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While working through Atiyah/MacDonald for my final exams I realized the following:

The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals

$$I\cdot J:=\langle\{ab\mid a\in I, b\in J\}\rangle$$

The internal Hom is given by the ideal quotient; let's denote it $[I, J]:=(J:I)=\lbrace a\in A\mid aI\subset J\rbrace$. We have

$$IJ\subset K \Leftrightarrow I\subset [J,K].$$

Furthermore the category(poset) of submodules $U(M)$ of an A-Module $M$ is enriched over the ideal category via

$$[U,V]:=\lbrace a\in A\mid aU\subset V\rbrace.$$

Where does this enrichment come from? More specifically: As in algebraic geometry we are dealing with the set of prime ideals ( wich is more or less the subcategory

$$\mathrm{Int^{op}}/A\subset\mathrm{CRing^{op}}/A$$

of ring-maps $A\to B$ with $B$ an integral domain), I'd like to see the category of Ideals as some "flattened" version of $\mathrm{CRing^{op}}/A$ by taking the kernel. Or more generally i'd like to see like to see it as a "flattened" version of the category of modules (by taking the annihilator?). So:

Question: Is it possible to build the enriched categories $I(A)$ and $U(M)$ out of $\mathrm{CRing^{op}}/A$ and $\mathrm{Mod}_A/M$ in a nice way? So: Can we for example write $I\cdot J$ down as a kernel of some $A\to B$? Or $\mathrm{Ann}(M)\cdot \mathrm{Ann}(N)$ just by using $N,M$ and constructions in/on $\mathrm{Mod}_A$?

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    $\begingroup$ Very interesting question. $I(A)$ is equivalent to the full subcategory of the comma-category which consists of regular epimorphisms $A \to ?$. I've already asked here (mathoverflow.net/questions/69037/…) how we can describe the ideal product under this equivalence. $\endgroup$ Oct 1, 2011 at 10:20
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    $\begingroup$ "The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals" That's how I'll explain commutative algebra to my son! $\endgroup$
    – fosco
    Feb 2, 2015 at 11:38

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