Strictly totally ordered semigroups - Looking for references - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:32:43Z http://mathoverflow.net/feeds/question/105851 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105851/strictly-totally-ordered-semigroups-looking-for-references Strictly totally ordered semigroups - Looking for references Salvo Tringali 2012-08-29T15:08:52Z 2012-08-29T22:40:50Z <p>Let $\mathfrak A = (A, \cdot)$ be a semigroup (written multiplicatively). We say that $\mathfrak A$ is <em>linearly orderable</em> if there exists a total order $\le$ on $A$ such that $ac &lt; bc$ and $ca &lt; cb$ for all $a,b,c \in A$ with $a &lt; b$ (note strict inequalities).</p> <p>Some examples of linearly orderable semigroups are: the real numbers with the usual addition; the positive integers divisible only for the members of a given set $S$ of (natural) primes with the usual multiplication; every abelian torsion-free cancellative semigroup (see Proposition 2 below); the polynomials in finitely many variables with nonnegative real coefficients with the usual Cauchy multiplication; the upper [lower] triangular matrices with positive real entries and the usual row-by-column multiplication; the free monoid on an alphabet $X$; subsemigroups and direct products of the previous ones.</p> <p>Now, denote by $\mathfrak A^{(1)}$ the (canonical) unitization of $\mathfrak A$. As a byproduct of something that I've just finished to write, I happened to prove the following:</p> <p><strong>Proposition 1.</strong> $\mathfrak A$ is linearly orderable (if and) only if the same holds true with $\mathfrak A^{(1)}$. </p> <p><strong>Proposition 2.</strong> Every abelian torsion-free (*) cancellative semigroup $\mathfrak A$ is linearly orderable.</p> <p>Proposition 2 has a kind of (trivial) converse: Every linearly orderable semigroup is torsion-free and cancellative (indeed, something stronger can be proved; i.e., none of the elements of the semigroup has finite order unless the semigroup is unital and such an element is the identity).</p> <p>I am reasonably sure that both results are nothing new, but I wasn't able to find any reference. In particular, I checked <a href="http://www.ams.org/journals/bull/1958-64-06/S0002-9904-1958-10221-9/S0002-9904-1958-10221-9.pdf" rel="nofollow">Clifford's 1958 survey</a>, but this seems focused more on totally ordered semigroups (there referred to simply as <em>ordered semigroups</em>) than on linearly ordered semigroups (there called <em>strictly ordered semigroups</em>). On another hand, I am aware of a 1913 result by F.H. Levi (<em>Arithmetische Gesetze im Gebiete diskreter Gruppen,</em> Rend. Circ. Mat. Palermo, Vol. 35 (1913), pp. 225-236), where it is proved that every torsion-free abelian <em>group</em> is linearly orderable (as a group). On another hand, I have no clue about Proposition 1. Then, here are my questions:</p> <blockquote> <blockquote> <p><strong>Question 1.</strong> Do you know of any paper, book, comic strip (<a href="http://mathworld.wolfram.com/FoxTrotSeries.html" rel="nofollow">I'm damned serious</a>) with a published proof of Propositions 1 and/or 2?</p> <p><strong>Question 2.</strong> Any hint on how to retrieve Levi's original paper? It seems impossible to find it, and there is no copy of it in my local library.</p> </blockquote> </blockquote> <p>Thank you in advance.</p> <p>Salvo.</p> <p>(*) To avoid misunderstandings due to terminology, I say that a semigroup $\mathfrak A$ is torsion-free if an element $a$ has finite order, that is, $a^m = a^n$ for some $m,n \in \mathbb N^+$ with $m \ne n$, if and only if $a$ is idempotent.</p> <hr> <p><em>Extra contents.</em> For what it is worth, my proof of Proposition 2 does not really add any significant new idea; it is based on Levi's result and use nothing but well-known basic facts: (i) $\mathfrak A$ embeds in its unitization $\mathfrak A^{(1)}$; (ii) $\mathfrak A$ is abelian/cancellative/torsion-free iff the same holds true with $\mathfrak A^{(1)}$; (iii) the inverse image of a linearly orderable semigroup under a semigroup embedding is linearly orderable; (iv) every subsemigroup of a linearly orderable semigroup is itself linearly orderable; (v) as a consequence of (i)-(iv), we can assume wlog that $\mathfrak A$ is an abelian torsion-free cancellative monoid and construct its Grothendieck group, say $\mathfrak A_\mathcal{G}$; (vi) $\mathfrak A_\mathcal{G}$ is torsion-free (and obviously abelian); (vii) $\mathfrak A$ embeds in $\mathfrak A_\mathcal{G}$, by cancellativity; (viii) we can use (iii), (iv), (vii) and Levi's result to conclude. Nonetheless, I think that it <em>may</em> deserve a little place in the paper (e.g., as a reference for future work). But I would feel better if I could have a pointer to a previously published proof. It goes the same with Proposition 1.</p> <p><em>Motivation (if it matters; if not, ignore it all).</em> Freiman, Herzog and coauthors <a href="http://www.math.tau.ac.il/~grisha/SDOGF.pdf" rel="nofollow">have recently proved</a> some results on sum-sets/produc-sets in linearly ordered groups (which they refer to simply as ordered groups), accordingly extending some parts of Freiman's previous work on small doubling on integers; see G. Freiman, M. Herzog, P. Longobardi, and M. Maj, <em>Small doubling in ordered groups,</em> J. Austral. Math. Soc., to appear. Even more recently, <a href="http://hal.archives-ouvertes.fr/docs/00/72/62/76/PDF/Salvatore_Tringali-Small_doubling_in_ordered_semigroups.pdf" rel="nofollow">I myself</a> extended some of their results from the setting of linearly ordered groups to linearly ordered semigroups. And this is where my questions come from.</p> http://mathoverflow.net/questions/105851/strictly-totally-ordered-semigroups-looking-for-references/105875#105875 Answer by Benjamin Steinberg for Strictly totally ordered semigroups - Looking for references Benjamin Steinberg 2012-08-29T20:08:17Z 2012-08-29T20:08:17Z <p>Corollary 3.4 of Gilmer's book on commutative semigroup rings states a commutative semigroup is totally orderable (in your sense) iff torsion-free and cancellative. </p>