?

Abstract Algebra

Free Version

Upgrade subject to access all content

Easy

Module Over a Ring: Definition

ABSALG-GC1VF5

Let $R$ be a ring (with identity) $1$.

A (unitary) left $R$-module $M$ is an abelian group $(M,+)$ together with an action of $R$ on $M$:

$R\times M\rightarrow M$, $\quad (r,m)\mapsto rm$

...such that:

$r(m+n)=rm+rn$, $\quad(r+s)m=rm+sm$, $\quad r(sm)=(rs)m$, $\quad 1m=m$

...for all $r,s\in R$ and $m,n\in M$. The definition of a right module is similar, with the action of $R$ on the right.

Very often the words "with identity" for $R$ and "unitary" for $M$ are omitted and tacitly assumed.

Which of the following $M$ is a unitary left $R$-modules?

Select ALL that apply.

A

$M$ is an abelian group with the $R={\mathbb Z}$-module structure induced by addition in $M$

B

$R$ is the ring of $n\times n$ matrices whose last row consists only of zeros and $M$ is the group of $n\times n$ matrices under addition. The action of $R$ on the left is given by matrix multiplication.

C

$R$ is a field and $M$ is an $R$-vector space

D

$R={\mathbb Z}$ and $M=\left\{\frac{p}{q}\mid p,q\in {\mathbb Z}, q=5^a, a\ge0\right\}$ with the action $\left(r,\frac{p}{q}\right)\mapsto \frac{rp}{q}, \;r\in {\mathbb Z}, \;\frac{p}{q}\in M$

E

$R={\mathbb Z}$ and $M=\left\{\frac{p}{q}\mid p,q\in {\mathbb Z}, q=5^a, a\ge0\right\}$ with the action $\left(r,\frac{p}{q}\right)\mapsto \frac{2rp}{q}, \;r\in {\mathbb Z}, \;\frac{p}{q}\in M$