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Let $\phi: A \to B$ an inclusion of commutative rings, $I \subset A$ an ideal of $A$, $J \subset B$ an ideal of $B$.
There are two natural operations on the ideals of $A$ and $B$, contraction $J^c := J \cap A = \phi^{-1}(J)$ and extension $I^e:= \phi(I) \cdot B$.

Trivially we have relations $(J^c)^e \subset J$ and $I \subset (I^e)^c$. Natural question is when these inclusions are even equalities?

The 2nd case has rather satisfying answer: Namely there is a rather nice class of ring extensions for which $I = (I^e)^c$ if $\phi:A \to B$ always holds: the faithfully flat extensions, see eg here.
#EDIT: But attention, as Martin Brandenburg indicated in a comment beeing faithfully flat is not equivalent to have equality $I = (I^e)^c$ for all ideals of $A$, but still it's a "nice" class of ring extensions which carry this feature. This also motivates:

Question: What about first case, under which non trivial conditions on $\phi$ holds $(J^c)^e = J$?
Are there some interesting class(es) of ring extensions which have this property in similar spirit faithfully flat extensions share the property forcing $I = (I^e)^c$.
For exactly, there are for example the "boring" case of quotients $B=A/I$. Another trivial example as Lukas Heger remarked in comments are field extensions. So beeing injective is not an exclusive criterion.

But can one say about which common features such classes of ring extensions would consequently share which imply this property in nearly similar vein/analogy as in case of faithfully flat extensions towards $I = (I^e)^c$, even though faithfully flat is stronger then having this property?

EDIT#:( As Lukas Heger remarked this part is wrong; I don't want to delete it completely as this at least inducates what I roughly looking for when I'm asking for their common features; these should play the role analogous to that faithfully flat extensions play with respect to condition $I = (I^e)^c$ ) [Asking differently, can it for example be showed that if $\phi$ is injective, the inclusion $(J^c)^e \subset J$ can never be an equality if $\phi$ is not an iso? (This would also confirm the intuition that quotient maps $A \to A/I$ can never be faithfully flat except $I =0$; In that sense that quotient maps are "opposite" to faithfully flat maps). Philosophy: This would give dual notion of faithfully flat maps. This question generalizes this one. ]

user267839
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    Clearly every field extension satisfies this. Thus there are injective examples which are not isos. – Lukas Heger Nov 13 '24 at 20:08
  • @LukasHeger: oh right, that's a good observation. But still, is it possible to say more about common properties the class of extensions for which $(J^c)^e = J$ holds. Do they consequently have some other distinguished features in common? (Your example shows that it was a bad way to expect them to be "opposite" to faithfully flat extnsions, but still I expect that they may also take some special role in ring theory...) – user267839 Nov 13 '24 at 21:00
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    The property that ideals of A are contractions is not equivalent to being faithfully flat. Hence I would suggest the change the wording, both in the title and the body of the post. See also the recent question https://math.stackexchange.com/questions/4967187/algebras-with-an-extension-contraction-property – Martin Brandenburg Nov 14 '24 at 01:07
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    We should not assume that the map is injective, it's not flexible enough. I assume you also know that the property (every ideal of B is an extension) holds for every localization. Also these ring homs are closed under composition. Not sure if we can classify them. Same as in the linked question. My first guess was that they are the ring epimorphisms. – Martin Brandenburg Nov 14 '24 at 01:08

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