Question of category of rings

Question

How to verify that $\mathbb{Z}[x_1, x_2],$ with the evident morphisms satisfies the universal property for the coproduct of two copies of $\mathbb{Z}[x]$ in the category of commutative rings. Further why it does not satisfy it in the category of Rings.

Answer 1

Using the yoneda lemma

$$\begin{align} \mathbf{CRing}(\mathbb{Z}[x] \amalg \mathbb{Z}[y], R) &\cong \mathbf{CRing}(\mathbb{Z}[x], R) \times \mathbf{CRing}(\mathbb{Z}[y], R) \\&\cong R \times R \cong R^2 \\&\cong \mathbf{CRing}(\mathbb{Z}[x,y] , R) \end{align}$$

So the yoneda lemma says $\mathbb{Z}[x] \amalg \mathbb{Z}[y] \cong \mathbb{Z}[x,y]$.

Using free rings

If $F$ is the free ring functor, then

$$ \mathbb{Z}[x,y] \cong F(\{x, y \}) \cong F(\{x\} \amalg \{y\}) \cong F(\{x\}) \amalg F(\{y\}) \cong \mathbb{Z}[x] \amalg \mathbb{Z}[y] $$

The free ring functor preserves coproducts (in fact, all colimits) because it is a left adjoint. (its right adjoint is the forgetful functor that sends a ring to its set of elements)

Noncommutative rings

For noncomutative rings, the problem is that

$$\mathbf{Ring}(\mathbb{Z}[x,y], R) \cong \left\{ (a,b) \in R^2 \mid ab = ba \right\} $$

or equivalently,

$$ \mathbb{Z}[x,y] \cong F(\{x,y\}) / \langle xy - yx \rangle $$