A Module is Irreducible if and only if It is a Cyclic Module With Any Nonzero Element as Generator

Module Theory problems and solutions

Problem 434

Let $R$ be a ring with $1$.
A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$.
(It is also called a simple module.)

(a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.

(b) Determine all the irreducible $\Z$-modules.

 
FavoriteLoadingAdd to solve later

Sponsored Links

Proof.

(a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.

$(\implies)$ Suppose that $M$ is an irreducible module.
Let $a\in M$ be any nonzero element and consider the submodule $(a)$ generated by the element $a$.

Since $a$ is a nonzero element, the submodule $(a)$ is non-zero. Since $M$ is irreducible, this yields that
\[M=(a).\] Hence $M$ is a cyclic module generated by $a$. Since $a$ is any nonzero element, we conclude that the module $M$ is a cyclic module with any nonzero element as its generator.


$(\impliedby)$ Suppose that $M$ is a cyclic module with any nonzero element as its generator.
Let $N$ be a nonzero submodule of $M$. Since $N$ is non-zero, we can pick a nonzero element $a\in N$. By assumption, the non-zero element $a$ generates the module $M$.

Thus we have
\[(a) \subset N \subset M=(a).\] It follows that $N=M$, and hence $M$ is irreducible.

(b) Determine all the irreducible $\Z$-modules.

By the result of part (a), any irreducible $\Z$-module is generated by any nonzero element.
We first claim that $M$ cannot contain an element of infinite order. Suppose on the contrary $a\in M$ has infinite order.

Then since $M$ is irreducible, we have
\[M=(a)\cong \Z.\] Since $\Z$-module $\Z$ has, for example, a proper submodule $2\Z$, it is not irreducible. Thus, the module $M$ is not irreducible, a contradiction.

It follows that any irreducible $\Z$-module is a finite cyclic group.
(Recall that any $\Z$-module is an abelian group.)
We claim that its order must be a prime number.

Suppose that $M=\Zmod{n}$, where $n=ml$ with $m,l > 1$.
Then
\[(\,\bar{l}\,)=\{l+n\Z, 2l+n\Z, \dots, (m-1)l+n\Z\}\] is a proper submodule of $M$, and it is a contradiction.
Thus, $n$ must be prime.

We conclude that any irreducible $\Z$-module is a cyclic group of prime order.

Related Question.

Here is another problem about irreducible modules.

Problem. 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.

For a proof of this problem, see the post “A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$.“.


FavoriteLoadingAdd to solve later

Sponsored Links

More from my site

  • 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 […]
  • Fundamental Theorem of Finitely Generated Abelian Groups and its applicationFundamental Theorem of Finitely Generated Abelian Groups and its application In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem. Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]
  • Can $\Z$-Module Structure of Abelian Group Extend to $\Q$-Module Structure?Can $\Z$-Module Structure of Abelian Group Extend to $\Q$-Module Structure? If $M$ is a finite abelian group, then $M$ is naturally a $\Z$-module. Can this action be extended to make $M$ into a $\Q$-module?   Proof. In general, we cannot extend a $\Z$-module into a $\Q$-module. We give a counterexample. Let $M=\Zmod{2}$ be the order […]
  • 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 […]
  • 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 […]
  • Ascending Chain of Submodules and Union of its SubmodulesAscending 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 […]
  • 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.   […]

You may also like...

1 Response

  1. 06/10/2017

    […] the post “A module is irreducible if and only if it is a cyclic module with any nonzero element as generator” for a […]

Leave a Reply

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

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

Close