This is an extension of [this](https://mathoverflow.net/questions/14739/how-can-i-define-the-product-of-two-ideals-categorically) 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](https://mathoverflow.net/questions/68980/is-there-a-way-to-define-a-prime-ideal-object-via-diagrams-in-the-category-of-rin)) 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](https://mathoverflow.net/questions/55931/invariance-of-zx-under-a-self-equivalence-of-the-category-of-commutative-ring). 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](https://mathoverflow.net/questions/49259/when-is-product-of-two-ideals-equal-their-intersection))? The underlying space is just the union, but the structure sheaf not. For $Z'=Z$ this process is called thickening. 



Categorical definition of the ideal product within the category of rings
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.


