MHB Help with B&K Exercise 1.2.11 Pg 33: Bimodules and Endomorphisms

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with PART 2 of Exercise 1.1.11 (Chapter 1: Basics, page 33) concerning bimodules and endomorphisms... ...

Exercise 1.1.11 reads as follows:
https://www.physicsforums.com/attachments/3073I need help to get started on showing that there is a ring homomorphism from R to End(M) (where M is a left E module) whose kernel is the annihilator Ann(M) of M?

Can someone please help?

Peter

 

[Euge]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with PART 2 of Exercise 1.1.11 (Chapter 1: Basics, page 33) concerning bimodules and endomorphisms... ...

Exercise 1.1.11 reads as follows:
https://www.physicsforums.com/attachments/3073I need help to get started on showing that there is a ring homomorphism from R to End(M) (where M is a left E module) whose kernel is the annihilator Ann(M) of M?

Can someone please help?

Peter

Hi Peter,

To attempt to do this problem, a few things need to be settled. You have an unknown, ${}_EM$, since it's not stated in the problem how $E$ acts on $M$. Let's make the following action: $e\cdot m := e(m)$ for all $e\in E$ and $m\in M$. In other words, $E$ acts on $M$ by evaluations. The action is well-defined since

$\displaystyle (e' \cdot e) \cdot m = e'(e(m)) = e'(e\cdot m) = e'\cdot (e\cdot m)$

and

$\displaystyle 1\cdot m = 1(m) = m$

for all $e,\, e'\in E$ and $m\in M$.

Given $r\in R$, the map $f_r : M \to M$ sending $m$ to $mr$ is a left $E$-module homomorphism. To see this, take $r\in R$. Given $m,\, m'\in M$ and $e\in E$, we have

$f_r(m + m') = (m + m')r = mr + m'r$

and (since $e$ is a right $R$-homomorphism)

$f_r(em) = (em)r = [e(m)]r = e(mr) = e f_r(m)$.

Now we can consider the map $\lambda : R \to \text{End}({}_EM)$ given by $\lambda(r) = f_r$. Try working with this and see what you get.

 

[Math Amateur]

Hi Peter,

To attempt to do this problem, a few things need to settled. You have an unknown, ${}_EM$, since it's not stated in the problem how $E$ acts on $M$. Let's make the following action: $e\cdot m := e(m)$ for all $e\in E$ and $m\in M$. In other words, $E$ acts on $M$ by evaluations. The action is well-defined since

$\displaystyle (e' \cdot e) \cdot m = e'(e(m)) = e'(e\cdot m) = e'\cdot (e\cdot m)$

and

$\displaystyle 1\cdot m = 1(m) = m$

for all $e,\, e'\in E$ and $m\in M$.

Given $r\in R$, the map $f_r : M \to M$ sending $m$ to $mr$ is a left $E$-module homomorphism. To see this, take $r\in R$. Given $m,\, m'\in M$ and $e\in E$, we have

$f_r(m + m') = (m + m')r = mr + m'r$

and (since $e$ is a right $R$-homomorphism)

$f_r(em) = (em)r = [e(m)]r = e(mr) = e f_r(m)$.

Now we can consider the map $\lambda : R \to \text{End}({}_EM)$ given by $\lambda(r) = f_r$. Try working with this and see what you get.

Thanks Euge ... but ... just a simple clarification ... ...

You write:

" ... ... Let's make the following action: $e\cdot m := e(m)$ for all $e\in E$ and $m\in M$. ... .."

... which is fine ..

But ... then you write:

" ... ... The action is well-defined since

$\displaystyle (e' \cdot e) \cdot m = e'(e(m)) = e'(e\cdot m) = e'\cdot (e\cdot m)$ ... ... "

I am slightly confused as to how this follows from the definition $$m := e(m)$$ since this definition does not seem to deal with $$e' \cdot e$$ ... ... indeed, this seems to involve a multiplication of two ring elements e, e' and the action does not deal with this ... ...

Can you clarify?

(hope my question makes sense!)

Peter

 

[Euge]

Multiplication in $E$ is composition of functions.

 

[Math Amateur]

Multiplication in $E$ is composition of functions.

Yes, true ... indeed ... so yes, that clarifies it somewhat for me ...

... but should we write $$e' \cdot e$$ which looks like $$e'$$ acting on $$e$$ when it is a ring multiplication ... that is a composition of functions not an evaluation ... ?

Peter

 

[Euge]

It doesn't really matter. A ring is a left module over itself with an action given by the ring multiplication. So there's no ambiguity between $e' \cdot e$ and $e'\circ e$:

$\displaystyle (e' \cdot e) \cdot m = (e'\circ e) \cdot m = (e'\circ e)(m) = e'(e(m))$.

 

[Math Amateur]

It doesn't really matter. A ring is a left module over itself with an action given by the ring multiplication. So there's no ambiguity between $e' \cdot e$ and $e'\circ e$:

$\displaystyle (e' \cdot e) \cdot m = (e'\circ e) \cdot m = (e'\circ e)(m) = e'(e(m))$.

oh ... yes, indeed ... well, that definitely clarifies the matter ... thank you!

Peter