For $R$ a commutative ring, the tensor product of $R$-algebras is the coproduct in the category of commutative $R$-algebras. In the noncommutative case it is no longer the coproduct in the category of associative $R$-algebras, but it does satisfy a universal property, as given on Wikipedia. Is this some sort of colimit? If not, is there are a straightforward description of this universal property via a functor (right?) adjoint to the tensor product, as is the case for tensor products of modules?


There is a tensor-hom adjunction for the tensor product of algebras, but it exists at the level of the Morita 2-category, rather than the 1-category of algebras.

Namely, the Morita 2-category has objects $k$-algebras, and the category of morphisms $A \to B$ is the category $\text{Mod}(A^{op} \otimes B)$ of $(A, B)$-bimodules, where composition is given by tensor product. (All tensor products are over $k$.) The Morita 2-category has an internal hom $[A, B] = A^{op} \otimes B$, and its left adjoint is the tensor product in the sense that we have natural identifications

$$[A \otimes B, C] \cong A^{op} \otimes B^{op} \otimes C \cong [A, B^{op} \otimes C] \cong [A, [B, C]].$$

In the Morita 2-category the coproduct of two algebras $A$ and $B$ is $A \times B$ (rather than the free product), and tensor product distributes over this.


Tensor products of noncommutative rings fail to distribute over finite coproducts. For instance, $$\mathbb{Z}[x]\otimes\left(\mathbb{Z}[y]\coprod \mathbb{Z}[z]\right)=\mathbb{Z}[x]\otimes \mathbb{Z}\langle y,z\rangle$$ can be described as the ring generated by elements $x$, $y$, and $z$ mod the relations that $x$ commutes with $y$ and $z$. On the other hand, $$(\mathbb{Z}[x]\otimes \mathbb{Z}[y])\coprod(\mathbb{Z}[x]\otimes\mathbb{Z}[z])=\mathbb{Z}[x,y]\coprod\mathbb{Z}[x',z]$$ is the ring generated by $x$, $y$, $x'$, and $z$ mod the relations that $x$ commutes with $y$ and $x'$ commutes with $z$. The canonical map from the second ring to the first identifies $x$ and $x'$ together as $x$, and is not an isomorphism.

It follows that the functor $\mathbb{Z}[x]\otimes -$ on noncommutative rings cannot have a right adjoint, and I would expect that similarly $A\otimes -$ cannot have a right adjoint for almost all rings $A$. (Note that this is actually also true for commutative rings by a similar example; the right adjoint to $A\otimes -$ only works for modules.)

The functor of two variables $-\otimes-:\mathrm{Ring}\times\mathrm{Ring}\to\mathrm{Ring}$ also fails to preserve coproducts and hence does not have a right adjoint (this time, in contrast with the case of commutative rings). Indeed, $(A\coprod C)\otimes (B\coprod D)$ differs from $(A\otimes B)\coprod (C\otimes D)$ because in the first ring $A$ and $C$ both commute with both $B$ and $D$, whereas in the second ring $A$ commutes with $B$ and $C$ commutes with $D$, but $A$ does not commute with $D$ and $C$ does not commute with $B$.

As for representing tensor products as a colimit, I'm not sure what you have in mind by that. The only obvious colimit you can take when you have two objects and no additional data is a coproduct, and you have already observed that the tensor product is not the coproduct. You can describe the tensor product as the coproduct modulo relations that say that elements from the two factors commute with each other, and you can write modding out these relations as a certain coequalizer, but it will not be particularly nice (because in order to even describe the objects involved in the diagram, you need to refer to the elements of your two rings).


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