Simple Modules and Maximal Right Ideals .... Bland, Proposition 6.1.7 ....

[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:https://www.physicsforums.com/attachments/6388
https://www.physicsforums.com/attachments/6389In the above text from Bland, in the proof of (1) we read the following:

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

 

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

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

We have from a previous post on which I received help ... ... that if \(\displaystyle \mathfrak{m}\) is a maximal submodule of a module \(\displaystyle M\) then \(\displaystyle M / \mathfrak{m}\) is a simple module ... ... BUT ... ... we can view a maximal right ideal as a maximal submodule of a ring \(\displaystyle R\) viewed as a right module over itself ... thus \(\displaystyle \mathfrak{m}\) is a maximal right ideal then \(\displaystyle R / \mathfrak{m}\) 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 \(\displaystyle R\) because of Bland's Corollary 1.2.4 which states that every ring \(\displaystyle R\) has at least one maximal right idea (maximal left ideal, maximal ideal).

A lingering question for me is ... why does Bland bother with \(\displaystyle S\) in the above proof ...

Hope someone can help ...

Peter

 

[Euge]

Hi Peter,

Suppose $S$ is a simple $R$-module. To avoid trivialities we assume $S$ is nonzero. Let $x$ be a nonzero element of $S$. The module $Rx = S$ by simplicity of $S$. The $R$-mapping $R \to S$ given by $r \mapsto rx$ is surjective with kernel $\operatorname{Ann}(x)$, so by the first isomorphism theorem, $R/\operatorname{Ann}(x) \approx S$.

Now it suffices to prove that $\mathfrak{m}:=\operatorname{Ann}(x)$ is a maximal ideal of $R$. Since $x \neq 0$, $1\notin \mathfrak{m}.$ Let $J$ be an ideal of $R$ such that $\mathfrak{m}\subset J \subsetneq R$. It must be shown that $J = \mathfrak{m}$. Let $r\in J$. If $r\notin \mathfrak{m}$, then $rx \neq 0$, whence $R(rx) = S$ by simplicity of $S$. Since $x\in S$, then $R(rx) = S$ implies $r'(rx) = x$ for some $r'\in R.$ Hence, $(1 - r'r)x = 0$, i.e., $1 - r'r\in \mathfrak{m}$. So $1 - r'r\in J$, and as $r\in J$, then $r'r\in J$. Closure under addition in $J$ yields $1\in J$, contradicting the assumption that $J\subsetneq R$. This proves $J = \mathfrak{m}$ and $\mathfrak{m}$ is maximal.

Conversely, suppose such a maximal ideal $\mathfrak{m}$ exists. Then $R/\mathfrak{m}$ is a field, so it is a simple $R$-module. Since $S$ is isomorphic to the simple $R$-module $R/\mathfrak{m}$, then $S$ is simple.

 

[Math Amateur]

Hi Peter,

Suppose $S$ is a simple $R$-module. To avoid trivialities we assume $S$ is nonzero. Let $x$ be a nonzero element of $S$. The module $Rx = S$ by simplicity of $S$. The $R$-mapping $R \to S$ given by $r \mapsto rx$ is surjective with kernel $\operatorname{Ann}(x)$, so by the first isomorphism theorem, $R/\operatorname{Ann}(x) \approx S$.

Now it suffices to prove that $\mathfrak{m}:=\operatorname{Ann}(x)$ is a maximal ideal of $R$. Since $x \neq 0$, $1\notin \mathfrak{m}.$ Let $J$ be an ideal of $R$ such that $\mathfrak{m}\subset J \subsetneq R$. It must be shown that $J = \mathfrak{m}$. Let $r\in J$. If $r\notin \mathfrak{m}$, then $rx \neq 0$, whence $R(rx) = S$ by simplicity of $S$. Since $x\in S$, then $R(rx) = S$ implies $r'(rx) = x$ for some $r'\in R.$ Hence, $(1 - r'r)x = 0$, i.e., $1 - r'r\in \mathfrak{m}$. So $1 - r'r\in J$, and as $r\in J$, then $r'r\in J$. Closure under addition in $J$ yields $1\in J$, contradicting the assumption that $J\subsetneq R$. This proves $J = \mathfrak{m}$ and $\mathfrak{m}$ is maximal.

Conversely, suppose such a maximal ideal $\mathfrak{m}$ exists. Then $R/\mathfrak{m}$ is a field, so it is a simple $R$-module. Since $S$ is isomorphic to the simple $R$-module $R/\mathfrak{m}$, then $S$ is simple.

Thanks for the help Euge ... much appreciated ...

But ... need a clarification ...

You write:

" ... ... Now it suffices to prove that $\mathfrak{m}:=\operatorname{Ann}(x)$ is a maximal ideal of $R$. Since $x \neq 0$, $1\notin \mathfrak{m}$ ... ... "

Can you explain why if $x \neq 0$ then we have that $1\notin \mathfrak{m}$ ...Hope you can help with this point ...

Peter

 

[Euge]

It's because $1x = x$.

 

[Math Amateur]

It's because $1x = x$.

oh! of course ...

... thanks Euge

Peter

 

[steenis]

Conversely, suppose such a maximal ideal $\mathfrak{m}$ exists. Then $R/\mathfrak{m}$ is a field, so it is a simple $R$-module.

This is not true for noncommutative rings, see Rotman -Advanced Modern Algebra, part 1, 3rd edition 2017, page 280. To prove this statement, you need the Correspondence Theorem for Rings / Modules.

 

[Euge]

This is not true for noncommutative rings, see Rotman -Advanced Modern Algebra, part 1, 3rd edition 2017, page 280. To prove this statement, you need the Correspondence Theorem for Rings.

An old post, but I was only referring to commutative rings, not noncommutative rings.

 

[steenis]

Ok, but Peter was referring to right $R$-modules over a ring $R$, not necessarily commutatutive.