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Stream: learning: questions

Topic: ideal extension (commutative algebra)

Hugo Jenkins (Mar 30 2024 at 06:33):

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

$I \mapsto I^e :=$ smallest ideal of $S$ containing $\phi(I)$

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

Question: is there any conceptual reason to expect this works? If $R\to S$ is a flat ring map, we can describe $I^e$ as $I\otimes_R S$, the extension of scalars of the $R$-module $I$. But not in general. Also, $\phi^{-1}(J)$ is definitely not the restriction of scalars to $R$ of the $S$-module $J$. 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, $I^e \subset J \Rightarrow I \subset \phi^{-1}(J)$ is tautological, and the reverse implication holds because $J$ is a substructure containing the set-theoretic image $\phi(I)$.

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.

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 $\phi: R \to S$ and an ideal $I \subset R$ such that the smallest ideal of $S$ containing $\phi(I)$ is not $I \otimes_R S$?

Brendan Murphy (Mar 30 2024 at 20:19):

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

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

Hugo Jenkins (Apr 02 2024 at 05:31):

This helps, thanks a lot

Brendan Murphy said: