Let $f: A \rightarrow B$ be a ring homomorphism. They symbols $c$ and $e$ are contraction and extension of an ideal. One of the result says that $\mathfrak{b}^{ce} \subset \mathfrak{b} $. I feel that the equality should hold since $\mathfrak{b}^{ce} = (f^{-1}(\mathfrak{b}) )^e = B f (f^{-1}(\mathfrak{b})) = B \mathfrak{b} = \mathfrak{b} $ (since $\mathfrak{b}$ is an ideal of $B$).
This is from chapter-1 of Atiyah and Macdonald- Commutative algebra book, proposition 1.17.
Best Answer
No, because if $f$ is not surjective it is still possible that $\mathfrak b\cap f(A)=\mathfrak a\cap f(A)$ for two distinct ideals of $B$. Consider for instance the map $f:R\to R[T]$, $f(x)=x$, $\mathfrak b=(T)$ and $\mathfrak a=0$.