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Let $R^d :=k[X_1,..., X_d] $ (resp. $R^e :=k[Y_1,..., Y_e] $) be polynomial rings in $d$ (resp. $e$) indeterminants over a field $k$ (maybe, it's ok if we more generally consider $k$ to be a Noetherian ring).

Let $I=\langle f_1,..., f_n \rangle $ be an ideal (generated by $f_i$'s) of $R^e \otimes_k R^f \cong k[X_1,..., X_e, Y_1,..., Y_f]$.

Consider the canonical inclusion $j:k[X_1,..., X_e] \to k[X_1,..., X_e, Y_1,..., Y_f]$ and the "contraction" of ideal $I^{c(e)}:= k[X_1,..., X_e] \cap I =j^{-1}(I)$.

Is there a well studied theory adressing the question & developing criteria when $I^{c(e)}$ "gives the original ideal $I$ back", ie when $I^{c(e)} \cdot k[X_1,..., X_e, Y_1,..., X_f]=I$?

Geometrically, this corresponds to scenario when for closed subscheme $V(I) \subset \Bbb A_k^{e+d}$ for canon projection $p_e: \Bbb A_k^e \times \Bbb A_k^d \to\Bbb A_k^e$ we get $V(I) =p_e^{-1}(p_e(V(I))$

user267839
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  • What is $R^f$? $k[Y_1,\cdots,Y_f]$ presumably, but there are some typos so it's hard to tell -- e.g. $k[X_1,\cdots,X_e,Y_1,\cdots,\mathbf{X_f}]$. – walkar Nov 13 '24 at 14:04
  • Also, your question has an answer here; https://math.stackexchange.com/a/179071/98077, you are simply not using a field extension. Over a general Noetherian ring, I am not sure if the same answer goes through. – walkar Nov 13 '24 at 14:05
  • @walkar: sorry for not making it explicit. Indeed, $R^d $ I defined to be $ k[X_1,..., X_d]$ and sequent definition $R^f$ follows this "logic/pattern". Yes thanks, there some typos, will try to remedy them! – user267839 Nov 13 '24 at 14:11
  • @walkar: I doubt that the answer given there adresses my problem: Firstly there the point of interest is if one can reconstruct the ideal living in the ring "down", the "smaller ring embedded in bigger here", there $k[x,y] \subset \overline{k}[x,y]$, in my case $R^e \subset R^{e+f}$, my object of interest is the reconstructibility of the ideal living in the ring "above". – user267839 Nov 13 '24 at 15:40
  • Also, the central problem in my question the "lost of infomation" when we contract some indeterminants restricting $I$ from $R^{e+f}$ to $R^e$, so I think the discussion in linked thread adresses a rather different issue – user267839 Nov 13 '24 at 15:42
  • I guess I was confused; I though the ideal $I$ started in $R^e$ -- in your question, $I$ starts in $R^{f+e}$? – walkar Nov 13 '24 at 16:52
  • @walkar: exactly – user267839 Nov 13 '24 at 16:54
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    Clearly the ideal $I$ has the property you desire if and only if it is extended from $R^e$. What kind of condition are you looking for? – Mohan Nov 17 '24 at 20:13
  • @Mohan: Yes, using your observation it suffice to develop techniques how to decide when a given ideal $I$ in $R^{e+f}$ comes from extension of an ideal from $R^e$. Are there some say "standard techniques" to approach this problem? – user267839 Nov 17 '24 at 20:31

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