The ring homomorphic image of an ideal is an ideal

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.24
Solution:
(1) Let $x,y \in \varphi^\ast[J]$. Now $0 \in J$ and $\varphi(0) = 0$, so that $0 \in \varphi^\ast[J]$. By hypothesis, $$\varphi(x-y) = \varphi(x) - \varphi(y) \in J,$$ so that $\varphi^\ast[J]$ is closed under subtraction.
Now let $r \in R$. We have $$\varphi(rx) = \varphi(r)\varphi(x) \in J$$ since $J$ is an ideal of $S$; thus $rx \in \varphi^\ast[J]$. Similarly, $xr \in \varphi^\ast[J]$. Thus $\varphi^\ast[J]$ is an ideal of $R$.
The final deduction is immediate since $\iota^\ast[J] = J \cap R$.
(2) Note that $0 \in I$, so that $\varphi(0) \in \varphi[I]$. Now let $x,y \in \varphi[I]$. Now $x = \varphi(r)$ and $y = \varphi(s)$ for some $r,s \in I$. Since $$x-y = \varphi(r)-\varphi(s) = \varphi(r-s) \in \varphi[I],$$ $\varphi[I]$ is closed under subtraction.
Now let $s \in S$. Since $\varphi$ is surjective, there exists $t \in R$ with $\varphi(t) = s$. Now $$sx = \varphi(t) \varphi(r) = \varphi(tr) \in \varphi[I]$$ since $I$ is an ideal; thus $sx \in \varphi[I]$. Similarly, $xs \in \varphi[I]$. Thus $\varphi[I]$ is an ideal of $S$.
For the counterexample, consider the inclusion $\iota : \mathbb{Z} \rightarrow \mathbb{Q}$. $\mathbb{Z}$ is an ideal of itself but not of $\mathbb{Q}$ since $2 \cdot \frac{1}{3} = \frac{2}{3}$ is not an integer.