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Does there exist a non-trivial commutative monoid that does not have non-trival idempotents yet has a trivial Grothendieck group?

To partially explain my motivation, a cute little monoid! Consider a commutative monoid $M = \{0, k, a\},\ a + a = a + k = k + k = k$. It is very trivial, but at the same time interesting, because its Grothendieck group is trivial, yet $a$ is not an idempotent (which illustrates how idempotents can 'drag down' other elements to the kernel of the canonical map $x \mapsto x - 0$), and $a$ is also not a sum of an invertible and an idempotent element.

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  • $\begingroup$ It was a misguided comment; sorry for the confusion. $\endgroup$ – Arturo Magidin Feb 2 '12 at 2:02
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Yes. Let $\mathbb{Z}_+ = \{1,2,3,\ldots\}$, let $M = \{(0,0)\}\cup (\mathbb{Z}_+\times\mathbb{Z}_+)$, and consider the monoid structure on $M$ defined by $$ (x_1,y_1) + (x_2,y_2) \;=\; \begin{cases}(x_2,y_2) & \text{if }x_1<x_2 \\ (x_1,y_1+y_2) & \text{if }x_1 = x_2 \\(x_1,y_1) & \text{if }x_1 > x_2\end{cases} $$ Then the Grothendieck group for this monoid is trivial, since $(x,y) + (x+1,y) = (x+1,y)$ for all $(x,y)\in\mathbb{N}\times\mathbb{N}$. However, the monoid itself is non-trivial and commutative, and has no nontrivial idempotent elements.

Incidentally, this monoid can also be described in the following way: let $\mathbb{N}[x]$ be the monoid of all polynomials with natural number coefficients under addition, and let $\sim$ be the congruence relation on $\mathbb{N}[x]$ defined by $p(x)\sim q(x)$ if and only if $p(x)$ and $q(x)$ have the same leading term. Then the quotient $\mathbb{N}[x]/{\sim}$ is isomorphic to the monoid $M$ defined above.

Though I do not have a proof, I would guess that any example must be infinite. The example above isn't even finitely-generated.

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    $\begingroup$ Suppose $M$ is a finite monoid without nontrivial idempotents. If $x\in M$ is not $1$, then the set $\{x, x^2,x^3,\dots\}$ is finite and a little work shows it contains an idempotent. The idempotent must be $1$, by hypothesis, so $x$ is in fact invertible in $M$. We thus see that $M$ is a group. $\endgroup$ – Mariano Suárez-Álvarez Feb 3 '12 at 7:31

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