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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

2 Answers 2

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) http://www.google.co.uk/url?q=http://citeseerx.ist.psu.edu/viewdoc/download%3Fdoi%3D10.

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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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