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We shall first define a functor
$$
F:mathsfMon^textoptomathsfCRing,
$$
where $mathsfMon^textop$ is the category opposite to the category of monoids and $mathsfCRing$ is the category of commutative rings with one.

Let $mathsfCSRing$ be the category of commutative semirings with one, and $L:mathsfCSRingtomathsfCRing$ the left adjoint to the forgetful functor (in particular $L(mathbb N)=mathbb Z$).

Our functor $F$ will be the composite $Lcirc F'$, where $F':mathsfMon^textoptomathsfCSRing$ is defined as follows:

Let $M$ be a monoid, $M$-$mathsfSet_textfin$ the category of finite $M$-sets, and $$
Ssubset Mtext-mathsfSet_textfin
$$
a skeleton. Set $0:=varnothingin S$, let $1in S$ be the terminal object, and for $X,Yin S$ let $X+Yin S$ be the coproduct of $X$ and $Y$, and $XYin S$ the product of $X$ and $Y$.

It is straightforward to check that the formula $F'(M):=S$ defines a functor $F':mathsfMon^textoptomathsfCSRing$, and we can set $F:=Lcirc F'$.

As $F$ sends the trivial monoid to $mathbb Z$, we have a natural morphism $F(M)tomathbb Z$. 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 $mathbb Z$?

The answer is Yes if

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

or if

$bullet G$ is finite.

(See below.)

Clearly, if $M$ is a group $G$, and $f:Gtohat G$ 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 $X,Yin I$ we have
$$
XY=sum_Zin I c_XY^Z Z,
$$
where each $(c_XY^Z)_Zin I$ is a finitely supported family of nonnegative integers. We get
$$
F(M)simeqmathbb Zleft[(T_X)_Xin Iright]/mathfrak a,
$$
where the $T_X$ are indeterminates and $mathfrak a$ is the ideal generated by the
$$
T_XT_Y-sum_Zin I c_XY^Z T_Z.
$$
In particular the element $1in F(M)$ and the images $t_X$ of the $T_X$ form a $mathbb Z$-basis of $F(M)$, and the natural morphism $F(M)tomathbb Z$ 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 $mathbb Zsubset F(G)$. 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 $(t_X)_Xin I$.

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

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

More precisely:

(b) for all $Xin I$ we have
$$
X^2=nX+sum_Y>Xn_YY
$$
with $n,n_Yinmathbb N$.

To prove (b) note that, for all $x_1,x_2in X$, 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 $H<H_i$ 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 $F(M)simeqmathbb Z$. We also have $F(mathbb Q)simeqmathbb Z$.

The ring $F(mathbb Z)$ admit a $mathbb Z$-basis $1,t_2,t_3,dots$ with
$$
t_it_j=(iland j) t_ilor j,
$$
where $iland j$ and $ilor j$ 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
$$
t_i^2=2t_i,quad u^2=4u,quad t_iu=2u,quad t_it_j=utext for ine j.
$$

First, a remark: your construction is known as the Burnside ring (usually considered only in the case of a group) which I imagine will aid you in finding more information about it.

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 $mathbbZ[x_2,x_3,x_5,dots]$ 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 $mathbbZ$ by the work you have done.

Incidentally, there is a quicker way to see $F(G)$ is integral over $mathbbZ$ 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 $mathbbZ$-module and thus is integral over $mathbbZ$.