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How to construct/describe the coproduct of two - not necessarily commutative - rings $R$ and $S$?

This in category $\mathbf{Ring}$ having as objects rings with a unit and as arrows unitary ringhomomorphisms.

I thought of firstly constructing monoid $M$ as coproduct of the underlying monoids $U(R)$ and $U(S)$ where $U:\mathbf{Ring}\rightarrow\mathbf{Mon}$ denotes the forgetful functor, and then secondly taking the ring $\mathbb{Z}\left[M\right]$ free over monoid $M$, but still have my doubts. If the rings have finite coprime characteristics then the coproduct should be the trivial ring, so something is wrong.

Can you give me a description of the coproduct (including its injections)? Thank you in advance.

drhab
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  • just for comparison, http://math.stackexchange.com/questions/143098/coproduct-in-the-category-of-noncommutative-associative-algebras – Noix07 Aug 05 '14 at 14:43

1 Answers1

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The coproduct of rings is somewhat similar to the coproduct of monoids or groups (aka free products), but that you use $\otimes_\mathbb{Z}$ instead of $\times$ (in fact, both are special cases of a general construction which works for monoid objects in cocomplete monoidal categories).

The coproduct of two rings $R,S$ is constructed as follows: Let $|R|$ and $|S|$ denote the underlying $\mathbb{Z}$-modules. Then consider the direct sum of tensor products

$\mathbb{Z} \oplus |R| \oplus |S| \oplus \bigl(|R| \otimes_\mathbb{Z} |S|\bigr) \oplus \bigl(|S| \otimes_\mathbb{Z} |R|\bigr) \oplus (|R| \otimes_\mathbb{Z} |S| \otimes_\mathbb{Z} |R|) \oplus \bigl(|S| \otimes_\mathbb{Z} |R| \otimes_\mathbb{Z} |S|\bigr) \oplus \dotsc$

Now we mod out the relations $x_1 \otimes \dotsc \otimes x_n \equiv x_1 \otimes \dotsc \otimes x_n \otimes 1 \equiv 1 \otimes x_1 \otimes \dotsc \otimes x_n$ and $ \dotsc \otimes x_i \otimes 1 \otimes x_{i+1} \otimes \dotsc \equiv \dotsc \otimes x_i x_{i+1} \otimes \dotsc $. The quotient has a multiplication induced by $$(x_1 \otimes \dotsc \otimes x_n ) \cdot (y_1 \otimes \dotsc \otimes y_m) := $$ $$\left\{\begin{array}{ll} x_1 \otimes \dotsc \otimes x_n \otimes y_1 \otimes \dotsc \otimes y_m & x_n \in R, y_1 \in S \text{ or } x_n \in S, y_1 \in R \\ x_1 \otimes \dotsc \otimes x_n y_1 \otimes \dotsc \otimes y_m & x_n,y_1 \in R \text{ or } x_n,y_1 \in S\end{array}\right.$$ We obtain a ring $R \sqcup S$ with obvious homomorphism $R \rightarrow R \sqcup S \leftarrow S$. It is the coproduct: Given $f : R \to T$ and $g : S \to T$, we define $h : R \sqcup S \to T$ by mapping, for example $x_1 \otimes x_2 \otimes x_3 \in |R| \otimes |S| \otimes |R|$ to $f(x_1) \cdot g(x_2) \cdot f(x_3)$. This is clearly a homomorphism of abelian groups on the infinite direct sum, but it respects the relations and therefore extends to a homomorphism on the quotient. It is checked that this is a homomorphism of rings; the unique one satisfying $h|_R = f$ and $h|_S =g$.

The construction of $R \sqcup S$ is somewhat complicated, but notice that its elements are just formal sums of products taken from $R$ or $S$, for example $r_1 + s_1 \cdot r_2 - r_3 \cdot s_2 \cdot r_4$. Actually this is how one comes up with the construction.

As with all algebraic structures, coproducts can also be described using generators and relations: If $R \cong \mathbb{Z}\langle X \rangle / I$ and $S \cong \mathbb{Z} \langle Y \rangle / J$ (for sets of variables $X,Y$ and ideals $I,J$), then $R \sqcup S = \mathbb{Z} \langle X \sqcup Y \rangle / (I+J)$. This is much easier than the construction above, but less concrete, especially when we don't have canonical presentations.

Your idea, using the forgetful functor $U : \mathsf{Ring} \to \mathsf{Mon}$, can also be made to work: $R \sqcup S = \mathbb{Z}[U(R) \sqcup U(S)]/I$, where the ideal $I$ is generated by the relations $(r+r') \cdot 1 \equiv r \cdot 1 + r' \cdot 1$ and $1 \cdot (s+s') \equiv 1 \cdot s + 1 \cdot s'$. These relations exactly guarantee that the canonical monoid homomorphisms $U(R) \to U(\mathbb{Z}[U(R) \sqcup U(S)]) \leftarrow U(S)$ lift to ring homomorphisms $R \to \mathbb{Z}[U(R) \sqcup U(S)]/I \leftarrow S$.

