Linearly Dependent Module Elements / Module Homomorphism and Linearly Independency
Problem 415
(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.
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Definition.
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.
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