Linearly Dependent Module Elements / Module Homomorphism and Linearly Independency

Module Theory problems and solutions

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.
 
LoadingAdd to solve later

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.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Short Exact Sequence and Finitely Generated ModulesShort 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 HomomorphismsNilpotent 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 DivisorsTorsion 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 AnnihilatorFinitely 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 […]
  • A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$.A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$. Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module. Prove that the $R$-module $M$ is irreducible if and only if $M$ is isomorphic to $R/I$, where $I$ is a maximal ideal of $R$, as an $R$-module.     Definition (Irreducible module). An […]
  • Basic Exercise Problems in Module TheoryBasic 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 IdealSubmodule 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 RingAnnihilator 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 […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Module Theory
Module Theory problems and solutions
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...

Close