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.
Best Answer
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 $U$.