MHB Bland - Rings and their Modules 2011

[steenis]

If the moderators agree, I have made a thread on the book of Bland - Rings and their Modules 2011.

 

[steenis]

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 ?

 

[Euge]

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.

 

[Math Amateur]

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 $$f \in \mbox{Hom} _R (M,N)$$ ... 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 ... ?

 

[steenis]

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
$f(xr)=f(x)r$

 

[steenis]

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.