= Module
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= Module
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= Module over a ring
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A left $R$-module consists of an <abelian group> $(M,+)$ and <scalar multiplication> $R\times M\to M$, written $(r,m)\mapsto rm$, where $R$ is a <ring> with <identity element> $1$. For every $r,s\in R$ and $m,n\in M$, require
$$
r(m+n)=rm+rn,\qquad (r+s)m=rm+sm,\qquad (rs)m=r(sm),\qquad 1m=m.
$$
The <module> elements can be added to one another and multiplied by scalars; a product of two <module> elements is not part of the definition. These conventions agree with https://kconrad.math.uconn.edu/blurbs/linmultialg/moduleintro.pdf[Keith Conrad's Introductory Notes on Modules, Definition 1.2]{ref}. For a <noncommutative ring>, the side matters: a right <module> instead has $(mr)s=m(rs)$. Over a <commutative ring>, the two conventions agree.
Over a <field> $k$, these are exactly the <vector space axioms>. A nonzero $a\in k$ has a <multiplicative inverse>, so $am=0$ implies $m=a^{-1}(am)=0$. Over the <integers>, $2$ has no <multiplicative inverse>: $2x=1$ has no solution in $\mathbb Z$. In the <module> $\mathbb Z/6\mathbb Z$, division can also lose uniqueness: $2\overline1=2\overline4=\overline2$. A <ring> element which is a <unit> still acts invertibly on every <module>; the difficulty lies with scalars which are not <units>.
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