On the coproduct of $K$-rngs

Question

A $rng$ is a ring without unit $1$. For an integral domain $K$, a $K$-$rng$ is then defined as $K$-bimodule which is also a $rng$. Being equationally defined, $K$-$rng$s generate a variety $V$ on which we can consider free $K$-$rng$'s (whose elements are simply polynomials on $K$ with indeterminates the elements of the basis, but without a unit $1$) and free products $*$ (which can be seen as the coproduct in the category given by $V$).

I'm reading Pope's article: https://core.ac.uk/download/pdf/82779675.pdf

where it is performed (at pg.51) the following construction: we take some basis $\{x_n:n<\omega\}$ of some free $K$-$rng$ $F$, and we define $y_n=x_n-x_{n+1}^2$, for every $n<\omega$. It is then claimed that for every $n<\omega$ the $K$-$rng$ $B_n=\langle y_0,\ldots,y_n\rangle$ is free and there is some substructure $C$ of $F_{n+1}=\langle x_0,\ldots,x_{n+1}\rangle$ s.t. $F_{n+1}=B_n* C$.

I think that the general argument should go as follows: first, we show by induction that every $B_n$ is free (this follows from the fact that $B_{n+1}=B_n*\langle y_{n+1}\rangle$). Then, we take the ideal $I$ generated by $B_n$ in $F_n$ and construct a short exact sequence: $$ B_n\hookrightarrow F_{n+1}\twoheadrightarrow F_{n+1}/I$$ From which we get, up to isomorphism, that $F_{n+1}=B_n \oplus(F_{n+1}/I)$. Since every expression in the $y_0,\ldots,y_n$ has indeterminate $x_{n+1}$ of exponent $\geq 2$, $x_{n+1}$ does not belong to $I$ and $F_{n+1}/I$ is not trivial. Moreover, $F_{n+1}/I$ is isomorphic to $\langle x_{n+1}\rangle$ by the map $x_{n+1}\mapsto x_{n+1}+I$. So, if the direct sum is the coproduct in $V$ we get the thesis.

So, my questions are:

1-Is this the case? I.e. the coproduct in $V$ is the direct sum in the usual sense for modules?

2-Do you think that my argument is correct?

Thank you in advance for your attention.

Answer 1

The "direct sum" does not provide coproduct of $K$-rngs. Notice that we have $xy=0$ in the direct sum when $x,y$ are elements that belong to different rngs. This is a relation that should not hold in a coproduct because it will break the universal property.

For $K = \mathbb{Z}$ we have the category of rngs. Here, coproducts can be described as follows: The initial object (=empty coproduct) is $0$. The coproduct of two rngs $R,S$ is based on the abelian group (tensor products are over $\mathbb{Z}$) $$R \, \oplus \, S \, \oplus \, (R \otimes S) \, \oplus \, (S \otimes R) \, \oplus \, (R \otimes S \otimes R) \, \oplus \, (S \otimes R \otimes S) \, \oplus \, \dotsc$$ with the "obvious" multiplication. For example, $(r \otimes s) \cdot (r') := (r \otimes s \otimes r')$, and $(r \otimes s) \cdot (s' \otimes r') := (r \otimes ss' \otimes r')$.

If $(R_i)_{i \in I}$ is any family of rngs, their coproduct is based on the abelian group $$\bigoplus_{\large i_1,\dotsc,i_m \in I, ~ i_s \neq i_{s+1}} R_{i_1} \otimes \cdots \otimes R_{i_m}.$$

If $K$ is any ring, then coproducts of $K$-rngs can be understood in the same way, just replace $\otimes$ by $\otimes_K$, the tensor product of $K$-bimodules.

And actually, the same construction can be used to construct coproducts of non-unital algebra objects inside any monoidal category with coproducts which distribute over the tensor product. (It does not have to be symmetric.) For rngs we take the monoidal category $(\mathbf{Ab},\otimes)$, for $K$-rngs we take the monoidal category $(\mathbf{BiMod}_K,\otimes_K)$. We may also apply it to $(\mathbf{Set},\times)$ to get coproducts of semigroups.