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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?

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If it is a commutative monoid, it is also a module: With $a\in M$ we have an additive inverse $(-1)\cdot a\in M$. – Hagen von Eitzen Dec 30 '12 at 22:39
up vote 3 down vote accepted

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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Thank you. I remember that $(-1)(m+n) = (-1)n+(-1)m$, but I forget the proof. Could you remind me? – Mike Izbicki Dec 31 '12 at 1:12
@Mike: $m + n + (-1)n + (-1)m = m + 0 + (-1)m = m + (-1)m = 0$. – Qiaochu Yuan Dec 31 '12 at 3:04

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