Adjoint to the forgetful functor $U:\mathbf{Ring}\to\mathbf{AbGrp}$ So, the next, and hopefully last, question in my growing list of questions about adjoints to forgetful functors concerns the left adjoint to functor $U:\mathbf{Ring}\to\mathbf{AbGrp}$.
My approach so far has been to take the free monoid $F(A)$ over the set under the abelian group A, then the free abgroup $AbF(A)$ of $F(A)$, and finally identify all elements $'ab+ab'$ with $a(b+b)$ and $(a+a)b$. 
But it does feel like I've gone too far in forgetting the abelian structure when making the free monoid. Perhaps we can take a free monoid over the group immediately and define it to be consistent with addition so that $a(b+c)d= abd+acd$. 
I found that a similar approach was necessary when defining the free group over a monoid, but the situation is very different here, and I lack the skill to confidently pursue my own solutions.
Thanks in advance for any answers.
 A: To give some more details about my comment: We let $F(A) = \bigoplus_{n\ge 0} A^{\otimes n}$ with the obvious addition and the multiplication induced by the maps $A^{\otimes n} \times A^{\otimes m} \to A^{\otimes (n+m)}$ on the summands. Then $F(A)$ is a ring. Given a Ring $R$ and an $\mathbf{AbGrp}$-morphism $f\colon A \to UR$ we define $\bar f\colon F(A) \to R$ by $\bar f(a_1 \otimes \cdots \otimes a_n) := f(a_1)\cdots f(a_n)$. Then $\bar f$ is a $\mathbf{Ring}$-morphism (just check by calculating). Obviously $ \bar f|_{A} = f$ (here we use $A = A^{\otimes 1} \subseteq F(A)$), so $f \mapsto \bar f$ is one-to-one. To see that it is onto, we let $g\colon F(A) \to R$ be a $\mathbf{Ring}$-morphism and $f := g|_A$. We now have
$$
  g(a_1\otimes \cdots \otimes a_n) 
   = g(a_1)\cdots g(a_n)
   = f(a_1) \cdots f(a_n)
   = \bar f(a_1 \otimes \cdots \otimes a_n)
$$
So $g = \bar f$. So $f \mapsto \bar f$ gives a bijection $\mathbf{AbGrp}(A, UR) \to \mathbf{Ring}(FA, R)$ which is natural. So $F$ is left adjoint to $U$.
A: When $R$ is a ring, the forgetful functor $\mathrm{Alg}(R) \to \mathrm{Mod}(R)$ has a left adjoint, namely the well-known tensor algebra of a module. If you want the left adjoint for the forgetful functor $\mathrm{CAlg}(R) \to \mathrm{Mod}(R)$, we have to mod out commutators in the tensor algebra and arrive at the well-known symmetric algebra. For $R=\mathbb{Z}$ you get the situation of your question.
If you want to avoid tensor products, you can do the following: Let $M$ be an $R$-module. Then the tensor algebra $T(M)$ is the free $R$-algebra generated by symbols $\underline{m}$, subject to the relations $\underline{m+n}=\underline{m}+\underline{n}$ and $r \cdot \underline{m} = \underline{r \cdot m}$ for all $m,n \in M$ and $r \in R$. Thus, elements of $T(M)$ are noncommutative polynomials in $M$, for example $m \cdot n + m \cdot n \cdot m$ (usually one writes $m$ instead of $\underline{m}$; this is OK since one can show that $M \to T(M)$ is injective). In order to construct the symmetric algebra, you take the free commutative $R$-algebra etc.. For example the above polynomial becomes identified with $m \cdot n + m \cdot m \cdot n$.
There is a general procedure how to produce left adjoint functors for forgetful functors between algebraic categories; I've already illustrated this in your former questions: How to make a monoid commutative or into a group. If you haven't seen the general pattern, try another example: Find the left adjoint of the forgetful functor $B \hookrightarrow \textbf{Mon}$, where $B$ is the category of "boolean monoids"; a monoid $M$ is called boolean if $m^2=m$ for all $m \in M$.
