Bland - Rings and their Modules 2011

In summary: Hi Euge,Thanks for your input. I am not sure I understand your counterexample though. Are you saying that Bland's claim is incorrect and that the given conditions do not hold for all $R$-modules $M$ and $N$? Or is there a specific $R$-module where they do not hold? I am not familiar with the notation $[b\cdot i]$, could you explain that?Also, I think I made a mistake in my summary. The given conditions $f(xa) = f(x)a$ and $[r\bullet f](x) = f(xr)$ do not imply that $[r\bullet f](xa) = [r\bullet f](x
  • #1
steenis
312
18
If the moderators agree, I have made a thread on the book of Bland - Rings and their Modules 2011.
 
Physics news on Phys.org
  • #2
I want to take you to Example 1 of section 1.5 p.33: "$\mbox{Hom} _R (M,N)$ as a left R-module.

Let $M$ and $N$ be two right R-modules. R is not necessarely commutative.
On 9th line of this example, Bland claims that $\mbox{Hom} _R (M,N)$ is a left R-module if we define $[r\bullet f](x)=f(xr)$ for $r\in R$ and $x\in M$. To prove that $\mbox{Hom} _R (M,N)$ is a left R-module, we have to prove conditions (1), (2), (3) and (4) of Definition 1.4.1 p.26 (adapted for left R-modules). Bland proves (3) as follows, $r\in R$, $s\in R$, $x\in M$:

$[s\bullet (r\bullet f)[(x)$ = $[r\bullet f](xs)$ = $[f]((xs)r)$ = $[f](x(sr))$ = $[(sr)\bullet f](x)$

To prove conditions (1), (2), and (4) is easy.

However, he omits to prove that if $f$ is an R-map then $[r\bullet f]$ is an R-map for $r\in R$, i.e., if $f\in \mbox{Hom} _R (M,N)$ then $[r\bullet f] \in \mbox{Hom} _R (M,N)$.

And that is my problem.

Given for $r\in R$, $a\in R$, and $x\in M$ is $[r\bullet f](x)=f(xr)$ and $f(xa)=f(x)a$, we have to prove that $[r\bullet f](xa)$ = $[r\bullet f](x)a$. Here is a start:

$[r\bullet f](xa)$ = $[f]((xa)r)$ = $[f](x(ar))$ = ? = $[r\bullet f](x)a$

Who can fill in the missing steps ?
 
  • #3
Hi steenis,

If the given conditions are correct, then there is problem with Bland's claim here. If $R$ is commutative, then indeed $\operatorname{Hom}_R(M,N)$ is a left $R$-module with the given $R$-action. However, if $R$ is non-commutative, then the result need not hold. For suppose $R$ is a non-commutative, unital ring. We may view $R$ as a right $R$-module in the usual way. Consider the element $i \in \operatorname{Hom}_R(R,R)$ given by $i(r) = r$. Then $[b\cdot i](a) = i(ab) = ab$, but $[b\cdot i](1)a = i(1b)a = i(b)a = ba\neq ab = [b\cdot i](a)$. Hence, $[b\cdot i]\notin \operatorname{Hom}_R(R,R)$.

By the way, I've noticed that you and Peter are working on issues between Bland's and Rotman's texts. I will address some of those questions if I have time.
 
  • #4
steenis said:
I want to take you to Example 1 of section 1.5 p.33: "$\mbox{Hom} _R (M,N)$ as a left R-module.

Let $M$ and $N$ be two right R-modules. R is not necessarely commutative.
On 9th line of this example, Bland claims that $\mbox{Hom} _R (M,N)$ is a left R-module if we define $[r\bullet f](x)=f(xr)$ for $r\in R$ and $x\in M$. To prove that $\mbox{Hom} _R (M,N)$ is a left R-module, we have to prove conditions (1), (2), (3) and (4) of Definition 1.4.1 p.26 (adapted for left R-modules). Bland proves (3) as follows, $r\in R$, $s\in R$, $x\in M$:

$[s\bullet (r\bullet f)[(x)$ = $[r\bullet f](xs)$ = $[f]((xs)r)$ = $[f](x(sr))$ = $[(sr)\bullet f](x)$

To prove conditions (1), (2), and (4) is easy.

