This is an extension of this question. Let $I,J$ be ideals of a ring $R$; every ring is commutative and unital here. Is it possible to define $R \to R/(I*J)$ out of $R \to R/I$ and $R \to R/J$ in categorical terms within the category of rings? To be more precise: Is there a formula in the language of category theory $\phi$ with a parameter of type "category" and three parameters of type "morphism", such that $\phi(\text{Ring},R \to R/I,R \to R/J,R \to R/K)$ is true if and only if $K = I*J$?

It is easy to do so for the ideal sum and the ideal intersection. Namely $K=I + J$ is characterized by $R/K = R/I \otimes_R R/J$ via the natural maps, the tensor product being the coproduct of $R$-algebras. And $K=I \cap J$ is characterized by the fact that $R/K$ is the universal regular quotient of $R$ such that $R \to R/I \times R/J$ factors through it; this is a fancy way of saying that $I \cap J$ is the kernel of $R \to R/I \times R/J$. But somehow it is quite difficult for the ideal product. Note that the linked question above shows that there will be no characterization just using regular quotients of $R$.

Although $I*J$ is the image of the natural morphism $I \otimes J \to R$, this takes place in the category of $R$-modules and thus leaves the given category of rings. Perhaps unitalizations are useful, but I cannot get rid of the factors $I,J$ in the canonical morphism $\tilde{I} \otimes \tilde{J} \to R$. Another idea is the following: $I*J \subseteq I \cap J$ and it suffices to characterize (the quotient of) $I \cap J / I*J$. This is an $R$-module isomorphic to $\text{Tor}_1(R/I,R/J)$. But again a priori this leaves the category of rings.

There is a categorical definition of prime ideals (see here) and thus also of radical ideals. Since we can also define intersections and inclusions, we also have $\text{rad}(I*J) = \text{rad}(I \cap J)$ as a categorical information, but of course this does not suffice to recover $I*J$.

EDIT: There is a somewhat nonsense positive answer: The ring $\mathbb{Z}[x]$ can be defined categorically, see here. Actually we also get the coring structure, including the multiplication resp. addition $\mathbb{Z}[x] \to \mathbb{Z}[x,y]$ and zero $\mathbb{Z}[x] \to \mathbb{Z}$. Then you can define the underlying set $|R|=\hom(\mathbb{Z}[x],R)$ categorically and also $|I|$ as the equalizer of two maps $|R| \to |R/I|$, the one being twisted by the zero morphism $\mathbb{Z}[x] \to \mathbb{Z} \to \mathbb{Z}[x]$. Now $|I*J| \subseteq |R|$ is the subset defined as the union of the images of the maps $(|R|^2)^n \mapsto |R|$ induced by $\mathbb{Z}[x] \to \mathbb{Z}[x_1,y_1,...,x_n,y_n], x \mapsto x_1 y_1 + ... + x_n y_n$, which comes from the coring structure. But now $R \to R/I$ is characterized by $|I|$, so we win.

Thus I want to change my question: Is there a more direct categorical definition of the ideal product, not going through $\mathbb{Z}[x]$ and thereby just imitating the element definition?

More generally (and here the above element definition does not work): If $X$ is a scheme and $Z \to X, Z' \to X$ are two closed subschemes, what is a categorical characterization of the closed immersion $Z'' \to X$ corresponding to the ideal product? What is the geometric meaning of it (this was partially discussed here)? The underlying space is just the union, but the structure sheaf not. For $Z'=Z$ this process is called thickening.

