# 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.

• Can't you just take $F(A) = \bigoplus_{n\ge 0} A^{\otimes n}$ as the free ring over $A$? Apr 11, 2012 at 11:16

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 $U$.

• Thanks a million, im not particularly fond of the tensor functor but i guess there is no way out of it. Apr 11, 2012 at 17:44
• This verification is a little bit cumbersome. It is better to write down maps $\mathrm{Hom}(A,U(R)) \cong \mathrm{Hom}(F(A),R)$ in both directions, show that they are well-defined (using the universal property of the tensor product, of course), and inverse to each other. (When there is a canonical bijection, you should not write down a map and show that it is injective and surjective, when the inverse is visible) Apr 12, 2012 at 18:09

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 $m \in M$.

• Your help is much appreaciated and dont worry, the pattern is begining to emerge. To make a boolean monoid you simply take the quotient $M/Eq$ where $Eq$ is the smallest congurence relation with the pairs $(m^2,m)$ for all $m\in M$ Further it does not have a right adjoint since coproducts arent preserved. But now im wondering if there is an approach similiar to the one above to make lattices. But dont answer unless its no. Apr 13, 2012 at 0:56
• Yes, exactly. Can you also describe the congruence relation explicitly? Apr 13, 2012 at 10:36
• It would be nice if boolean monoids where commutative. If so we get that any words written with the same letters are equal, regardless of order. If not we get $x\sim y$ iff $x=(\prod_{i\in I}m_i)^k$ and $y=(\prod_{i\in I}m_i)^l$ for some index $I$ and $k,l \in \mathbb{N}$ Apr 13, 2012 at 14:23
• I dont know if your are still following this thread but just incase. Suppose we are only dealing with commutative boolean monoids, we then get hom-sets which themself are boolean monoids and boolean "tensor products" as thier adjoints. Since we have coproducts it should be possible to do a free functor from the boolean monoids to the boolean "rings". The whole things seems very reasonable to me the question is if this line actually goes anywhere? I know next to nothing about boolean monoids or algebras so im sort of lost. If you could give me quike yes or no that would be great Apr 13, 2012 at 17:11