Let $(S,*)$ be a semigroup admitting a distinguished element $0$ such that $z*s = s*z = z$, for all $s \in S$. Moreover, let $(\mathbb{G},\cdot)$ be a commutative group. Consider an action $$ \mathbb{G} \times S \to S, ~~~~~~~~~~~ (k,s) \mapsto k.s, $$ satisfying, for all $s,t \in S$, and $k,l \in \mathbb{G}$,

  1. $~~~ k.(l.s) = (k\cdot l).s$,

  2. $~~~ k.(s*t) = (k.s)*t = s*(k.t)$,

  3. $~~~~ 1_{\mathbb{G}}.s = s$,

  4. $~~~~ 0.s = z.$

Does such an object have a name, or is is easily seen to be equivalent to a standard structure? If such things are studied, what can one say about them?

Such a commutative semigroup admits an equivalence relation $$ s \simeq t, \text{ if there exists a } k \in \mathbb{G}, \text{ such that } s = k.t. $$ Does the quotient have $S\,/\!\simeq$ have a name. For example, might one call it the projectivization of $S$?

If $S$ is a monoid does anything extra interesting happen?

EDIT: Based on the comments of M. Sapir, the definition has been refined.

  • $\begingroup$ Do you have examples of such actions in mind? $\endgroup$ Mar 9, 2018 at 14:38
  • $\begingroup$ the examples are a bit technical, so will probably not help so much. $\endgroup$ Mar 9, 2018 at 17:05
  • $\begingroup$ If S is a polynomial ring or function field over K, then these examples are frequent in various studies related to algebra. This isn't like finding a field in the endomorphism monoid, but the fact that most maps are units gives that these maps are bijective, and if you can prove that they are structure morphisms, then you have K is a subfield of an automorphism group. I would take K to be the two or three element field, and find a semigroup where one of the maps obeys your relations and is not a semigroup morphism. Gerhard "Spellcheck Thinks Semigroup Is Delicious" Paseman, 2018.03.09. $\endgroup$ Mar 9, 2018 at 17:06
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    $\begingroup$ What is $0\cdot s$? It seems to be an element $x$ of $S$ such that $k\cdot x = x$ for every $k$. Moreover for every $z\in S$, $x*z$ is also such an element. So $K$ fixes a whole right ideal of $S$ pointwise. Then I would look at the identity element $1$ in $S$ and the element $v=0\cdot 1$. We have $v*z=(0\cdot 1)*z=0\cdot z$ for every $z\in S$, and so on. The definition seems artificial because the addition operation of $K$ does not play any role. $\endgroup$
    – user6976
    Mar 9, 2018 at 19:13
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    $\begingroup$ After the March 10th edit, several things about this question don't make any sense. In the first sentence, is the zero element of the semigroup called $0$ or $z$? In axiom (3), presumably $1_{\mathbb{K}}$ should be the identity element of $\mathbb{G}$. Axiom (4) just doesn't make sense, since $\mathbb{G}$ doesn't have an element called $0$. Did you mean to assume that $g.z = z$ for all $g\in \mathbb{G}$? Later, you write "such a commutative semigroup..." Did you mean to assume $S$ is commutative to begin with? $\endgroup$ Oct 30, 2018 at 17:00

1 Answer 1


I'm afraid my general algebra brain has turned to mush. I cannot give you a good reference to anything that points to items in the literature. I cannot even suggest a good search term right now. Let me make some other observations which might help.

Fix a similarity type which has a binary operation , and a unary operation for each field element k. Then you get a variety which extends a semigroup variety by these unary maps which are not quite semigroup morphisms. So if you have a nontrivial semigroup for a field K, you get a bunch more with more than one element in the semigroup.

In addition, you get a closure under a kind of disjoint union. For semigroups S and T and Z in the variety, where Z is the one element semigroup, take the disjoint union and extend the semigroup operation to be zero (so in Z) everywhere where two elements do not both belong to S, or to T, or to Z. There may be another construction dual to this that involves adjoining a unit to a semigroup and making it a monoid, but I am not thinking of it.

I am sure something like this has been considered. Other than field action on a semigroup, I don't know what search terms to use. Embedding a field in a clone perhaps?

Gerhard "Spellcheck Still Finds Semigroups Delicious" Paseman, 2018.03.09.

  • $\begingroup$ Oh, if you have a monoid for S, then t being 1 gives that each map is a right action by some element called k.1. (And also a left action.) So you can say more, but I lack the qualifications to elaborate. Gerhard "Calling All Professional Delicious Theorists" Paseman, 2018.03.09. $\endgroup$ Mar 9, 2018 at 17:52

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