# From monoids to commutative rings

We shall first define a functor $$F:\mathsf{Mon}^{\text{op}}\to\mathsf{CRing},$$ where $$\mathsf{Mon}^{\text{op}}$$ is the category opposite to the category of monoids and $$\mathsf{CRing}$$ is the category of commutative rings with one.

Let $$\mathsf{CSRing}$$ be the category of commutative semirings with one, and $$L:\mathsf{CSRing}\to\mathsf{CRing}$$ the left adjoint to the forgetful functor (in particular $$L(\mathbb N)=\mathbb Z$$).

Our functor $$F$$ will be the composite $$L\circ F'$$, where $$F':\mathsf{Mon}^{\text{op}}\to\mathsf{CSRing}$$ is defined as follows:

Let $$M$$ be a monoid, $$M$$-$$\mathsf{Set}_{\text{fin}}$$ the category of finite $$M$$-sets, and $$S\subset M\text{-}\mathsf{Set}_{\text{fin}}$$ a skeleton. Set $$0:=\varnothing\in S$$, let $$1\in S$$ be the terminal object, and for $$X,Y\in S$$ let $$X+Y\in S$$ be the coproduct of $$X$$ and $$Y$$, and $$XY\in S$$ the product of $$X$$ and $$Y$$.

It is straightforward to check that the formula $$F'(M):=S$$ defines a functor $$F':\mathsf{Mon}^{\text{op}}\to\mathsf{CSRing}$$, and we can set $$F:=L\circ F'$$.

As $$F$$ sends the trivial monoid to $$\mathbb Z$$, we have a natural morphism $$F(M)\to\mathbb Z$$. In other words, it would be better to view $$F$$ as a (contravariant) functor from monoids to commutative rings over $$\mathbb Z$$. In particular $$F(M)$$ contains $$\mathbb Z$$.

Question 1. Is the Krull dimension of $$F(M)$$ always equal to one?

Question 2. If the monoid $$M$$ is a group $$G$$, is $$F(G)$$ always integral over $$\mathbb Z$$?

$$\bullet$$ the finite index subgroups of $$G$$ are normal,

or if

$$\bullet\ G$$ is finite.

(See below.)

Clearly, if $$M$$ is a group $$G$$, and $$f:G\to\hat G$$ is the morphism to the profinite completion, then $$Ff:F(\hat G)\to F(G)$$ is an isomorphism.

The ring $$F(M)$$ can be described as follows:

Say that an $$M$$-set is indecomposable if it cannot be written as a disjoint union of sub-$$M$$-sets in a nontrivial way, and let $$I$$ be the set of indecomposable objects of cardinality at least two in our skeleton $$S$$. For all $$X,Y\in I$$ we have $$XY=\sum_{Z\in I}\ c_{XY}^Z\ Z,$$ where each $$(c_{XY}^Z)_{Z\in I}$$ is a finitely supported family of nonnegative integers. We get $$F(M)\simeq\mathbb Z\left[(T_X)_{X\in I}\right]/\mathfrak a,$$ where the $$T_X$$ are indeterminates and $$\mathfrak a$$ is the ideal generated by the $$T_XT_Y-\sum_{Z\in I}\ c_{XY}^Z\ T_Z.$$ In particular the element $$1\in F(M)$$ and the images $$t_X$$ of the $$T_X$$ form a $$\mathbb Z$$-basis of $$F(M)$$, and the natural morphism $$F(M)\to\mathbb Z$$ sends $$t_X$$ to zero.

Also note that if $$M$$ is a group $$G$$, and $$N$$ a normal subgroup of index $$i<\infty$$, then we have $$(t_{G/N})^2=it_{G/N}$$, and $$t_{G/N}$$ is integral over $$\mathbb Z\subset F(G)$$. This justifies the first claim after Question 2.

To prove the second claim after Question 2, recall that $$I$$ is the set of indecomposable objects of cardinality at least two in the skeleton $$S$$, and that $$F(G)$$ is generated by a family $$(t_X)_{X\in I}$$.

Assume first that the monoid $$M$$ is a (possibly infinite) group $$G$$. Order $$I$$ by setting $$X\le Y$$ if there is a surjective morphism $$Y\to X$$. We claim

(a) for all $$X\in I$$ the ring $$F(G)$$ is integral over the subring generated by the $$t_Y$$ with $$Y>X$$.

