 0 \}$, so that $S^{-1}A = \mathbb{Q}$. Then $\mathbb{Z}$ is of course an ideal of itself, but its image is not an ideal of $\mathbb{Q}$.

Only prime ideals which are disjoint from S have 1 to 1 corresponds with prime ideals in $S^{-1}A$. There are no general mapping between ideals. BTW, I got problem 3.