Martin Brandenburg
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  • Is what you call 'algebraic structure' the same thing as 'algebraic system (of a given type)' mentioned in CWM on page 120? By the way, you are of great help to me on this site. Not only in this question. Tibi gratias ago! – drhab Jan 03 '14 at 14:44
  • Yes, exactly. CWM keeps this quite short (although very concise), you can find more about algebraic structures in texts about universal algebra. – Martin Brandenburg Jan 03 '14 at 14:46
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    I don't think $R$ and $S$ embed into your coproduct in the sense of injective maps. Try $S = 0$ and $R \neq 0$. – darij grinberg Jan 03 '14 at 15:25
  • @Darij: You are right, thank you. – Martin Brandenburg Jan 03 '14 at 15:27
  • @MartinBrandenburg: I'm guessing the same construction works for $R$-algebras: $$A\ast B= \frac{R\oplus A\oplus B\oplus (A!\otimes_R!A)\oplus (A!\otimes_R!B)\oplus (B!\otimes_R!A)\oplus (B!\otimes_R!B)\oplus \ldots}{ \langle\ldots!\otimes! a!\otimes! a'!\otimes!\ldots!-!\ldots!\otimes! aa'!\otimes\ldots,: \ldots!\otimes! b!\otimes! b'!\otimes!\ldots!-!\ldots!\otimes! bb'!\otimes!\ldots,: 1_R!-!1_A,: 1_R!-!1_B\rangle}$$ and $s\cdot t:=!s!\otimes!t$ for $s,t!\in!A!\ast!B$, right? If $A$ and $B$ are graded, how is $A!\ast!B$ graded? – Leo Mar 13 '15 at 19:41
  • I'm assuming you meant alternating summands $A!\otimes!B!\otimes!A!\otimes!B!\otimes!A!\otimes!B!\otimes!\ldots$ and $B!\otimes!A!\otimes!B!\otimes!A!\otimes!B!\otimes!A!\otimes!\ldots$. But aren't you missing relations $1_R-1_A$ and $1_R-1_B$? Or is the copy of $R$ perhapy unnecessary, since we have $1_A=1_A\otimes1_B=1_B$? – Leo Mar 13 '15 at 19:54
  • @Martin, what if $R$ and $S$ are assumed to have nonzero intersection $T$? Is the definition of coproduct detailed enough so that it recognizes this fact and reduces the computation when an element of $R$ meets an element of the copy of $T$ which lies inside $S$, or is it more crude and the two copies of $T$ (one inside $R$, one inside $S$) are in fact two cartesianly separated objects? – Jose Brox Jun 25 '15 at 12:40
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    It doesn't make sense to talk about $R \cap S$. And yes, $R$ and $S$ are completely isolated. – Martin Brandenburg Jun 25 '15 at 12:59
  • @MartinBrandenburg (Sorry, didn't see your answer!!). From a ring-theoretic perspective I think it makes sense to talk about $R\cap S$ when we have an ambient ring as universe (e.g $R$ upper triangular matrices, $S$ lower triangular matrices, $D=R\cap S$ diagonal matrices, all inside a matrix ring). This still bugs me: the ring $R\coprod S/I$, where $I$ is the ideal that mods out $R\cap S$ (i.e., $x\oplus 0=0\oplus x$ when $x\in R\cap S$ and derived identities), seems more natural to me, fusing the two rings where the two overlap. Do you know if this construction has any name? Or has problems? – Jose Brox Nov 07 '17 at 17:19
  • I find this description very useful. However, and moving to the case of algebras, I do not see how this could be the same of the tensor product of commutative algebras. It looks more like the tensor algebra rather than the tensor product of algebras. – gibarian Oct 15 '20 at 07:44
  • @gibarian [I do not see how this could be the same of the tensor product of commutative algebras.]---> Not at the level of the construction but at the level of universal solution, changing the ambient category. – Duchamp Gérard H. E. Feb 19 '21 at 07:13
  • @darijgrinberg Embedding (resp. universal solution) works for augmented algebras (resp. within the category of augmented algebras). – Duchamp Gérard H. E. Feb 19 '21 at 07:23
  • Isn't it simpler to say that if $A,B$ are two $R$-algebras (they could be $\mathbb Z$-algebras so that covers the case of rings), you can define their coproduct as a quotient of $$ A \sqcup B = \bigoplus_{n \ge 0} (A \oplus B)^{\otimes n} / rels $$ where $(A \oplus B)^{\otimes 0} = R$ and all the tensor products are taken as $R$-modules? If we have a third $R$-algebra $C$ with $R$-algebra morphisms $A,B \to C$, we get a map of $R$-modules $A \oplus B \to C$, extend it to $(A \oplus B)^{\otimes n} \to C$ by taking tensor powers of the same map, and then their direct sum $A \sqcup B \to C$. – Patrick Da Silva Aug 28 '23 at 18:50
  • The ideal $rels$ (ran out of characters) would be basically defined by the relations which are trivially holding when we map all those elements into $C$, such as the ones already mentioned, so you get generators of the form $$ a_1 \otimes a_2 - a_1a_2, \quad a_1, a_2 \in A \cdots \text{ same with B } $$ or $$ a \otimes 1 - a, \quad b \otimes 1 - b, \quad a \in A, b \in B $$ – Patrick Da Silva Aug 28 '23 at 18:57
  • @JoseBrox from the past, the concept you were looking for is the pushout. – Jose Brox Feb 02 '24 at 22:17