When $f(I)S=S$ for each ideal $I$ of $R$? Let $R$ and $S$ be commutative rings (with $1$) and $f : R\to S$ be a ring homomorphism. For an ideal $I$ of $R$, set $I^e:=\langle f(I)S\rangle$ (called the extension of $I$ to $S$).

Question 1. What (non-trivial) conditions can be posed on rings or the homomorphism to have $I^e=S$ for any non-zero ideal $I$ of $R$?
Question 2. Are there special rings and homomorphisms that have this property?

 A: There are generally two cases to consider. If you define everything in a very clever way, they are actually just one case, but I think it is helpful to distinguish here.
Case 1: $R$ contains zerodivisors. In this case, let $x\in R$ be a zerodivisor, i.e. $x\ne 0$ and there is $0\ne y\in R$ with $xy=0$. Since you require that the ideal generated by both $f(x)$ and $f(y)$ is equal to $S$, you have $1\in\langle f(x)\rangle$ and $1\in\langle f(y)\rangle$, so there are $a,b\in S$ with $1=af(x)=bf(y)$. This implies 
$$0=a\cdot b\cdot 0=a\cdot b\cdot f(xy)=a\cdot f(x) \cdot b\cdot f(y)=1\cdot 1 = 1$$
in $S$, so $S$ is the zero ring and $f$ is the zero map.
Case 2: $R$ is an integral domain and we assume that $f$ is not the zero map. In this case, let $K=\{ \frac ab \mid a,b\in R, b\ne0\}$ be the field of fractions of $R$. Then, there is an injective homomorphism $F:K\to S$ such that $F|_R=f$. In particular, $f$ is injective in this case and $S$ is a ring extension of $K$. To see this, take any nonzero element $x\in R$. Then, $1\in \langle f(x)\rangle$ by assumption, so $f(x)$ is a unit in $S$. The universal property of the localization yields the existence of $F:K\to S$ with $F|_R=f$. Since $K$ is a field, $F$ must be injective: This follows because $\ker(F)$ is an ideal of $K$, and the only ideals of $K$ are $\{0\}$ and $K$ itself. Since $f$ is not the zero map, $\ker(F)=\{0\}$.
In other words, the only kinds of interesting morphisms with this property will be morphisms where $R$ is an integral domain and that map from $R$ to a ring extension of $K$.
