Jacobson Radical and Rad(M) - Bland Corollary 6.1.3 ....

[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 Corollary 6.1.3 ... Corollary 6.1.3 (including the preceding Proposition) reads as follows:

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My questions are as follows:
Question 1

In the proof of Corollary 6.1.3 above we read:

"... ... Since \(\displaystyle R\) is generated by \(\displaystyle 1, J(R) \neq R\). ... ...
My question is as follows: why, given that \(\displaystyle R\) is generated by \(\displaystyle 1\), is it true that \(\displaystyle J(R) \neq R\) ... ... ?Question 2

Bland seems to argue that if we accept that \(\displaystyle J(R) \neq R\), then the Corollary is proved ... ... that is that

\(\displaystyle J(R) \neq R \Longrightarrow \text{ Rad}(M) \neq M\) ... ...

But ... why would this be true ...?Hope someone can help ... ... Peter
===========================================================================In order to give forum readers the notations, definitions and context of the above post, I am providing the first two pages of Chapter 6 of Bland ... ... as follows ... ... :
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View attachment 6310

 

[Euge]

Happy New Year, Peter!

Question 1

In the proof of Corollary 6.1.3 above we read:

"... ... Since \(\displaystyle R\) is generated by \(\displaystyle 1, J(R) \neq R\). ... ...
My question is as follows: why, given that \(\displaystyle R\) is generated by \(\displaystyle 1\), is it true that \(\displaystyle J(R) \neq R\) ... ... ?

Since maximal right ideals of $R$ don't contain $1$ by definition, then $J(R)$, the intersection of all maximal right deals, cannot contain $1$. That's why $J(R) \neq R$.

Question 2

Bland seems to argue that if we accept that \(\displaystyle J(R) \neq R\), then the Corollary is proved ... ... that is that

\(\displaystyle J(R) \neq R \Longrightarrow \text{ Rad}(M) \neq M\) ... ...

But ... why would this be true ...?

The paragraph proceeding Corollary 6.1.3 is not a proof of the Corollary.

 

[Math Amateur]

Happy New Year, Peter!Since maximal right ideals of $R$ don't contain $1$ by definition, then $J(R)$, the intersection of all maximal right deals, cannot contain $1$. That's why $J(R) \neq R$.The paragraph proceeding Corollary 6.1.3 is not a proof of the Corollary.

Thanks for clearing up my confusion, Euge ... yes, obvious now ...Presumably the proof of Corollary 6.1.3 is simply that M, being finitely generated, has at least one maximal submodule, which is a proper submodule (as would any other submodules that exist in \(\displaystyle M\)) ... and so ... the intersection of the maximal submodules of \(\displaystyle M\) would again be a proper submodule ... ... so \(\displaystyle \text{ Rad}(M) \neq M\) ... ...

Is that correct?

 

[Math Amateur]

Thanks for clearing up my confusion, Euge ... yes, obvious now ...Presumably the proof of Corollary 6.1.3 is simply that M, being finitely generated, has at least one maximal submodule, which is a proper submodule (as would any other submodules that exist in \(\displaystyle M\)) ... and so ... the intersection of the maximal submodules of \(\displaystyle M\) would again be a proper submodule ... ... so \(\displaystyle \text{ Rad}(M) \neq M\) ... ...

Is that correct?

Peter

Hopefully someone can clarify the following:
Regarding the conditions \(\displaystyle J(R) = R\) and \(\displaystyle J(R) = 0\) ... ... ... ... I am still a little confused ...Now, regarding \(\displaystyle \text{ Rad}(M)\) ... Bland is clear when it comes to \(\displaystyle \text{ Rad}(M) = M\) ... indeed Bland's definition of \(\displaystyle \text{ Rad}(M)\) is as follows:"... ... If \(\displaystyle M\) is an \(\displaystyle R\)-module, then the radical of \(\displaystyle M\), denoted by \(\displaystyle \text{ Rad}(M)\), is the intersection of the maximal submodules of \(\displaystyle M\). If \(\displaystyle M\) fails to have maximal submodules, then we set \(\displaystyle \text{ Rad}(M) = M\). ... ...
Presumably \(\displaystyle \text{ Rad}(M) = \{ 0 \}\) when there exist maximal submodules in \(\displaystyle M\), but their intersection is \(\displaystyle \{ 0 \}\)
Now with \(\displaystyle J(R)\) Bland does not explicitly give a condition for which \(\displaystyle J(R) = R\) ... indeed his definition of the Jacobson radical is as follows:"... ... The Jacobson radical of \(\displaystyle R\), denoted \(\displaystyle J(R)\), is the intersection of maximal right ideals of \(\displaystyle R\). If \(\displaystyle J(R) = 0\), then R is said to be a Jacobson semisimple ring ... ..."Is it correct to assume that if there are no maximal right ideals in \(\displaystyle R\), then \(\displaystyle J(R) = R\) ... ...

