Bimodules and Endomorphisms - Berrick and Keating - Exercise 1.2.11 page 33

[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 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/3056Can someone please help me get started on showing that statements (a) and (b) are equivalent?

Help would be appreciated.

Peter

 

[Deveno]

Suppose there is a ring homomorphism $\eta:S \to \text{End}(M_R)$

Define a (left) $S$-action on $M_R$ by:

$s\cdot m = \eta(s)(m)$.

Since $\eta(s) \in \text{End}(M_R)$, we have:

$s\cdot(m+m') = \eta(s)(m+m') = \eta(s)(m) + \eta(s)(m') = s\cdot m + s\cdot m'$.

Since $\eta$ is a ring-homomorphism:

$\eta(s+s') = \eta(s) + \eta(s')$, hence:

$(s+s')\cdot m = \eta(s+s')(m) = (\eta(s) + \eta(s'))(m) = \eta(s)(m) + \eta(s')m = s\cdot m + s'\cdot m$.

Another consequence of $\eta$ being a ring-homomorphism is that:

$\eta(ss') = \eta(s) \circ \eta(s')$, so:

$(ss')\cdot m = \eta(ss')(m) = (\eta(s)\circ \eta(s'))(m) = \eta(s)(\eta(s')(m)) = s\cdot(s'\cdot m)$

and if $S$ is a unital ring:

$1_S\cdot m = \eta(1_S)(m) = 1_{M_R}(m) = m$, so that $M_R$ is a left $S$-module (and since it started as a right $R$-module, this means it is an $S,R$-bimodule).

To get you started in the other direction, show that if $M_R$ is an $S,R$-bimodule, that if we define:

$\eta(s) = s\cdot(-)$

this gives a ring homomorphism $\eta: S \to \text{End}(M_R)$.

**************

Note that in the special case $R = S = F$, a field, we can identify these two actions, and we have a vector space.

 

[Math Amateur]

Suppose there is a ring homomorphism $\eta:S \to \text{End}(M_R)$

Define a (left) $S$-action on $M_R$ by:

$s\cdot m = \eta(s)(m)$.

Since $\eta(s) \in \text{End}(M_R)$, we have:

$s\cdot(m+m') = \eta(s)(m+m') = \eta(s)(m) + \eta(s)(m') = s\cdot m + s\cdot m'$.

Since $\eta$ is a ring-homomorphism:

$\eta(s+s') = \eta(s) + \eta(s')$, hence:

$(s+s')\cdot m = \eta(s+s')(m) = (\eta(s) + \eta(s'))(m) = \eta(s)(m) + \eta(s')m = s\cdot m + s'\cdot m$.

Another consequence of $\eta$ being a ring-homomorphism is that:

$\eta(ss') = \eta(s) \circ \eta(s')$, so:

$(ss')\cdot m = \eta(ss')(m) = (\eta(s)\circ \eta(s'))(m) = \eta(s)(\eta(s')(m)) = s\cdot(s'\cdot m)$

and if $S$ is a unital ring:

$1_S\cdot m = \eta(1_S)(m) = 1_{M_R}(m) = m$, so that $M_R$ is a left $S$-module (and since it started as a right $R$-module, this means it is an $S,R$-bimodule).

To get you started in the other direction, show that if $M_R$ is an $S,R$-bimodule, that if we define:

$\eta(s) = s\cdot(-)$

this gives a ring homomorphism $\eta: S \to \text{End}(M_R)$.

**************

Note that in the special case $R = S = F$, a field, we can identify these two actions, and we have a vector space.

Thanks for the help Deveno ... working through you post in detail now ...

Peter

 

[Euge]

Adding to Deveno's answer, for the first part you need to also verify that $s(mr) = (sm)r$ for all $s\in S$, $m\in M$, and $r\in R$, using $\eta$. To do this, take any $s\in S$, $m\in M$, and $r\in R$. Then since $\phi(s)$ is $R$-balanced,

$s(mr) = \eta(s)(mr) = [\eta(s)(m)]r = (sm)r$.

 

[Deveno]

Adding to Deveno's answer, for the first part you need to also verify that $s(mr) = (sm)r$ for all $s\in S$, $m\in M$, and $r\in R$, using $\eta$. To do this, take any $s\in S$, $m\in M$, and $r\in R$. Then since $\phi(s)$ is $R$-balanced,

$s(mr) = \eta(s)(mr) = [\eta(s)(m)]r = (sm)r$.

Good catch. The two actions need to be compatible.

 

[Math Amateur]

Suppose there is a ring homomorphism $\eta:S \to \text{End}(M_R)$

Define a (left) $S$-action on $M_R$ by:

$s\cdot m = \eta(s)(m)$.

Since $\eta(s) \in \text{End}(M_R)$, we have:

$s\cdot(m+m') = \eta(s)(m+m') = \eta(s)(m) + \eta(s)(m') = s\cdot m + s\cdot m'$.

Since $\eta$ is a ring-homomorphism:

$\eta(s+s') = \eta(s) + \eta(s')$, hence:

$(s+s')\cdot m = \eta(s+s')(m) = (\eta(s) + \eta(s'))(m) = \eta(s)(m) + \eta(s')m = s\cdot m + s'\cdot m$.

Another consequence of $\eta$ being a ring-homomorphism is that:

$\eta(ss') = \eta(s) \circ \eta(s')$, so:

$(ss')\cdot m = \eta(ss')(m) = (\eta(s)\circ \eta(s'))(m) = \eta(s)(\eta(s')(m)) = s\cdot(s'\cdot m)$

and if $S$ is a unital ring:

$1_S\cdot m = \eta(1_S)(m) = 1_{M_R}(m) = m$, so that $M_R$ is a left $S$-module (and since it started as a right $R$-module, this means it is an $S,R$-bimodule).

