# 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.

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## 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

- $(r+s)m=rm+sm$
- $(rs)m=r(sm)$
- $r(m+n)=rm+rn$

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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