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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.

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Can't you just take $F(A) = \bigoplus_{n\ge 0} A^{\otimes n}$ as the free ring over $A$? –  martini Apr 11 '12 at 11:16
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2 Answers

up vote 4 down vote accepted

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 $U$.

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Thanks a million, im not particularly fond of the tensor functor but i guess there is no way out of it. –  user25470 Apr 11 '12 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) –  Martin Brandenburg Apr 12 '12 at 18:09
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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 $m \in M$.

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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. –  user25470 Apr 13 '12 at 0:56
Yes, exactly. Can you also describe the congruence relation explicitly? –  Martin Brandenburg Apr 13 '12 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 $k,l \in \mathbb{N}$ –  user25470 Apr 13 '12 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 –  user25470 Apr 13 '12 at 17:11
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