If the monoid algebra $R[M]$ is finitely generated, then $M$ is a finitely generated monoid.

Question

Consider a commutative, cancellative, torsion-free monoid $M$ and a commutative ring $R.$ If the monoid algebra $R[M]$ is finitely generated as an $R$-algebra, then $M$ is finitely generated.

I am following the proof of the above fact (Proposition 2.7 in Bruns and Gubeladze's Polytopes, Rings, and K-Theory), and the authors' exposition leaves something to be desired.

Explicitly, the argument is as follows.

We will assume that $f_1, \dots, f_n$ generate $R[M]$ as an $R$-algebra. By definition of $R[M],$ there exist finitely many symbols $x^{m_i}$ and scalars $r_{m_i}$ such that $f_i = \sum r_{m_i} x^{m_i}.$ Consider the finite set $G$ that consists of the elements $m_i$ in $M.$ We have that $M' = \mathbb Z_+ G$ is a finitely generated monoid. We claim that $M = M'.$ Certainly, we have that $M \supseteq M',$ hence it suffices to prove the inclusion $\subseteq.$ Observe that any $R$-linear combination of the monomials $f_1^{a_1} \cdots f_n^{a_n}$ with $a_i \in \mathbb Z_+$ can be written as an $R$-linear combination of monomials $x^a$ for some element $a$ of $M'.$ Considering that $f_1, \dots, f_n$ generate $R[M]$ as an $R$-algebra, it follows that each of the symbols $x^b$ for $b$ in $M$ can be written as an $R$-linear combination of some $f_1^{a_1} \cdots f_n^{a_n}$ with $a_i \in \mathbb Z_+,$ and thus, each of the symbols $x^b$ is an element of $R[M'].$

From here, Bruns and Gubeladze conclude that "this implies that $M = M'$;" however, I fail to see why this should be true. I would appreciate any assistance or suggestions. Thank you in advance.

Answer 1

Claim. Consider a commutative ring $R$ and a free $R$-module $X$ with a basis $B.$ Given any subset $B’$ of $B,$ we have that $X' \cap B = B',$ where $X'$ is the $R$-submodule of $X$ that is spanned by $B'.$

Proof. Certainly, we have that $B' \subseteq X' \cap B$ since every element of $B'$ is contained in $B$ and the elements $b' = 1_R \cdot b'$ of $B'$ are all contained in $X'.$ Conversely, given any element $x$ of $X' \cap B,$ we have that $x = r_1 \cdot b_1 + \cdots + r_n \cdot b_n$ for some elements $r_i$ of $R$ and $b_i$ of $B'$ and $x = b = 1_R \cdot b$ for some element of $B.$ Observe that $b = r_1 \cdot b_1 + \cdots + r_n \cdot b_n$ is a linear combination of elements of $B.$ But the expression of any element in $X$ as an $R$-linear combination is uniquely determined by the scalars $r_i$ and basis elements $b_i,$ hence we must have that $x = b = b_i$ for some index $i,$ i.e., $x$ is in $B'.$ We conclude therefore that $X' \cap B = B'.$ QED.

By definition, we have that $R[M]$ is the free $R$-module with a basis consisting of the monomials $x^m$ for each element $m$ of $M.$ We may view $M$ as a subset of $R[M]$ via the injective map $M \to R[M]$ that sends $m \mapsto x^m.$ We have shown that for each element $b$ of $M,$ we have that $x^b$ is in $R[M'],$ from which it follows by our identification that $M \subseteq R[M'].$ By the claim above, we have that $M = R[M'] \cap M = M'.$

Answer 2

Let us expound on the last line as follows to strengthen our intuition of the result.

Considering that $f_{1}, f_{2}, \cdots f_{n}$ generate $R[M]$ as an $R$-algebra, it follows that each of the symbols $X^{b}$ for $b \in M$ can be written as an $R$-linear combination of some $f_{1}^{a_{1}} \cdots f_{n}^{a_{n}}$ with $a_{i} \in \mathbb{Z}_{+}$...

This line recalls what it takes to be a system of generators for $R[M]$ as an $R$-algebra to collect sufficient arguments to prove the next point.

and thus, each of the symbols $X^{b}$ is an element of $R[M′]$.

Here, the author has used that fact that any linear combination of $f_{i}^{a_{i}}$ can be written as a linear combination of monomials $X^{a}$ with $a \in M'$. Now, fix some $b \in M$. Since

• $X^{b}$ can be written as an $R$-linear combination of monomials $f_{1}^{a_{1}} \cdots f_{n}^{a_{n}}$ with $a_{i} \in \mathbb{Z}_{+}$, and

• any $R$-linear combination of $f_{i}^{a_{i}}$ can be written as a linear combination of monomials $X^{a}$ with $a \in M'$

then, $X^{b}$ can be written as a linear combination of monomials $X^{a}$ with $a \in M'$.

From here, we can see that $X^{b} \in R[M']$. Hence, we have $b \in M'$ by definition of $R[M']$. At this point, we have shown that whenever we fix $b \in M$, we also have $b \in M'$.

Hence, we get the desired inclusion $M \subseteq M'$.

P.S. Sorry, took me a while to find this PDF link http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.126.2385&rep=rep1&type=pdf to check in context what the author might have assumed the reader must already know at the point of presentation. Noticed that the version that I found was an incomplete version of the monograph. :)

Update: Corrected commas as products in $f_{1}^{a_{1}} \cdots f_{n}^{a_{n}}$. Thanks, @Carlo! :)