# Is $\mathfrak{b}^{ce} = \mathfrak{b}$ where $c$ and $e$ are contraction and extension of an ideal.

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.

• For some important cases, the answer is yes, e.g. if $A \rightarrow B$ is surjective or is a localization (or more generally a flat epimorphism). Similarly, $I^{ec} = I$ does not generally hold for ideals $I$ of $A$, but also in some important cases it does, e.g. if $A \rightarrow B$ is faithfully flat (or more generally, universally injective). May 13, 2020 at 17:37

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 $$\mathfrak a=0$$.
• Just wanted to where I went wrong in my answer : $f(f^{-1} (\mathfrak{b})) \subset \mathfrak{b}$ (equality need not hold necessarily), when $f$ is not a surjective function.
• @MUH Exactly. For surjective functions $f(f^{-1}(\mathfrak b))=\mathfrak b$.