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This tag is used if a reference is needed in a paper or textbook on a specific result.

5 votes

Dihedral extensions and the Ankeny–Artin–Chowla conjecture

It appears that my former officemate Andreas Reinhart (University of Graz, Austria) has disproven the Ankeny–Artin–Chowla conjecture: more precisely, Andreas has found that $$ d := 331914313984493$$ i …
LSpice's user avatar
  • 12.9k
4 votes
1 answer
217 views

True or false? Every left or right cancellative, duo semigroup is cancellative

A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, …
8 votes
1 answer
314 views

Does every cancellative duo semigroup embed into a group?

Prompted by the comments to a recent answer by YCor to a related question (here), I'd like to ask the following: Q. Does every cancellative duo semigroup embed into a group? A (multiplicatively writ …
3 votes
Accepted

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative ...

Sorry for answering my own question, but YCor's construction in a related thread (here) gave me a lightbulb moment. Hopefully, it's not a broken lightbulb. The answer to the question asked in the OP …
Salvo Tringali's user avatar
7 votes
2 answers
459 views

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative ...

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the ta …
8 votes
2 answers
541 views

If a semigroup embeds into a group, then is it a subdirect product of groups?

The title has it all: Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups? If yes, then $S$ is a subdirect product of subdirectly irreducible groups …
3 votes
0 answers
78 views

Ordering the elements of a semigroup by $a \le b$ iff $a=b$ or $b=ab=ba$

Let $S$ be a semigroup, written multiplicatively. The binary relation $\le$ on (the underlying set of) $S$, whose graph consists of all pairs $(a,b) \in S \times S$ such that $a = b$ or $b = ab = ba$, …
9 votes
1 answer
408 views

On the origin of a fundamental theorem of additive number theory

Given $a, b \in \mathbb Z$, set $[\![a,b]\!] := \{x \in \mathbb Z: a \le x \le b\}$. A basic result in additive number theory goes as follows: If $A$ is a finite subset of $\mathbb N$ with $0 \in A$ …
12 votes
Accepted

On the origin of a fundamental theorem of additive number theory

I shared the link to this thread with several colleagues, inviting them to contribute to the discussion. Notably, I received a response from Melvyn Nathanson himself, most of which is reproduced below …
Salvo Tringali's user avatar
8 votes
2 answers
560 views

Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?

A result commonly, and probably erroneously, attributed to W. Sierpiński is that every non-atomic, countably additive, nonnegative measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on …
2 votes
0 answers
80 views

An alternative definition for finitely generated (and principal) ideals in a semigroup

Let $S$ be a semigroup. An ideal (of $S$) is a subset $I$ of $S$ such that $SI$ and $IS$ are both contained in $I$. The non-empty ideals constitute a subsemigroup, $\mathfrak I(S)$, of the power semig …
5 votes
0 answers
160 views

$S$ and $T$ globally isomorphic semigroups, with $S$ (commutative and) cancellative, iff $S$...

Denote by $\mathcal P(S)$ the semigroup obtained by equipping the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication induced by $S$: …
2 votes
0 answers
90 views

A recursive description of the smallest divisor-closed subsemigroup containing a set

Let $S$ be a semigroup and $\widehat{S}$ be its unitization, i.e., the monoid obtained from $S$ by adjoining an identity element if necessary (so that $\widehat{S} = S$ when $S$ is already a monoid). …
2 votes
0 answers
213 views

Squares whose differences are squares

EDIT. I've just noticed a thread from 2011 in the "Related" column on the right (click me), where a closely related question is being discussed (the main difference seems to be that, in Question 2 bel …
3 votes
1 answer
166 views

Every homomorphism between (rational) Puiseux monoids is multiplication by a non-negative ra...

Let a (rational) Puiseux monoid be a non-trivial submonoid of the non-negative rational numbers under (the usual operation of) addition. It is not difficult to show that, if $f \colon H \to K$ is a (m …

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