(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 $\{f(x_1), \dots, f(x_n)\}$ is linearly independent, then the set $\{x_1, \dots, x_n\}$ is also linearly independent.

Let $R$ be a ring and $M$ be a left $R$-module.
A finite subset $\{v_1, \dots, v_n\}$ of $M$ is said to be linearly independent if whenever
\[r_1v_1+\cdots r_n v_n=0\]
holds, then we have
\[r_1=\dots=r_n=0.\]

Proof (a) any two distinct elements of the module $R$ are linearly dependent.

Let $x, y$ be distinct elements of the module $R$.

As $R$ is a commutative ring, we have
\begin{align*}
(-y)\cdot x+x\cdot y=0
\end{align*}
This equality should be read as a linear combination of elements $x$ and $y$ in the module $R$ with coefficients $-y$ and $x$ in $R$.
Since $x$ and $y$ are distinct, at least one of them is non-zero.
Hence this equality yields that $x, y$ are linearly dependent.

Proof (b) the set $\{x_1, \dots, x_n\}$ is linearly independent.

Suppose that we have a linear combination
\[r_1x_1+\cdots r_n x_n=0\]
for $r_1, \dots , r_n \in R$.
Then we have
\begin{align*}
&0=f(0)\\
&=f(r_1x_1+\cdots r_n x_n)\\
&=r_1f(x_1)+\cdots r_n f(x_n).
\end{align*}
since $f$ is a $R$-module homomorphism.

It follows from the linearly independence of elements $f(x_1), \dots, f(x_n)$ that we obtain
\[r_1=\cdots r_n=0.\]
Hence the elements $x_1, \dots, x_n$ are linearly independent.

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.
[…]

Nilpotent Ideal and Surjective Module Homomorphisms
Let $R$ be a commutative ring and let $I$ be a nilpotent ideal of $R$.
Let $M$ and $N$ be $R$-modules and let $\phi:M\to N$ be an $R$-module homomorphism.
Prove that if the induced homomorphism $\bar{\phi}: M/IM \to N/IN$ is surjective, then $\phi$ is surjective.
[…]

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 […]

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 […]

Basic Exercise Problems in Module Theory
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 […]

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 […]

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 […]