Semiring construction from a monoid Reading throug the paper Definable sets up to definable bijections in Presburger groups by Raf Cluckers and Immanuel Halupczok, I have come across a construction (Definition 2.2.1) that builds a commutative semiring out of a commutative monoid $M$. 
At first, I thought it was talking about the set of all maps $M\to \mathbb{N}$ with finite support where + is given by addition of maps and $\times$ is the Cauchy product
$$
(f \times g)(z) = \sum_{a+b = z} f(a)g(b) \cdot \textbf{1}_z 
$$
where $\textbf{1}_z$ is the map sending $z \mapsto 1$ and every other element to $0$. The construction continues by identifying the elements
$$
\begin{array}{c}
\textbf{1}_a + \textbf{1}_b = \textbf{1}_{a+b}\\
\textbf{1}_0 = 0\\
\textbf{1}_g = 1\\
\end{array}
$$
where $0$ and $1$ are the neutral elements of the resulting semiring and $g$ is a preselected element of $M$.
The authors provide an example (after Theorem 2.3.4) of such a construction applied to the monoid $\mathbb{N}_\infty$. 
When I follow this construction, I get that the elements of the resulting semiring are equivalence classes represented by
$$
\mathbb{N}_+ \cup \{\infty\}
$$
while they assert that they should be given by
$$
\mathbb{N} \cup \{\infty, \infty^2, \infty^3,\dots\}
$$
Is there another construction of a semiring where after doing the identification this is the resulting semiring? 
They name the construction as reduced symmetric algebra, but reading about symmetric algebras I can't figure out what is the mathematical object they are constructing.
 A: I think the construction as given in the article is correct. First of all, given a set $G$, the free commutative semiring over $G$ is the semiring of commutative polynomials $\mathbb{N}[(X_g)_{g \in G}]$.
Let $G$ be a commutative monoid and let $s: (\mathbb{N},+) \to G$ be a monoid homomorphism. Now, the semiring defined in the article is the quotient $SG$ of $\mathbb{N}[(X_g)_{g \in G}]$ by the relations $X_0 = 0$, $X_{s(1)} = 1$ and $X_g + X_h = X_{g+h}$.
Suppose first that $G = \mathbb{N}$. Then, one gets by induction the relations $X_n = n$ for all $n$ and hence $SG = \mathbb{N}$. If now $G = \mathbb{N} \cup \{\infty\}$, one gets the further relations $X_n + X_\infty = X_\infty$, whence $n + X_\infty = X_\infty$, for all $n \in \mathbb{N}$ and $X_\infty + X_\infty = X_\infty$. It follows that in this case, $SG$ is the quotient of $\mathbb{N}[X_\infty]$ by the relations $n + X_\infty = X_\infty$ and $X_\infty + X_\infty = X_\infty$. It follows that $kX_\infty = X_\infty$ for all $k > 0$. Moreover
$X_\infty^2 = X_\infty(1 + X_\infty) = X_\infty  + X_\infty^2$.
Consider now a polynomial $c_{i_1}X_\infty^{i_1} + \ldots + c_{i_n}X_\infty^{i_n}$, with nonzero coefficients and $i_1 < \dotsm < i_n$. Then
$$c_{i_1}X_\infty^{i_1} + \dotsm + c_{i_n}X_\infty^{i_n} = X_\infty^{i_1} + \dotsm + X_\infty^{i_n} = X_\infty^{i_n}$$
and thus $SG = \mathbb{N} \cup \{X_\infty^k \mid k \in \mathbb{N}\}$.
