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

## Best Answer

The universal property of the polynomial ring $K[x]$ is that

$$\hom_K(K[x], R) \simeq |R|,$$

where $\hom$ is taken in the category of $K$-algebras, and $|R|$ is the underlying set of $R$. The bijection is determined by looking at the image of the free variable $x$. In other words, a free variable is free to go where it wants.

Another way to say this is that $K[x]$

representsthe forgetful functor from $K$-algebras to sets.In MacLane's language, the element $x \in |K[x]|$ is the "universal element" for the forgetful functor.