The Proposition is
Let $A\subseteq B$ be integral domains, $B$ integral over $A$. Then $B$ is a field iff $A$ is a field.
The proof is easy. I want to generalize this proposition. I want to prove that
Let $A\to B$ be integral domains, $B$ integral over $A$. Then $B$ is a field iff $A$ is a field.
One side is easy: Suppose $A$ is a field, let $y\in B, y\not = 0$. Since $B$ is integral over $A$, let $y^{n}+f(a_{1})y^{n-1}+\cdots+f(a_{n})=0$ $(a_{i}\in A)$ be an equation of integral dependence for $y$ of smallest possible degree. Then $a_n\not= 0$, so $y^{-1}=-f(a_{n}^{-1})(y^{n-1}+\cdots+f(a_{n-1}))\in B$, hence $B$ is a field. However, I do not know how to prove the other side: if $B$ is a field, then $A$ is a field.
Could you tell me how to prove it?
I think that this is not true: the canonical projection $$\mathbb Z \to \mathbb Z /p \mathbb Z$$ is a counter example.
Note that in the case $A$ is a field, your mapping must be one-to-one and hence $A$ is isomorphic to $f(A) \subseteq B$, which reduces the problem to the above one. This is why that implication works.
For the converse you run into troubles if $\ker(f) \neq \{0 \}$.