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Motivated by this question: Let $\mathsf{Int}$ be the category of integral domains with ring homomorphisms (perhaps only injective ring homomorphisms, if you need this). Is there a functor $\mathsf{Int} \times \mathsf{Int} \to \mathsf{Int}$ which does not factor through one of the two projections? Thus, this should be a functorial construction of an integral domain associated to two integral domains $R,S$.

In order to exclude boring examples, such as the free commutative ring on the product of the underlying sets of $R,S$, or the free commutative $R$-algebra on the underlying set of $S$, let us impose further conditions, for example that the functor does not factor through $\mathsf{Set}$, $\mathsf{Ab}$, $\mathsf{Ring}$ or similar categories, or products of these categories. I am not sure if this already excludes boring examples (namely those, which only use underlying structures of $R,S$).

Bonus: It would be nice, if possible, to have associativity and/or commtativity up to isomorphism, and perhaps even $\mathbb{Z}$ as a unit. This leads to the following question: Does $\mathsf{Int}$ admit any (symmetric) monoidal structure?

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Martin Brandenburg
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    One variant which might be interesting would be to ask whether $\mathsf{Int}$ admits a monoidal structure. I would guess that it doesn't admit a closed symmetric monoidal structure; proving this might be an interesting start. – tcamps Apr 14 '15 at 13:48
  • The follow-up question on MO: https://mathoverflow.net/questions/457027 – Martin Brandenburg Jan 01 '24 at 23:16

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I hesitate to say it --

Take any functor $\mathsf{Int} \times \mathsf{Int} \to \mathsf{Set}$ which doesn't factor through a projection (e.g. take the underlying set of the product ring), and then follow it with the free ring functor $\mathsf{Set} \to \mathsf{Int}$ (the free ring on a set is a domain).

Ugh. I hope there's something nicer.

If you want to know what I'm blabbing on to Martin about in the comments, take a look at the history.

tcamps
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  • Yes, this works. Sorry, but this means that my question wasn't stated precisely enough. I have edited it. – Martin Brandenburg Apr 14 '15 at 13:00
  • I hope that this procedure is OK. If not, let me know, because then I will simply accept your answer and ask the refined question elsewhere. – Martin Brandenburg Apr 14 '15 at 13:09
  • In principle I think the restriction is reasonable. But if the answer is that no such functor exists, one will have to be very clear about what "related categories" are if one wants to prove it. – tcamps Apr 14 '15 at 13:29
  • What should be the field od fractions of $R \times S$?? – Martin Brandenburg Apr 14 '15 at 13:31
  • Ah, yes, it will be trivial. You formally adjoin an inverse to every nonzero element. One consequence is that every idempotent will be identified with 1. We then have $1 = (1,1) = (1,0) + (0,1) = 1+1 = 2$ and hence $1=0$. So that doesn't work. – tcamps Apr 14 '15 at 13:41
  • If you invert $(1,0)$ and $(0,1)$, then you will invert their product, i.e. $0$, which means that the result is $0$. – Martin Brandenburg Apr 14 '15 at 14:04
  • You say "invert all non-zero elements", but this is not functorial. Ok, it is functorial with respect to injective maps, but if $R \to S$, $R' \to S'$ are injective, then the induced map $R \otimes S \to R' \otimes S'$ does not have to be injective. Also, $R \otimes S$ may be zero (hence, every localization, too), which is not allowed for integral domains. – Martin Brandenburg Apr 14 '15 at 14:05
  • Sorry, I deleted my comment as soon as I saw that issue. I hate it when people do that to me. – tcamps Apr 14 '15 at 14:19
  • @MartinBrandenburg Actually, what's an example of a coproduct of injections between domains which is not an injection? Maybe all of these problems would be solved if we simply make the trivial ring an honorary domain. – tcamps Apr 14 '15 at 14:32
  • Your claim about tensor products of injective maps is quite wrong. See the notion of "flatness" and "Tor-functor". Your conception of the tensor product is wrong. – Martin Brandenburg Apr 14 '15 at 15:36
  • Thanks. I thought too briefly about the adjointness properties of $A \otimes -$ and mistakenly thought they were on my side. One issue with my "proof" is that the relations $0 \otimes b = 0$ and $a \otimes 0 = 0$ need to be put in separately. This could be remedied, but the real issue is that I'm implicitly inducting on the number of "moves" needed to get from $\sum_i f(a_i) \otimes g(b_i)$ to $\sum_j f(a_j) \otimes g(b_j)$ and I make the erroneous assumption that at every point along the way, the sum $\sum_k r_k \otimes s_k$ will be of the form $\sum_k f(a_k) \otimes g(b_k)$. – tcamps Apr 14 '15 at 16:23
  • Well, in any case, it's a bad idea to work with elements of the tensor product and its explicit construction. The universal property is all what one will need. Forever. – Martin Brandenburg Apr 14 '15 at 17:20
  • Sorry for my rude comments back then. I will accept your answer now and ask the question about monoidal structures on mathoverflow. – Martin Brandenburg Oct 24 '23 at 06:47