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(This question was asked a long time ago on MSE but got no answer so far...)

Let $E$ be an additively written cancellable commutative monoid with no non-trivial units. We furnish $E$ with the order defined by "$x\leq y$ if and only if there exists $z\in E$ with $y=x+z$", so that it is an ordered monoid. If we consider a subset $F\subseteq E$, we can ask whether it is noetherian as an ordered set, i.e., whether every nonempty subset of $F$ has a maximal element.

It is not hard to see that if $E$ is finitely generated, then the noetherian subsets of $E$ are precisely the finite ones. So, we could ask for the converse: If the noetherian subsets of $E$ are precisely the finite ones, is then $E$ finitely generated?

I think this is not true, but was not able to come up with a counterexample. So, here is my question:

What is an example of a monoid as above, but not finitely generated, whose noetherian subsets are precisely the finite ones?

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If a monoid has the property (P) that all noetherian subsets are finite then it is finitely generated:

  1. If a partially ordered set has property (P), then any subset has a minimal element: otherwise there is an infinite descending sequence in the subset, and the elements of a descending sequence form a noetherian subset.

  2. Assume a monoid has property (P) and is not finitely generated. Let $x_1$ be a minimal element of the monoid, let $x_2$ be a minimal element not in the monoid generated by $x_1$, let $x_3$ be a minimal element not in the monoid generated by $x_1$ and $x_2$, etc. The $x_i$ are pairwise incomparable because if $x_i \lt x_j$ then: if $i>j$ then $x_i$ contradicts the minimality of $x_j$; if $i<j$ with $x_j = x_i + y$ for some $y$ then $y$ contradicts the minimality of $x_j$ ($y$ can't be in the monoid generated by $x_1$,...,$x_{j-1}$ else so would be $x_j$). Thus {$x_1, x_2, ...$} would be an infinite noetherian subset, contradicting property (P).

If the monoid is totally ordered then it must be ℕ$x_1$: $n{x_1}<n{x_1}+y<(n+1){x_1}$ for some $n$ would imply $y<x_1$.

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  • $\begingroup$ Dear @David, thank you very much for this proof. $\endgroup$ – Fred Rohrer May 26 '15 at 12:18

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