Extension and Contraction
Let $f: R\rightarrow S$ be a ring homomorphism. Let $I \triangleleft R, J \triangleleft S$. The extension $I^e$ of $I$ (with respect to $f$) is $\langle f(I) \rangle$. Then contraction $J^c$ of $J$ (with respect to $f$) is $f^{1}(J)$, which is an ideal in $R$.
Note that if $J$ is prime than so is $J^c$, though the same is not always true for extensions. For example, take the identity map $f:\mathbb{Z} \rightarrow \mathbb{Q}$. Then for any prime $p$, $p\mathbb{Z}$ is prime in $\mathbb{Z}$ but $(p\mathbb{Z})^e = \mathbb{Q}$ which is not prime in $\mathbb{Q}$.
In general there is no simple relationship between the prime ideals of $R$ and $S$. For example consider the identity map $f:\mathbb{Z}\rightarrow\mathbb{Z}[i]$ (see notes on <a href="../numberfield"> number fields</a>).
Other properties of extension and contraction:

$I\subset I^{e c}, J\supset J^{c e}$

$J^{c} = J^{c e c}, I = I^{e c e}$

Let $\mathcal{C}$ be the set of contracted ideals in $R$ and $\mathcal{E}$ be the set of extended ideals in $S$. Then $\mathcal{C} = \{ K\triangleleft R  K^{e c} = K\}$, $\mathcal{E} = \{ L\triangleleft S L^{c e} = L\}$, and $K \mapsto K^{e}$ for all $K \in \mathcal{C}$ defines a bijection $\mathcal{C} \rightarrow \mathcal{E}$ whose inverse is $L \mapsto L^{c}$ for all $L \in \mathcal{E}$.
Let $I_1,I_2\triangleleft R$ and $J_1,J_2 \triangleleft S$. Then

$(I_1 + I_2)^e = I_1^e + I_2^e$, $(J_1 + J_2)^c \supset J_1^c + J_2^c$

$I_1 \cap I_2)^e \subset I_1^e \cap I_2^e$, $(J_1 \cap J_2)^c = J_1^c \cap J_2^c$

$(I_1 I_2)^e = I_1^e I_2^e$, $(J_1 J_2)^c \supset J_1^c J_2^c$

$(I_1 :I_2)^e \subset (I_1^e :I_2^e)$, $(J_1 :J_2)^c \subset (J_1^c :J_2^c)$

$(\sqrt{I_1})^e \subset \sqrt{I_1^e}$, $(\sqrt{J_1})^c = \sqrt{J_1^c}$
The set $\mathcal{E}$ is closed under sum and product while the set $\mathcal{C}$ is closed under intersection, forming ideal quotients and taking radicals.