# I Simple Modules and Maximal Right Ideals ...

1. Feb 4, 2017

### Math Amateur

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with the proof of Proposition 6.1.7 ...

Proposition 6.1.7 and its proof read as follows:

In the above text from Bland, in the proof of (1) we read the following:

" ... ... Since $S$ is a simple $R$-module if and only if there is a maximal ideal $\mathfrak{m}$ of $R$ such that $R / \mathfrak{m} \cong S$ ... ... "

I do not follow exactly why the above statement is true ...

Can someone help me to see why and how, exactly, the above statement is true ...

Hope someone can help ...

Peter

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2. Feb 4, 2017

### Math Amateur

Just trying to clarify a few things regarding my question ...

We have from a previous post on which I received help ... ... that if $\mathfrak{m}$ is a maximal submodule of a module $M$ then $M / \mathfrak{m}$ is a simple module ... ... BUT ... ... we can view a maximal right ideal as a maximal submodule of a ring $R$ viewed as a right module over itself ... thus $\mathfrak{m}$ is a maximal right ideal then $R / \mathfrak{m}$ is a simple module ... is that correct?

Not sure how to piece together the rest of the proof of the statement above ... but we know that a maximal right ideal exists in $R$ because of Bland's Corollary 1.2.4 which states that every ring $R$ has at least one maximal right idea (maximal left ideal, maximal ideal).

A lingering question for me is ... why does Bland bother with $S$ in the above proof ...

Hope someone can help ...

Peter

3. Feb 5, 2017

### andrewkirk

Yes that sounds right. It covers only one of the two directions of the sentence in the OP though. I would elaborate the proof slightly as follows:

We use the theorem that a submodule $m$ of $M$ is maximal iff $M/m$ is simple, together with the fact that ideals of a ring $R$ can be treated as submodules of ${}_RR$, which is $R$ as a module over itself.

Forward Direction

Say there is a maximal ideal $m$ of $R$, then ${}_Rm$ must be a maximal submodule of ${}_RR$. Because if there is some proper submodule ${}_RQ$ of ${}_RR$, and ${}_Rm$ is a proper submodule of that, then $Q$ is a proper ideal of $R$ that properly contains $m$, so that $m$ cannot be a maximal ideal, which is a contradiction. Hence ${}_Rm$ is a maximal submodule of ${}_RR$, from which it follows from the above theorem that ${}_RR/{}_Rm$ is simple. If we further assume that ${}_RR/{}_Rm \cong S$ then $S$ must be simple since ${}_RR/{}_Rm$ is.

For the Reverse Direction we assume that $S$ is a simple $R$-module, and try to prove that there must be a maximal ideal $m$ of $R$ such that ${}_RR/{}_Rm\cong S$.

That looks harder, because we need to get ideals from modules, which is less obvious a process than getting modules from ideals. I will need to reflect on it.

4. Feb 6, 2017

### Math Amateur

Thanks for your help, Andrew ... appreciate it ...

Still reflecting on what you have written ...

Thanks again,

Peter

5. Feb 7, 2017

### Math Amateur

Andrew,

Can you help with how and why Bland can justifiably conclude that $\text{ann}_r( R / \mathfrak{m}) = \text{ann}_r(S)$ ... ... ?

Peter

6. Feb 7, 2017

### Staff: Mentor

This follows directly from $R/\mathfrak{m} \cong S$. Simply write the annihilator in front of it.
So the question remains, why $R/\mathfrak{m} \cong S$.

$\Longrightarrow :$ (see @andrewkirk 's post #3 above, or Ex. 1.3 in Bland)

If $\mathfrak{m} \subsetneq R$ is a maximal ideal, then there is simply no room left in $\{0\} = \mathfrak{m}/\mathfrak{m} \subsetneq R/\mathfrak{m}$ for an ideal of $R$ that contains $\mathfrak{m}$, so as an $R-$module $R/\mathfrak{m}$ has to be simple for it contains $\mathfrak{m}$ as zero element.

$\Longleftarrow :$

If $S$ is a simple $R-$module, then we chose a fixed element $t \in S - \{0\}$ and consider the mapping $\varphi : R \rightarrow S$ with $\varphi(r) := r\cdot t\;$. You can show, that this is an $R-$modul homomorphism. It is also surjective, because $S$ is simple. ($\{0\} \neq R\cdot t \subseteq S$ is a submodule. We have to either request $1 \in R$ here or that $R$ doesn't act trivially on $S$.)
Now by simple calculations $\ker \varphi = \textrm{ ann}_R (t)$ is an ideal of $R$ and $R/\ker \varphi \cong S$ because of the exact sequence $\{0\} \rightarrow \ker \varphi \rightarrow R \rightarrow R/ \ker \varphi \rightarrow \{0\}$

At last, $\ker \varphi$ has to be maximal, for otherwise $S$ wouldn't be simple (due to the isomorphism).

7. Feb 7, 2017

### andrewkirk

@fresh_42 Very nice indeed.
I would just add that, because exact sequences - at least in my text - are not covered in all texts on modules (I first came across them in algebraic topology), it may be more intuitive to present the last step via the first isomorphism theorem for modules (part 3 of the result in that linked wiki paragraph), which is central to any study of modules. You have shown that $\varphi$ is surjective, so that Im $\varphi=S$. Hence, by part 3 of the first module isomorphism theorem: ${}_RR/$ ker $\varphi\cong$ Im $\varphi=S$.

Also, just slightly elaborating (for my own benefit if for nobody else's) on the last step, if there is a proper submodule ${}_RN$ of ${}_RR$ that properly contains ker $\varphi$, then $\varphi ({}_RN)$ is a submodule of $S$ Then, by considering $\varphi(r)$ for $r\in {}_RN-$ ker $\varphi$, and noting that we cannot have $\varphi(r)=0$ because $r\notin$ ker $\varphi$, we see that $\varphi({}_RN)$ is nontrivial, contradicting the assumed simplicity of $S$.