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 $R$-module $M$ is called irreducible if $M$ is not the zero module and $0$ and $M$ are the only submodules of $M$.

An irreducible module
is also called a simple module.

Proof.

$(\implies)$ Suppose that $M$ is an irreducible $R$-module.
Then by definition of an irreducible module, $M$ is not the zero module.
Take any nonzero element $m\in M$, and consider the cyclic submodule $(m)=Rm$ generated by $m$.
Since $M$ is irreducible, we must have $M=Rm$.

Now we define a map $f:R\to M$ by sending $r\in R$ to $f(r)=rm$.
Then the map $f$ is an $R$-module homomorphism regarding $R$ is an $R$-module.

In fact, we have
\begin{align*}
&f(r+s)=(r+s)m=rm+sm=f(r)+f(s)\\
&f(rs)=(rs)m=r(sm)=rf(s)
\end{align*}
for any $r, s\in R$.
Since $M=Rm$, the homomorphism $f$ is surjective.
Thus, by the first isomorphism theorem, we obtain
\[R/I\cong M,\] where $I=\ker(f)$.

It remains to show that $I$ is a maximal ideal of $R$.
Suppose that $J$ is an ideal such that $I\subset J \subset R$.
Then by the third isomorphism theorem for rings, we know that $J/I$ is an ideal of the ring $R/I\cong M$, hence $J/I$ is a submodule.

Since $M$ is irreducible, we must have either $J/I=0$ or $J/I=M$.
This implies that $J=I$ or $J=M$.
Hence $I$ is a maximal ideal.

$(\impliedby)$ Suppose now that $M\cong R/I$ for some maximal ideal $I$ of $R$.
Let $N$ be any submodule of $R/I$. (We identified $M$ and $R/I$ by the above isomorphism.)
Then $N$ is an ideal of $R/I$ since $N$ is an abelian group and closed under the action of $R$, hence that of $R/I$.

Since $R$ is a commutative ring and $I$ is a maximal ideal of $R$, we know that $R/I$ is a field.
Thus, only the ideals of $R/I$ are $0$ or $R/I$.

Hence we have $N=0$ or $N=R/I=M$.
This proves that $M$ is irreducible.

Related Question.

A similar technique in the proof above can be used to solve the following problem.

See the post “A module is irreducible if and only if it is a cyclic module with any nonzero element as generator” for a proof.

The post A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$. first appeared on Problems in Mathematics.

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.

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.

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 $I$.“.

The post A Module is Irreducible if and only if It is a Cyclic Module With Any Nonzero Element as Generator first appeared on Problems in Mathematics.