= Solution
The <inverse image> $I=\theta^{-1}(J)$ is an <ideal>, by preservation of differences and multiplication. It is proper since $\theta(1)=1\notin J$. If $ab\in I$, then $\theta(a)\theta(b)\in J$; primality of $J$ gives $\theta(a)\in J$ or $\theta(b)\in J$, hence $a\in I$ or $b\in I$. Therefore \b[the <inverse image> is prime].
For maximality, take the inclusion $\mathbb Z\hookrightarrow\mathbb Q$ and $J=(0)$. This is a <maximal ideal> of the <field> $\mathbb Q$, but its <inverse image> is $(0)$ in $\mathbb Z$, strictly contained in the proper <ideal> $2\mathbb Z$, so it is not maximal. Preservation of identity matters for proper inverse images: without it the zero homomorphism could pull a proper <ideal> back to the whole <ring>.
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