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Stream: learning: questions

Topic: ideal extension (commutative algebra)


view this post on Zulip Hugo Jenkins (Mar 30 2024 at 06:33):

Here is a nice fact: given a map ϕ:RS\phi: R\to S of unital commutative rings, the extension operation

IIe:=I \mapsto I^e := smallest ideal of SS containing ϕ(I)\phi(I)

combines with the usual preimage to give a monotone Galois connection between the ideal posets of RR and SS. That is, the operations form an adjoint pair of functors between these two poset categories, and one has e.g. Iece=IeI^{ece} = I^{e}.

Question: is there any conceptual reason to expect this works? If RSR\to S is a flat ring map, we can describe IeI^e as IRSI\otimes_R S, the extension of scalars of the RR-module II. But not in general. Also, ϕ1(J)\phi^{-1}(J) is definitely not the restriction of scalars to RR of the SS-module JJ. So that is not where this adjunction comes from.

Other question: Is it always true that for substructures which aren't preserved by maps (maybe because they care about the ambient structure too much in some sense), one can "fix" usual the set-theoretic direct image/preimage Galois connection, by taking the generated substructure on the target side?

Edit: Of course you can do this for any substructures preserved by preimage, assuming you have a "substructure generated" in the sense of smallest substructure containing. Indeed, IeJIϕ1(J)I^e \subset J \Rightarrow I \subset \phi^{-1}(J) is tautological, and the reverse implication holds because JJ is a substructure containing the set-theoretic image ϕ(I)\phi(I).

view this post on Zulip Morgan Rogers (he/him) (Mar 30 2024 at 12:19):

You might enjoy thinking about how you can recover this Galois connection from the tensor-hom adjunction between R-modules and S-modules.

view this post on Zulip John Baez (Mar 30 2024 at 16:12):

I probably can't help with this question, so I'll just ask another question. What's an example of a commutative ring homomorphism ϕ:RS\phi: R \to S and an ideal IRI \subset R such that the smallest ideal of SS containing ϕ(I)\phi(I) is not IRSI \otimes_R S?

view this post on Zulip Brendan Murphy (Mar 30 2024 at 20:19):

Your question is a little ill formed John. The module IRSI \otimes_R S isn't generally an ideal of SS, in that the canonical map IRSRRSSI \otimes_R S \to R \otimes_R S \cong S isn't injective. This map being injective for all II is exactly what it means for SS to be flat over RR. Note however that the image of IRSRRSSI \otimes_R S \to R \otimes_R S \cong S is always the extension of II (the bilinear map I×SSI \times S \to S it corresponds to is multiplication, so the image is the set of sums of products of stuff in II and stuff in SS, which is the extension of II).

An example of a non flat morphism of rings is the projection from R=k[ε]/(ε2)R = k[\varepsilon]/(\varepsilon^2) to S=kS = k. If we let I=(ε)I = (\varepsilon) then IRSI \otimes_R S is nonzero but the map IRSSI \otimes_R S \to S is zero (this latter bit is easy, just note ε\varepsilon acts as zero on SS). To see IRS0I \otimes_R S \neq 0 observe SR/IS \cong R/I and so IRSIRR/II/I2II \otimes_R S \cong I \otimes_R R/I \cong I/I^2 \cong I (or less directly, II is a nonzero finitely generated module over a local ring and hence by Nakayama's lemma the tensor with the residue field is nonzero)

view this post on Zulip Hugo Jenkins (Apr 02 2024 at 05:31):

This helps, thanks a lot

Brendan Murphy said: