# How to construct the coproduct of two (non-commutative) rings

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

This in category $\mathbf{Rng}$ 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 $U:\mathbf{Rng}\rightarrow\mathbf{Mon}$ 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.

-
just for comparison, math.stackexchange.com/questions/143098/… – user39158 Aug 5 '14 at 14:43

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 $R \to \mathbb{Z}[U(R) \sqcup U(S)]/I \leftarrow S$.

-
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 3 '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 3 '14 at 14:46
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 3 '14 at 15:25
@Darij: You are right, thank you. – Martin Brandenburg Jan 3 '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? – Leon Mar 13 '15 at 19:41