... ... and \(\displaystyle J(R) = 0\) if there do exist maximum right ideals but their intersection is \(\displaystyle \{ 0 \}\)
Can someone please confirm that my thinking is correct and/or point out any shortcomings or errors ...

Hope someone can help ...

Peter

 

[mathbalarka]

The point is unital rings always have maximal ideals, so you can always take intersection amongst them and $J(R)$ would be well-defined. In fact $J(R)$ is always a proper ideal (because maximal ideals are proper and intersection of a bunch of proper things is always going to give you a proper thing).

 

[Math Amateur]

The point is unital rings always have maximal ideals, so you can always take intersection amongst them and $J(R)$ would be well-defined. In fact $J(R)$ is always a proper ideal (because maximal ideals are proper and intersection of a bunch of proper things is always going to give you a proper thing).

Thanks for the help, mathbalarka ...

Appreciate your guidance and help

Peter

 

[Euge]

Thanks for clearing up my confusion, Euge ... yes, obvious now ...Presumably the proof of Corollary 6.1.3 is simply that M, being finitely generated, has at least one maximal submodule, which is a proper submodule (as would any other submodules that exist in \(\displaystyle M\)) ... and so ... the intersection of the maximal submodules of \(\displaystyle M\) would again be a proper submodule ... ... so \(\displaystyle \text{ Rad}(M) \neq M\) ... ...

Is that correct?

Yes, that's correct.

Hopefully someone can clarify the following:
Regarding the conditions \(\displaystyle J(R) = R\) and \(\displaystyle J(R) = 0\) ... ... ... ... I am still a little confused ...Now, regarding \(\displaystyle \text{ Rad}(M)\) ... Bland is clear when it comes to \(\displaystyle \text{ Rad}(M) = M\) ... indeed Bland's definition of \(\displaystyle \text{ Rad}(M)\) is as follows:"... ... If \(\displaystyle M\) is an \(\displaystyle R\)-module, then the radical of \(\displaystyle M\), denoted by \(\displaystyle \text{ Rad}(M)\), is the intersection of the maximal submodules of \(\displaystyle M\). If \(\displaystyle M\) fails to have maximal submodules, then we set \(\displaystyle \text{ Rad}(M) = M\). ... ...
Presumably \(\displaystyle \text{ Rad}(M) = \{ 0 \}\) when there exist maximal submodules in \(\displaystyle M\), but their intersection is \(\displaystyle \{ 0 \}\)
Now with \(\displaystyle J(R)\) Bland does not explicitly give a condition for which \(\displaystyle J(R) = R\) ... indeed his definition of the Jacobson radical is as follows:"... ... The Jacobson radical of \(\displaystyle R\), denoted \(\displaystyle J(R)\), is the intersection of maximal right ideals of \(\displaystyle R\). If \(\displaystyle J(R) = 0\), then R is said to be a Jacobson semisimple ring ... ..."Is it correct to assume that if there are no maximal right ideals in \(\displaystyle R\), then \(\displaystyle J(R) = R\) ... ...

... ... and \(\displaystyle J(R) = 0\) if there do exist maximum right ideals but their intersection is \(\displaystyle \{ 0 \}\)
Can someone please confirm that my thinking is correct and/or point out any shortcomings or errors ...

Hope someone can help ...

Peter

Like I already explained, $J(R) \neq R$ as $R$ is a ring with unity. It is however possible for $J(R) = R$ if $R$ is a ring without unity.

A theorem of Melvin Henriksen states that a commutative ring has no maximal ideals if and only if $R = J(R)$ and $R^2 + pR = R$ for every integer prime $p$.

Regardless of whether or not $J(R) = 0$, if $R$ is a nonzero ring with unity, it has a maximal right ideal.

 

[Math Amateur]

Yes, that's correct.

Like I already explained, $J(R) \neq R$ as $R$ is a ring with unity. It is however possible for $J(R) = R$ if $R$ is a ring without unity.

A theorem of Melvin Henriksen states that a commutative ring has no maximal ideals if and only if $R = J(R)$ and $R^2 + pR = R$ for every integer prime $p$.

Regardless of whether or not $J(R) = 0$, if $R$ is a nonzero ring with unity, it has a maximal right ideal.

Thanks Euge ...

Appreciate your help, as always ...

Peter