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2 votes
1 answer
211 views

Terminology for a monoid $(H, \cdot)$ s.t. $ax=a$ or $xa =a$ only if $x$ is a unit

Let $(H, \cdot)$ be a (multiplicative) monoid. Is there any consolidated name for the following Property $\text{(P)}$, or for the class of monoids for which it is satisfied? $$\text{(P) If }\,xy = x\...
Salvo Tringali's user avatar
13 votes
3 answers
678 views

IBN for algebraic theories

Let us say that a finitary algebraic theory $\tau$ has IBN (invariant basis number) if the free functor $F : \mathsf{Set} \to \mathsf{Mod}(\tau)$ reflects the isomorphism relation: If $S,T$ are sets ...
Martin Brandenburg's user avatar
8 votes
2 answers
596 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,...
Salvo Tringali's user avatar
5 votes
3 answers
542 views

Congruences that aren't "finite from above"

Let $\mathfrak{A}=(A;...)$ be an algebra in the sense of universal algebra. Say that a congruence $\sim$ on $\mathfrak{A}$ is parafinite iff there is an equivalence relation $E\subseteq A^2$ with ...
Noah Schweber's user avatar
62 votes
5 answers
10k views

Does "finitely presented" mean "always finitely presented"? (Answered: Yes!)

Precisely, if an R-module M has a finite presentation, and Rk → M is some unrelated surjection (k finite), is the kernel necessarily also finitely generated? Basically I want to believe I can ...
Andrew Critch's user avatar
22 votes
3 answers
6k views

Why are ring actions much harder to find than group actions?

I admit freely that the following question is a bit of a fishing expedition inspired by this lovely "definition" of a module as found on Wikipedia: A module is a ring action on an abelian group. ...
Vidit Nanda's user avatar
  • 15.5k
11 votes
3 answers
939 views

What is the smallest variety of algebras containing all fields?

A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative inverse semigroup with respect to multiplication. The unique multiplicative ...
Thomas Klimpel's user avatar
9 votes
1 answer
712 views

Generalizing detropicalization

Given an identity in max,plus arithmetic, are there ways to turn it into an ordinary algebraic identity it other than by replacing addition by multiplication and replacing max by series-plus or by ...
James Propp's user avatar
  • 19.7k
9 votes
2 answers
661 views

Birkhoff's completeness theorem put into practice

Birkhoff's completeness theorem (see here, Theorem 14.19) states that an equation which is true in all models of an algebraic theory can be proven in equational logic. Question. Does the proof of ...
Martin Brandenburg's user avatar
7 votes
2 answers
488 views

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?

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 ...
Salvo Tringali's user avatar
5 votes
2 answers
974 views

Shape of axioms in algebraic structures

When defining algebraic structures (like monoids, groups, etc...), are there some constraints on the shape of the axioms, for the structure to have good properties that we implicitly use in many ...
Denis's user avatar
  • 1,341
4 votes
1 answer
1k views

Commutative associative rational binary operations

What are all the nondegenerate rational binary operations that are commutative and associative? (Examples: $(x,y) \mapsto x+y$, $xy+x+y$, $xy/(x+y)$.) Feel free to re-tag if you can think of ...
James Propp's user avatar
  • 19.7k
3 votes
1 answer
328 views

What does it mean for the surreal numbers/partizan games to be "universally embedding"?

In "On numbers and games", Conway writes that the surreal Numbers form a universally embedding totally ordered Field. Later Jacob Lurie proved that (the equivalence classes of) the partizan ...
FreakyByte's user avatar