More precisely:

(b) for all $$X\in I$$ we have $$X^2=nX+\sum_{Y>X}n_YY$$ with $$n,n_Y\in\mathbb N$$.

To prove (b) note that, for all $$x_1,x_2\in X$$, the stabilizer $$H$$ of $$(x_1,x_2)\in X^2$$ is the intersection of the stabilizers $$H_1$$ and $$H_2$$ of $$x_1$$ and $$x_2$$. Thus we have either $$H=H_1=H_2$$ and $$G(x_1,x_2)\simeq X$$ , or $$H for $$i=1,2$$, and $$G(x_1,x_2)>X$$ (more correctly $$Y>X$$ if $$Y$$ is the unique element of $$I$$ isomorphic to $$G(x_1,x_2)$$). This proves (b), and thus (a).

Clearly, if $$G$$ is finite, (a) implies that $$F(G)$$ is integral over $$\mathbb Z$$, which is the second claim after Question 2.

Here are some examples:

As already indicated, if $$M$$ is the trivial monoid, then $$F(M)\simeq\mathbb Z$$. We also have $$F(\mathbb Q)\simeq\mathbb Z$$.

The ring $$F(\mathbb Z)$$ admit a $$\mathbb Z$$-basis $$\{1,t_2,t_3,\dots\}$$ with $$t_it_j=(i\land j)\ t_{i\lor j},$$ where $$i\land j$$ and $$i\lor j$$ denote the gcd and the lcm of $$i$$ and $$j$$.

If $$M$$ is the monoid $$\{0,1\}$$ with the obvious multiplication, then the ring $$F(M)$$ admit a $$\mathbb Z$$-basis $$\{1,t_1,t_2,\dots\}$$ with $$t_it_j=t_{(i+1)(j+1)-1}.$$ If $$S_3$$ denotes the symmetric group on three letters, then the ring $$F(S_3)$$ admit a $$\mathbb Z$$-basis $$\{1,t_2,t_3,t_6\}$$ with $$t_2^2=2t_2,\quad t_3^2=t_3+t_6,\quad t_it_6=it_6,\quad t_2t_3=t_6.$$ If $$G$$ is the Klein four-group (that is, the non-cyclic group of order $$4$$), then the ring $$F(G)$$ admit a $$\mathbb Z$$-basis $$\{1,t_1,t_2,t_3,u\}$$ with $$t_i^2=2t_i,\quad u^2=4u,\quad t_iu=2u,\quad t_it_j=u\text{ for }i\ne j.$$

Question 1: No. Indeed, the monoid $$M=\{0,1\}$$ you mentioned is a counterexample. Let me write $$x_n$$ for what you have written $$t_{n-1}$$, so $$x_n$$ is the indecomposable $$M$$-set with $$n$$ elements (the $$n$$-element set with only one point in the image of $$0$$). Then these elements $$x_n$$ satisfy $$x_nx_m=x_{nm}$$. This makes it clear that actually $$F(M)$$ is just the polynomial ring $$\mathbb{Z}[x_2,x_3,x_5,\dots]$$ on the elements $$x_p$$ where $$p$$ is prime. This ring has infinite Krull dimension.
Question 2: Yes. To prove this, let $$X$$ be any finite $$G$$-set and let $$K$$ be the kernel of the action of $$G$$ on $$X$$. Note that $$K$$ is a finite index normal subgroup of $$G$$, and acts trivially on $$X^n$$ for all $$n$$. It follows that actually the subring of $$F(G)$$ generated by $$X$$ is isomorphic to the subring of $$F(G/K)$$ generated by $$X$$. Since $$G/K$$ is finite, this shows $$X$$ is integral over $$\mathbb{Z}$$ by the work you have done.
Incidentally, there is a quicker way to see $$F(G)$$ is integral over $$\mathbb{Z}$$ when $$G$$ is a finite group. Just note that $$F(G)$$ is generated as an abelian group by the $$G$$-sets $$G/H$$ where $$H$$ ranges over all subgroups of $$G$$. In particular, $$F(G)$$ is a finitely generated $$\mathbb{Z}$$-module and thus is integral over $$\mathbb{Z}$$.