# 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.

• What I read here is "I came across a construction... at first I thought it was this other construction..." so now I am confused if you are describing the other construction or the one you are confused about. Please clarify. Aug 24, 2017 at 19:37
• – user451844
Aug 24, 2017 at 19:40
• @rschwieb The construction I provide was my first guess, but it is not the correct one. I am looking for a different construction that provides the semiring with elements $\mathbb{N} \cup \{\infty,\infty^2,\dots\}$ when applied to the example given. Aug 24, 2017 at 19:41
• @MartinAzpillaga It is silly to ask us to reverse-engineer a construction to explain something you read in a paper which you haven't even cited for us. Someone may be able to do it, but otherwise you are not giving the rest of us much to go on. Aug 24, 2017 at 19:42
• @MartinAzpillaga You're saying that you're familiar with semigroup rings and you think that is not the construction implied? Aug 24, 2017 at 19:44

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 $SG = \mathbb{N} \cup \{X_\infty^k \mid k \in \mathbb{N}\}$.