# Basic Exercise Problems in Module Theory

## Problem 408

Let $R$ be a ring with $1$ and $M$ be a left $R$-module.

(a) Prove that $0_Rm=0_M$ for all $m \in M$.

Here $0_R$ is the zero element in the ring $R$ and $0_M$ is the zero element in the module $M$, that is, the identity element of the additive group $M$.
To simplify the notations, we ignore the subscripts and simply write
$0m=0.$ You must be able to and must judge which zero elements are used from the context.

(b) Prove that $r0=0$ for all $s\in R$. Here both zeros are $0_M$.

(c) Prove that $(-1)m=-m$ for all $m \in M$.

(d) Assume that $rm=0$ for some $r\in R$ and some nonzero element $m\in M$. Prove that $r$ does not have a left inverse.

## Definition of a module.

Let $R$ be a ring with $1$.
A set $M$ is a left $R$-module if it has a binary operation $+$ on $M$ under which $M$ is an abelian group, and it has a map $R\times M\to M$ (called an action of $R$ on $M$) denoted by $rm$, for all $r\in R$ and $m\in M$ satisfying the following axioms

1. $(r+s)m=rm+sm$
2. $(rs)m=r(sm)$
3. $r(m+n)=rm+rn$

for all $r,s \in R$ and $m,n\in M$.

If the ring $R$ has a $1$, we also impose the axiom:
4. $1m=m$

for all $m\in M$.

## Proof.

### (a) Prove that $0_Rm=0_M$ for all $m \in M$.

We have
\begin{align*}
0m&=(0+0)m\\
&=0m+0m && \text{by axiom 1}.
\end{align*}
Subtracting $0m$, we obtain $0m=0$ as required.

### (b) Prove that $r0=0$ for all $s\in R$.

We have
\begin{align*}
r0=&r(0+0)\\
&=r0+r0 && \text{by axiom 3}.
\end{align*}
Subtracting $r0$ from both sides, we obtain
$r0=0.$

### (c) Prove that $(-1)m=-m$ for all $m \in M$.

We compute
\begin{align*}
m+(-1)m&=1m+(-1)m && \text{by axiom 4}\\
&=(1+(-1))m && \text{by axiom 1}\\
&=0m\\
&=0 && \text{by part (a)}.
\end{align*}
This shows that $(-1)m$ is the additive inverse of $m$, which must be $-m$ by the uniqueness of the additive inverse of an abelian group.
Hence we obtain $(-1)m=-m$.

### Prove that $r$ does not have a left inverse.

Seeking a contradiction, assume that $r$ has a left inverse $s$. That is, we have $sr=1$.
Multiplying $rm=0$ by $s$ on the left, we have
\begin{align*}
s(rm)=s0=0
\end{align*}
by part (b).
The left hand side is
\begin{align*}
s(rm)&=(sr)m && \text{by axiom 2}\\
&=1m\\
&=m && \text{by axiom 4}.
\end{align*}
It follows that $m=0$ but by assumption $m$ is a nonzero element of $M$.
Thus this is a contradiction, and we conclude that $r$ does not have a left inverse.

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