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 \(\displaystyle m := e(m)\) since this definition does not seem to deal with \(\displaystyle e' \cdot e\) ... ... 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 \(\displaystyle e' \cdot e\) which looks like \(\displaystyle e'\) acting on \(\displaystyle e\) 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