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.
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.
Submodule Consists of Elements Annihilated by Some Power of an Ideal
Let $R$ be a ring with $1$ and let $M$ be an $R$-module. Let $I$ be an ideal of $R$.
Let $M'$ be the subset of elements $a$ of $M$ that are annihilated by some power $I^k$ of the ideal $I$, where the power $k$ may depend on $a$.
Prove that $M'$ is a submodule of […]
Linearly Dependent Module Elements / Module Homomorphism and Linearly Independency
(a) Let $R$ be a commutative ring. If we regard $R$ as a left $R$-module, then prove that any two distinct elements of the module $R$ are linearly dependent.
(b) Let $f: M\to M'$ be a left $R$-module homomorphism. Let $\{x_1, \dots, x_n\}$ be a subset in $M$. Prove that if the set […]
Annihilator of a Submodule is a 2-Sided Ideal of a Ring
Let $R$ be a ring with $1$ and let $M$ be a left $R$-module.
Let $S$ be a subset of $M$. The annihilator of $S$ in $R$ is the subset of the ring $R$ defined to be
\[\Ann_R(S)=\{ r\in R\mid rx=0 \text{ for all } x\in S\}.\]
(If $rx=0, r\in R, x\in S$, then we say $r$ annihilates […]
Torsion Submodule, Integral Domain, and Zero Divisors
Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$.
The set of torsion elements is denoted
\[\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.\]
(a) Prove that if $R$ is an […]
Short Exact Sequence and Finitely Generated Modules
Let $R$ be a ring with $1$. Let
\[0\to M\xrightarrow{f} M' \xrightarrow{g} M^{\prime\prime} \to 0 \tag{*}\]
be an exact sequence of left $R$-modules.
Prove that if $M$ and $M^{\prime\prime}$ are finitely generated, then $M'$ is also finitely generated.
[…]
Ascending Chain of Submodules and Union of its Submodules
Let $R$ be a ring with $1$. Let $M$ be an $R$-module. Consider an ascending chain
\[N_1 \subset N_2 \subset \cdots\]
of submodules of $M$.
Prove that the union
\[\cup_{i=1}^{\infty} N_i\]
is a submodule of $M$.
Proof.
To simplify the notation, let us […]
Finitely Generated Torsion Module Over an Integral Domain Has a Nonzero Annihilator
(a) Let $R$ be an integral domain and let $M$ be a finitely generated torsion $R$-module.
Prove that the module $M$ has a nonzero annihilator.
In other words, show that there is a nonzero element $r\in R$ such that $rm=0$ for all $m\in M$.
Here $r$ does not depend on […]