However, he omits to prove that if $f$ is an R-map then $[r\bullet f]$ is an R-map for $r\in R$, i.e., if $f\in \mbox{Hom} _R (M,N)$ then $[r\bullet f] \in \mbox{Hom} _R (M,N)$.

And that is my problem.

Given for $r\in R$, $a\in R$, and $x\in M$ is $[r\bullet f](x)=f(xr)$ and $f(xa)=f(x)a$, we have to prove that $[r\bullet f](xa)$ = $[r\bullet f](x)a$. Here is a start:

$[r\bullet f](xa)$ = $[f]((xa)r)$ = $[f](x(ar))$ = ? = $[r\bullet f](x)a$

Who can fill in the missing steps ?
Hi Steenis,

Just working through your post and checking Bland, Example 1 of section 1.5 p.33 ... ...

I can see why you want to prove if $f$ is an R-map then $[r\bullet f]$ is an R-map for $r\in R$ ... but I am having trouble completely following you ... ...

Perhaps you can help ...

I understand we have the given condition $[r\bullet f](x)=f(xr)$ ...

... BUT ...

where exactly does $f(xa)=f(x)a$ come from ... how does this arise exactly ... ?(presumably from \(\displaystyle f \in \mbox{Hom} _R (M,N)\) ... somehow ...? )Hope you can help ...

Peter
*** EDIT ***

Oh God! it is probably one of the conditions on f being an R-module homomorphism (R-Linear map) ... is that right ... ?
 
Last edited:
  • #5
Yes, you are right. By definition $f : M \longrightarrow N$ is an R-module homomorphism or R-linear map or R-map between right R-modules, if for all $x, y \in M$ and $r \in R$:
$f(x+y) = f(x) + f(y)$
and
$f(xr)=f(x)r$
 
  • #6
Euge said:
Hi steenis,

If the given conditions are correct, then there is problem with Bland's claim here. If $R$ is commutative, then indeed $\operatorname{Hom}_R(M,N)$ is a left $R$-module with the given $R$-action. However, if $R$ is non-commutative, then the result need not hold. For suppose $R$ is a non-commutative, unital ring. We may view $R$ as a right $R$-module in the usual way. Consider the element $i \in \operatorname{Hom}_R(R,R)$ given by $i(r) = r$. Then $[b\cdot i](a) = i(ab) = ab$, but $[b\cdot i](1)a = i(1b)a = i(b)a = ba\neq ab = [b\cdot i](a)$. Hence, $[b\cdot i]\notin \operatorname{Hom}_R(R,R)$.

By the way, I've noticed that you and Peter are working on issues between Bland's and Rotman's texts. I will address some of those questions if I have time.

Thank you, Euge, for your counterexample, very good. I am not good with examples and counterexamples. And of course, the more input the better.
 

FAQ: Bland - Rings and their Modules 2011

What is the purpose of the book "Bland - Rings and their Modules 2011"?

The purpose of this book is to provide a comprehensive and in-depth discussion of rings and their modules, which are important mathematical structures used in various fields of science and engineering.

Who is the target audience of this book?

This book is intended for graduate students and researchers in mathematics and related fields who have a basic understanding of abstract algebra and are interested in learning more about rings and their modules.

What topics are covered in this book?

This book covers a wide range of topics related to rings and their modules, including basic definitions and properties, homomorphisms and isomorphisms, submodules, quotient modules, direct sums and products, and modules over principal ideal domains.

Are there any prerequisites for reading this book?

It is recommended that readers have a solid understanding of abstract algebra, including groups, rings, and fields, as well as basic knowledge of linear algebra and vector spaces. Some familiarity with basic set theory and logic may also be helpful.

Does this book include any practical applications of rings and modules?

Yes, this book includes various examples and applications of rings and modules in fields such as number theory, algebraic geometry, and coding theory. It also discusses the connections between rings and modules and other mathematical structures, such as vector spaces and algebras.

Back
Top