MHB Proving $M$ is Cyclic: Simple $R$-Module & Isomorphism with Maximal Ideal $J$

[mathmari]

Hey!

Let $R$ be a commutative ring with unit and $M$ a $R$-module.

If $M$ is a simple $R$-module, i.e., the only $R$-submodule are $O$ and $M$, then $M$ is cyclic and isomorphic to $R/J$ where $J$ is a maximal ideal of $R$. Could you give me some hints how we could show that $M$ is cyclic? (Wondering)

 

[Euge]

Hi mathmari,

Let $x$ be a nonzero element of $M$ and consider the $R$-submodule $Rx$. Use simplicity of $M$ to prove $M = Rx$. This also proves $M$ is cyclic.

 

[mathmari]

Let $x$ be a nonzero element of $M$ and consider the $R$-submodule $Rx$. Use simplicity of $M$ to prove $M = Rx$. This also proves $M$ is cyclic.

Since $M$ is simple, we have that $Rx=O$ or $Rx=M$.
Since $x$ is nonzero and since we have that $x=1\cdot x\in Rx$, it cannot be that $Rx=O$, right? (Wondering)
Therefore, $Rx=M$.
That implies that $M$ is cyclic. We have that $R/J$ is a ring and since $J$ is a maximal ideal of $R$, we have that $R/J$ is a field.
To show that $M$ is isomorphic to $R/J$, do we have to consider the mapping $R\rightarrow M$ and show that it is an homomorphism and bijective and that the kernel is $J$ ? (Wondering)

 

[Euge]

Since $M$ is simple, we have that $Rx=O$ or $Rx=M$.
Since $x$ is nonzero and since we have that $x=1\cdot x\in Rx$, it cannot be that $Rx=O$, right? (Wondering)
Therefore, $Rx=M$.
That implies that $M$ is cyclic.

Yes, that's correct.

We have that $R/J$ is a ring and since $J$ is a maximal ideal of $R$, we have that $R/J$ is a field.
To show that $M$ is isomorphic to $R/J$, do we have to consider the mapping $R\rightarrow M$ and show that it is an homomorphism and bijective and that the kernel is $J$ ? (Wondering)

Not exactly. The $J$ is not yet determined. Show that the mapping $R \to M$ is a surjective $R$-homomorphism. By the first isomorphism, you can then claim that $M$ is isomorphic to $R/J$ where $J$ is the kernel of that mapping. So $R/J$ is simple. Finally, show that that if $K$ is an ideal of $R$ such that $J \subset K \subsetneq R$, then $J = K$. That would prove that $J$ is a maximal ideal.

 

[mathmari]

Show that the mapping $R \to M$ is a surjective $R$-homomorphism.

We consider the mapping $\phi: R\rightarrow M$ with $r\mapsto rm$.
We have that $\text{Im}\phi$ is a $R$-submodule of $M$. Since $M$ is simple it follows that $\text{Im}\phi=O$ or $\text{Im}\phi=M$.
To show that $\phi$ is surjective, we have to conclude that $\text{Im}\phi=M$, right? How could we conclude that? (Wondering)

 

[Euge]

To show that $\phi$ is surjective, we have to conclude that $\text{Im}\phi=M$, right?

Yes, that's right.

How could we conclude that? (Wondering)

Using the fact that $Rx = M$.

 

[mathmari]

Using the fact that $Rx = M$.

For $x=m$ we have that $M=Rm$, since $M$ is cyclic.
We have that $rm\in Rm$ and from the mapping we have that $rm\in \text{Im}\phi$, so $M=Rm\subseteq \text{Im}\phi$.
From that it follows that $\text{Im}\phi$ cannot be $O$. Therefore, it must be $\text{Im}\phi=M$.
Is this correct? (Wondering)

 

[Euge]

For $x=m$ we have that $M=Rm$, since $M$ is cyclic.
We have that $rm\in Rm$ and from the mapping we have that $rm\in \text{Im}\phi$, so $M=Rm\subseteq \text{Im}\phi$.
From that it follows that $\text{Im}\phi$ cannot be $O$. Therefore, it must be $\text{Im}\phi=M$.
Is this correct? (Wondering)

No, $x$ was already chosen. You can't substitute $m$ for $x$.

 

[mathmari]

No, $x$ was already chosen. You can't substitute $m$ for $x$.

So, at the mapping we choose $m=x$, so we have the mapping $r\mapsto rx$, or not? (Wondering)

 

[Euge]

You're still saying $x = m$, which isn't right. The element $x$ was introduced in post #2. Since $Rx = M$, for every $m\in M$, there corresponds an $r\in R$ such that $rx = m$, that is, $\varphi(r) = m$. Hence, $\varphi$ is onto.

 

[mathmari]

Show that the mapping $R \to M$ is a surjective $R$-homomorphism. By the first isomorphism, you can then claim that $M$ is isomorphic to $R/J$ where $J$ is the kernel of that mapping. So $R/J$ is simple.

We have shown that $\phi$ is onto.

We have that $\phi$ is an homomorphism since:
$$\phi (r_1+r_2)=(r_1+r_2)m=r_1m+r_2m=\phi (r_1)+\phi (r_2) \\ \phi (ar)=arm=a\phi (r)$$ Now it is left to show that the kernel of that mapping is the maximal ideal $J$, or not? (Wondering)
How do we show that? (Wondering)

Finally, show that that if $K$ is an ideal of $R$ such that $J \subset K \subsetneq R$, then $J = K$. That would prove that $J$ is a maximal ideal.

Why do we have to prove that? Do we not know from the exercise statement that $J$ is a maximal ideal? (Wondering)

 

[Euge]

Now it is left to show that the kernel of that mapping is the maximal ideal $J$, or not? (Wondering)
How do we show that? (Wondering)

No, since by definition (look at Post #4) $J$ is the kernel.

Why do we have to prove that? Do we not know from the exercise statement that $J$ is a maximal ideal? (Wondering)

You are to prove the exercise statement, not assume it.

 

[mathmari]

Finally, show that that if $K$ is an ideal of $R$ such that $J \subset K \subsetneq R$, then $J = K$. That would prove that $J$ is a maximal ideal.

How could we show that? Using the property that $M$ is simple? (Wondering)

 

[Euge]

How could we show that? Using the property that $M$ is simple? (Wondering)

Yes, that's right.

 

[mathmari]

Yes, that's right.

How exactly can we do that? I got stuck right now... Could you give me a hint? (Wondering)

 

[Euge]

Let $K$ be an ideal of $R$ such that $J\subset K \subsetneq R$. Then $K/J$ is a submodule of $R/J$. Since $R/J$ is simple (since $M$ is simple), what can you say about $K/J$?

 

[mathmari]

Let $K$ be an ideal of $R$ such that $J\subset K \subsetneq R$. Then $K/J$ is a submodule of $R/J$. Since $R/J$ is simple (since $M$ is simple), what can you say about $K/J$?

We have that $K/J=O$ or $K/J=R/J$, but since $K\subsetneq R$ we have that $K/J=O$. Is this correct? (Wondering),

 

[Euge]

Yes, that's absolutely right.

 

[mathmari]

Yes, that's absolutely right.

Does $K/J=O$ mean that $K=J$ ? (Wondering)

 

[Euge]

Correct!

 

[mathmari]

Ok... Thank you very much for your help! (Smile)