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