(All my rings are commutative, but not necessarily unital.)
I was playing around with the ring freely generated by an Abelian group, and it seems to me that the following holds: letting $U$ denote the functor that takes a ring and returns its underlying Abelian group, and letting $F$ denote its left-adjoint, it would seem that $F(U(R))$ is isomorphic to the ring of all polynomials with coefficients in $R$ and constant term equal to $0$.
For example, consider the ring $\mathbb{Z}/4\mathbb{Z} = \{0,1,2,3\}$. Now take its underlying Abelian group, and change notation by introducing an $x$ variable. So:
$$U(\mathbb{Z}/4\mathbb{Z}) = \{0,x,2x,3x\}$$
Then the ring freely generated by this Abelian group should have elements like $$2x+3x^2+x^5$$
as well as a $0$ element.
Question. Is it true that, for all rings $R$, $F(U(R))$ is always isomorphic to the ring of polynomials with coefficients in $R$ and constant term equal to $0$? If so, how can we prove this rigorously?
