A module $M$ is an Abelian group with the extra property that any element $m \in M$ can be multiplied by any element $r$ in a ring $R$. I want to relax this definition so that the $M$ need not be an Abelian group, but instead needs to only be a monoid. Does this structure have a standard name?
What do you call this generalization of a module?
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6If it is a commutative monoid, it is also a module: With $a\in M$ we have an additive inverse $(-1)\cdot a\in M$. – 2012-12-30
1 Answers
It's called a module. That is, $M$ must already be an abelian group. As Hagen observes in the comments, if $m \in M$ is an element then $(-1)m \in M$ is its additive inverse, so $M$ is a group. Now
$(-1)(m + n) = (-1)m + (-1)n$
by distributivity, but
$(-1)(m + n) = (-1)n + (-1)m$
in any group, from which it follows that $M$ is abelian. (In other words, a group is abelian if and only if taking inverses is a homomorphism.)
If $R$ is also relaxed so that it is a semiring, then $M$ can be a commutative monoid and then is referred to as a semimodule. I do not immediately see how to prove that $M$ must be commutative, but asking $M$ to be noncommutative is unnatural from the abstract point of view on what a semiring is, namely that a semiring is a thing that acts on commutative monoids. $M$ also has to be pretty close to commutative anyway, since similar to the above we have
$2(m + n) = 2m + 2n = (m + n) + (m + n)$
hence if $M$ is cancellative then it is commutative, and even if $M$ is not cancellative it has to be fairly close to be commutative even to be, say, a $\mathbb{Z}_{\ge 0}$-semimodule.
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1@Mike: $m + n + (-1)n + (-1)m = m + 0 + (-1)m = m + (-1)m = 0$. – 2012-12-31