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For the case of commutative algebras, I know that the coproduct is given by the tensor product, but how is the situation in the general case? (for associative, but not necessarily commutative algebras over a ring $A$). Does the coproduct even exist in general or if not, when does it exist? If it helps, we may assume that $A$ itself is commutative.

I guess the construction would be something akin to the construction of the free products of groups in group theory, but it would be nice to see some more explicit details (but maybe that would be very messy?) I did not have much luck in finding information about it on the web anyway.

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It's indeed something akin to the free product and is fairly messy. Pick a set of generators and relations over your base ring, then combine them. – Qiaochu Yuan May 9 '12 at 15:38
Good question. Any answers/links are still welcome. – Leon Dec 29 '12 at 2:21
Here I give the construction of the coproduct of rings, i.e. $\mathbb{Z}$-algebras. Exactly the same works for $A$-algebras, when $A$ is a commutative ground ring. – Martin Brandenburg Jun 24 '15 at 10:08
up vote 2 down vote accepted

In the book Rings with generalized identities by Beidar, Martindale and Mikhalev, Section 1.4 (from page 24 onwards) you can find the definition from commutative diagrams, the characterization with proof (which is the one given by Chindea Filip in his answer) and several properties.

The main purpose of the coproduct in this book is to serve as the "home" for generalized polynomials identities for prime and semiprime rings, which are, informally speaking, like polynomial identities in which in addition it is allowed to insert elements from the ring (or even from any of its usual rings of quotients). So, for example $xy-yx$ is a PI, while $pxy-yqx$ with $p,q$ fixed elements of the maximal ring of quotients is a GPI.

The mentioned part of the book is readable through Google Books:

Rings with generalized identities - Coproducts

and the book is downloadable elsewhere.

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The following is a link to an article which provides a partial answer, namely it gives (on page 8, without proof) the coproduct of two non-commutative algebras (over a field rather than a ring, I don't know the ring case)

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It is amazing that we still see compiled TeX files with lines that start with ¿From... !!! – Mariano Suárez-Alvarez Jun 24 '13 at 20:49
@MarianoSuárez-Alvarez: The reason for the ¿ is typically that the TeX file has been sent as an attachment via email. As far as I know, the archaic email standard has some strange replacement rules for text attachments, which introduces the ¿ in some circumstances. To avoid this, the tex file should be put into some container format like .tar, which yields a binary email attachment which won't be modified. – azimut Jun 24 '13 at 21:11
@azimut, I know: that is precisely why I find it amazing. I have even seen it in printed articles in reputed journals... – Mariano Suárez-Alvarez Jun 24 '13 at 21:15

Quillin and Cuntz desribe this in the book "Cyclic Homology in non-commutative geometry" on the first page, without proof.

The construction goes as follows:

$A\star B \cong A \oplus B \oplus (A \otimes B) \oplus (B \otimes A) \oplus (A \otimes A \otimes B) \oplus (A \oplus B \oplus A) \oplus (B \oplus A \oplus A) \oplus ...$ (and keeps cycling though tensors).
Its a construction known as the free product of associative algebras. Though it should be no canonical embedding of A (or B) into any of the tensor products above, since any arrow $A \rightarrow A \otimes B$ cannot be free of relations.

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I had little success in searching on the web for a rigorous construction in the ring ($A = \mathbb{Z}$) case some time ago and gave it up in the end. However this proved useful so I am going to settle this and write the full solution in the general case of algebras over a (commutative) ring $k$. The following construction was communicated to me by Prof. Gigel Militaru; as far as I understood the original source seems obscure.

So, start with a ground ring $k$ and a (possibly infinite) family of $k$-algebras $(A_i)_{i \in I}$ and let $T$ be the tensor algebra of the direct sum $M$ of all $A_i$ regarded as $k$-modules.

We will use the fact that the tensor algebra is the "smallest" algebra containing a module, in the sense that it comes equipped with a $k$-linear morphism $\alpha : M \to T$ such that for all $k$-algebras $A$ and $k$-linear $f : M \to A$, there is an unique $k$-algebra morphism $g : T \to A$ such that $g \circ \alpha = f$.

Now, let $\alpha_i = \alpha \circ j_i$, where $j_i : A_i \to M$ are the canonical monomorphisms and $\alpha$ is the inclusion just mentioned. Of course, $\alpha_i$ are not yet $k$-algebra maps, since $j_i$ are not multiplicative, so we need to make them so.

Let $I$ be the two-sided ideal generated by all the relations $\alpha_i(ab) - \alpha_i(a)\alpha_i(b)$ and $1_T - \alpha_i(1_{A_i})$. Let $A = T/I$ and $\gamma_i = \pi \circ \alpha_i$, where $\pi : T \to T/I$ is the canonical map. The claim is that $A$ together with all $\gamma_i$ is the coproduct.

It is obvious that $\gamma_i$ are $k$-algebra maps. Let $(B, \eta_i : A_i \to B)$ be a pair consisting of a $k$-algebra $B$ and $k$-algebra morphisms. These extend to a $k$-linear map $\eta : M \to B$, and by the indicated property of $T$, to a $k$-algebra map $\eta' : T \to B$. Since $\eta_i = \eta' \circ \alpha_i$, we can factor through $T/I$ to find the desired morphism for the universal property. Uniqueness follows from uniqueness in the three universal properties.

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