  • $\begingroup$ Notation: what is your I*J ? The ideal containing all the products ij where i in I and j in J ? $\endgroup$ – Zoran Skoda Jul 9 '11 at 12:44
  • $\begingroup$ ? en.wikipedia.org/wiki/Product_of_ideals#Ideal_operations $\endgroup$ – Martin Brandenburg Jul 9 '11 at 19:41
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    $\begingroup$ A very vague remark: perhaps this is related to the notion of commutator in varieties of algebraic systems, see, e.g., R. Freese and R. McKenzie, Commutator Theory for Congruence Modular Varieties, London Math. Soc. Lect. Note Ser. 125 (1987). If so, the "correct" notion is not just the product IJ of ideals, but their "commutator" IJ+JI. $\endgroup$ – Pasha Zusmanovich Aug 8 '11 at 10:47
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    $\begingroup$ Very interesting paper! I've just started to read it, but it seems to answer a variant of my question. Namely, it gives a (in fact categorical) characterization of the operation $(I,J) \mapsto IJ+JI$ on pairs of ideals in a varying noncommutative ring. Probably the same works in the commutative case, where we get $IJ$. The only difference to my question is that here we consider the operation globally, not just for a fixed ring. $\endgroup$ – Martin Brandenburg Aug 8 '11 at 14:17
  • $\begingroup$ I really should start reading this paper ... $\endgroup$ – Martin Brandenburg Oct 1 '11 at 10:25

The unitalization approach can be made to work.

Let $C_K = \{ (r,s) \in R \times R \mid r-s \in K \}$ be the congruence defined by an ideal $K$.

Then, we have three maps defined on $S = C_I \otimes_R C_J$ :

  • $\pi_0 : S \to C_I$ induced by the first projection $C_J \to R$ and the identity on $C_I$
  • $\pi_1 : S \to C_J$ induced by the first projection $C_I \to R$ and the identity on $C_J$
  • $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.

Letting $\Delta \subseteq R \times R$ be the image of the diagonal, define $T = \{ x \in S \mid \pi_0(x) \in \Delta \wedge \pi_1(x) \in \Delta \}$. I claim that $\mu(T) = C_{IJ}$.

On the level of $R$ modules, we have isomorphisms $C_K \cong R \oplus K$, such as $(r,s) \mapsto (r, s-r)$, and so we have

$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$

In this form, the maps $\pi_i$ become projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.

By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form $$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$ and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.

  • $\begingroup$ Your notation is very sloppy. Which of the two projections $p_i \colon C_J \to R$ do you use for $\pi_0$? How do you view $R$ as a subring of $C_J$ in the definition of $T$? Which 'evident isomorphism' $C_K \cong R \oplus K$ do you choose? You could take $(p_1,p_1-p_2)$, or $(p_2,p_1-p_2)$, or a sign variation thereof. $\endgroup$ – R. van Dobben de Bruyn Jan 3 '17 at 15:06
  • $\begingroup$ @R.vanDobbendeBruyn: Hrm. In my scratchwork I had used $R \oplus K$ throughout, and only changed to $C_K$ in the final revision, so the choices were indicated by the notation. In the end they only matter up to isomorphism, so I don't think it necessary to bog it down with the details of a particular choice, but I suppose it's short enough that it shouldn't hurt much to fill in more details. $\endgroup$ – Hurkyl Jan 3 '17 at 15:41
  • $\begingroup$ And I suppose that the point of this answer is that $C_K$ and all other objects involved are in fact rings (rather than mere modules). $\endgroup$ – R. van Dobben de Bruyn Jan 3 '17 at 16:54
  • $\begingroup$ @R.vanDobbendeBruyn: Right; the overall idea was inspired by Martin's suggested approach by unitalization. $\endgroup$ – Hurkyl Jan 3 '17 at 18:12

Define a "ideals" as sub-object that are (but isomorphisms) kernel of some morphim. The orders of ideals is in order bijection with the orders of the extremal quotients, that are (but isomorphism) just the surjective morphisms. This bijection is merely:

$I \mapsto (q_I: R \to R/I)$ and $q \mapsto Ker(q)$.

Now given two ideals $I, J \subset R$ their "product" $I\ast J$ is the ideal generated by these two, or the minimal ideal containing both $I$ and $J$, in other words it is $I \vee J$ in the order of the ideals of $R$. Then $I\ast J = Ker (q_{I, J})$ where $q_{I, J}: R \to Q$ come from the pushout of $q_I: R\to R/I$, $q_J: R\to R/J$, because this pushout is $q_I\vee q_J$ in the order of the extremal quotients.

Excuse my poor English.

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    $\begingroup$ No. As Martin pointed out in his question, the pushout/tensor product corresponds to the sum of ideals. $\endgroup$ – Gerrit Begher Sep 17 '14 at 7:34

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