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.

  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Aug 24, 2017 at 19:37
  • $\begingroup$ en.wikipedia.org/wiki/… ? $\endgroup$
    – user451844
    Aug 24, 2017 at 19:40
  • $\begingroup$ @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. $\endgroup$ Aug 24, 2017 at 19:41
  • 1
    $\begingroup$ @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. $\endgroup$
    – rschwieb
    Aug 24, 2017 at 19:42
  • $\begingroup$ @MartinAzpillaga You're saying that you're familiar with semigroup rings and you think that is not the construction implied? $\endgroup$
    – rschwieb
    Aug 24, 2017 at 19:44

1 Answer 1


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


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