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.

varyingnoncommutative 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