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.

  • $\begingroup$ 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). $\endgroup$ May 13, 2020 at 17:37

1 Answer 1


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 $\mathfrak a=0$.

  • $\begingroup$ 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. $\endgroup$
    – MUH
    May 13, 2020 at 17:32
  • $\begingroup$ @MUH Exactly. For surjective functions $f(f^{-1}(\mathfrak b))=\mathfrak b$. $\endgroup$
    – user239203
    May 13, 2020 at 17:35

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