# What is the name of the algebraic structure constructed with an abelian monoid and a field?

Take a vector space built from an abelian group $(V,+)$ whose elements are the vectors, a field $K$ whose elements are the scalars, and there is an operation (multiplication by scalars) that distributes on the group operation.

What is the name of the structure obtained if $(V,+)$ instead of being an abelian group is an abelian monoid?

Thanks and forgive my ignorance :)

Adam showed quite nicely that this isn't interesting when $K$ is a field or, as we can see looking at his argument, any ring with an identity $1$ which has an additive inverse.

We can try to find a nontrivial example of your concept by talking about semirings and semimodules. As semiring is like a regular ring, except we only require the underlying group to be a monoid (no inverses necessarily.) Usually the axiom that $0*r=0$ is added back in (since it's no longer derivable from the existence of additive inverses.)

A semimodule is an additive abelian monoid with a semiring acting on it in the same way a ring acts on a regular module. Again, we may have to add back in that $0*m=0$ to keep things nice.

For a nice example of a semiring, you can just consider the semiring of natural numbers $\Bbb N=\{0,1,2,\ldots\}$. It is a semimodule over itself.

In the same line of thought, you can consider the set of nonnegative rational numbers as a semifield, where there aren't any additive inverses (except for $0$) and yet everything has a multiplicative inverse, and it is closed under addition and multiplication, etc. They probably call semimodules over a semifield "semivector spaces".

There cannot be such a structure without the group information. If you use the same axioms replacing the group info with monoid info, you end up being able to show it's a vector space.

Proof: Let $v\in V$. Then as $1_F\cdot v=v$ we have that $v+(-1_F)v=(1_F-1_F)\cdot v=0\cdot v=0$. Because $V$ is closed under the field operation that $(-1_F)\cdot v$ is an additive inverse for $v$, hence $V$ is inherently an abelian group.

• Thanks a lot, I did not stop to think in this top-to-bottom perspective. Your observation is highly appreciated. I was in the search of name, a keyword, to open the doors for more reading, so I will mark rschwieb answer as "the answer" but yours is also extremely educative! – JuanPi Dec 5 '14 at 21:08