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Let $M$ be a commutative monoid with zero. Then the condition $M^* = M \setminus \{0\}$ is very similar to the condition for a commutative ring to be a field. This analogy is also used in the work "Schemes over $\mathbb{F}_1$ and Zeta functions" by Connes and Consani. However they don't give these monoids a name.

A very silly idea might be to call them "monoid fields".

Question. How are these monoids called in the literature? If there is no existing terminology yet, which one would you propose?

The answer by BS tells us that in the non-commutative case these are called groups with zero. My question deals with the commutative case. I would like to have a proper name, not just a combination such as "abelian group with zero" (which is confusing anyway).

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    $\begingroup$ It would help to elaborate on the condition M*. For all I know, it looks like a submonoid with (a different) zero. Gerhard "Ask Me About System Design" Paseman, 2012.02.09 $\endgroup$
    – Gerhard Paseman
    Commented Feb 9, 2012 at 17:44
  • $\begingroup$ what about "pointed abelian group"? $\endgroup$
    – Dan Petersen
    Commented Feb 15, 2012 at 14:52
  • $\begingroup$ @Dan: This is not compatible with the usage of pointed objects (ncatlab.org/nlab/show/pointed+object). $\endgroup$
    – Martin Brandenburg
    Commented Feb 15, 2012 at 15:11
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    $\begingroup$ @Martin: point taken. $\endgroup$
    – Dan Petersen
    Commented Feb 15, 2012 at 15:18
  • $\begingroup$ @MartinBrandenburg You referred me here in this question on math stackexchange a long time ago and I just stumpled upon it again by chance. I just wanted to inform you that I have started to privately call them “stereoids” and “division stereoids” half-jokingly. $\endgroup$
    – k.stm
    Commented Dec 16, 2017 at 20:13

3 Answers 3

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The usual term in semigroup theory for a group with adjoined zero is a group with zero. See The Algebraic Theory of Semigroups Volume I by Clifford and Preston.

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  • $\begingroup$ Thanks! This seems to be an established terminology, you can also find it in other places. But as you might guess this does not really satisfy me. Since I'm interested in commutative monoids with zero, I would have to call them "commutative groups with zero" or "abelian groups with zero". But this is confusing since abelian groups are usually written additively and the identity element is written as a zero. Also I would like to use a short name (similar to "field", which is very short). $\endgroup$
    – Martin Brandenburg
    Commented Feb 10, 2012 at 15:19
  • $\begingroup$ So the upshot is that I am open for neologisms for the commutative case. $\endgroup$
    – Martin Brandenburg
    Commented Feb 10, 2012 at 15:21
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    $\begingroup$ Some people say group with absorbing element. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 10, 2012 at 19:44
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One should take seriously the option of simply calling them "abelian groups with an adjoined zero".

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  • $\begingroup$ Isn't this just a bit clumsy, or rather a characterization? Namely, the functor $A \mapsto A^*$ establishes an isomorphism of categories between "field monoids" and abelian groups. $\endgroup$
    – Martin Brandenburg
    Commented Feb 9, 2012 at 22:44
  • $\begingroup$ (Edited to include the word "abelian"). $\endgroup$
    – James Cranch
    Commented Feb 9, 2012 at 23:01
  • $\begingroup$ I agree it's clumsy, but I'm not wholly convinced you should disguise the characterisation. "Abelian groups with zero" is not all that long compared to some phrases in mathematics. $\endgroup$
    – James Cranch
    Commented Feb 9, 2012 at 23:02
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    $\begingroup$ Of course, this functor is not quite an equivalence of categories, unless you demand that homomorphisms preserve zero. Otherwise, for example, there's an extra map between any two objects which sends everything (including the zero) to the identity. $\endgroup$
    – James Cranch
    Commented Feb 9, 2012 at 23:02
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    $\begingroup$ Because people who are new to this type of mathematics won't appreciate the right way of doing it unless someone takes the time to compare it to the wrong way now and again? $\endgroup$
    – James Cranch
    Commented Feb 12, 2012 at 14:06
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One possible alternative name for monoids with zero would be $\mathbb{F}_{1}$-algebras: just like $R$-algebras are monoids in $(\mathsf{Mod}_{R},\otimes_{R},R)$, monoids with zero are monoids in $(\mathsf{Mod}_{\mathbb{F}_{1}},\otimes_{\mathbb{F}_{1}},\mathbb{F}_{1})\mathbin{``="}(\mathsf{Sets}_*,\wedge,S^0)$.

Now, every object in $(\mathsf{Sets}_*,\wedge,S^0)$ comes with a natural diagonal morphism $X\to X\wedge X$ given by the composition $X\to X\times X\twoheadrightarrow X\times X/X\vee X$ and also a projection map $X\twoheadrightarrow S^0$.

Because of these maps, every monoid with zero $M$ satisfying $M^\times=M\setminus\{0\}$ will be an example of a "Hopf $\mathbb{F}_{1}$-algebra", a Hopf algebra object in $(\mathsf{Sets}_*,\wedge,S^0)$.

Not every such Hopf $\mathbb{F}_{1}$-algebra arises as such an $M$, but this seems like a notion that is quite close to what you are looking for.

Every Hopf $\mathbb{F}_{1}$-algebra arises in this way because the categories of comonoids in $(\mathsf{Sets},\times,*)$ and $(\mathsf{Sets}_*,\wedge,S^0)$ are equivalent, see Lemma 2.4 here. So your notion is equivalent to that of a Hopf $\mathbb{F}_{1}$-algebra, i.e. a Hopf algebra object in $(\mathsf{Sets}_*,\wedge,S^0)$.

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