To get you started in the other direction, show that if $M_R$ is an $S,R$-bimodule, that if we define:

$\eta(s) = s\cdot(-)$

this gives a ring homomorphism $\eta: S \to \text{End}(M_R)$.

**************

Note that in the special case $R = S = F$, a field, we can identify these two actions, and we have a vector space.

Thanks Deveno and Euge ... I appreciate your help ...

You write:

" ... ... to get you started in the other direction, show that if $M_R$ is an $S,R$-bimodule, that if we define:

$\eta(s) = s\cdot(-)$

this gives a ring homomorphism $\eta: S \to \text{End}(M_R)$. ... "

Can you explain what you mean by " ... define:

$\eta(s) = s\cdot(-)$ ... "

Peter

 

[Deveno]

Thanks Deveno and Euge ... I appreciate your help ...

You write:

" ... ... to get you started in the other direction, show that if $M_R$ is an $S,R$-bimodule, that if we define:

$\eta(s) = s\cdot(-)$

this gives a ring homomorphism $\eta: S \to \text{End}(M_R)$. ... "

Can you explain what you mean by " ... define:

$\eta(s) = s\cdot(-)$ ... "

Peter

Define the image of $s \in S$ to be the mapping $\eta(s) \in \text{End}(M_R)$ that sends $m \mapsto s\cdot m$ (often called "scalar multiplication").

This process, of identifying a function of two variables (in this case $(-\cdot-): S \times M_R \to M_R$) with a function of a single variable (the $s$) that outputs ANOTHER function of a single variable (in this case $\eta(s)$, seen as a "left-action") is known as currying, and is a common technique in mathematics:

underneath it all, we are invoking a set-isomorphism:

$M_R^{S \times M_R} \cong (M_R^{M_R})^S$

This let's us deal with an action (which involves TWO structures) "one structure at a time".

You should recognize the analogy here with a group action on a set, which is a similar kind of "product":

$G \times X \to X$.

We can view this as a kind of "scalar multiplication" where $X$ lacks any abelian group structure. We need "some" of the rules of modules to apply:

$g\cdot(g'\cdot x) = (gg')\cdot x$

$e_G\cdot x = x$

This is equivalent to saying we have a group-homomorphism:

$\sigma: G \to \text{Sym}(X)$ (the group of bijections on $X$; if $X$ is finite, these are called permutations of $X$ These are precisely the set-automorphisms of $X$.).

The symmetric group $S_n$ has a natural action on any set with $n$ elements, since:

$S_n \cong \text{Sym}(\{1,2,\dots,n\})$ (so $S_n$ has a representation as permutations which are precisely the elements of $S_n$ itself...in group-action theory, $S_n$ is an "invisible group", much like the standard basis for $\Bbb R^n$ is an "invisible basis").

Now, with an abelian group, not all endomorphisms are automorphisms. But we can form a similar notion of "monoid action" (which is actually a better analogy, since the multiplicative monoid of a ring with unity is typically NOT a group), which is a monoid-homomorphism:

$\phi: M \to X^X$ (the set $X^X$ forms what is called the monoid of transformations of $X$, that is: ALL functions $f:X \to X$).

This is pretty basic: what monoids essentially are, are "things we can do right after another". What groups essentially are, are "things we can do that can be un-done". The process of "undoing" is "inversion".

For example, with an object, say $a$, we can "add more $a$'s, to get things like:

$aa$
$aaaaaa$
$aaa$

We can "partially" undo this (subtracting $a$'s), as long as we have enough $a$'s to start with. But if we have:

$aa$

it is not immediately clear how to "take away 3 $a$'s".

What I've done above should remind you of something: free objects. Free monoids, for example, have their humble beginnings in something very basic: natural numbers. In a similar vein, free groups owe THEIR origins to something similar, integers.

I don't think I can stress enough, that these basic structures are essential for more abstract ones. If you have a doubt about modules, for example, one way to test your ideas is to see if they hold for square matrices with integer entries. While these form rather "special" modules, it is often easy to find simple counter-examples there. If you need a non-commutative ring $R$, these also make a good "test case" for $R$. A lot of sophisticated mathematics eventually boils down to some involved kind of arithmetic.

 

[Math Amateur]

Define the image of $s \in S$ to be the mapping $\eta(s) \in \text{End}(M_R)$ that sends $m \mapsto s\cdot m$ (often called "scalar multiplication").

This process, of identifying a function of two variables (in this case $(-\cdot-): S \times M_R \to M_R$) with a function of a single variable (the $s$) that outputs ANOTHER function of a single variable (in this case $\eta(s)$, seen as a "left-action") is known as currying, and is a common technique in mathematics:

underneath it all, we are invoking a set-isomorphism:

$M_R^{S \times M_R} \cong (M_R^{M_R})^S$

This let's us deal with an action (which involves TWO structures) "one structure at a time".

You should recognize the analogy here with a group action on a set, which is a similar kind of "product":

$G \times X \to X$.

We can view this as a kind of "scalar multiplication" where $X$ lacks any abelian group structure. We need "some" of the rules of modules to apply:

$g\cdot(g'\cdot x) = (gg')\cdot x$

$e_G\cdot x = x$

This is equivalent to saying we have a group-homomorphism:

$\sigma: G \to \text{Sym}(X)$ (the group of bijections on $X$; if $X$ is finite, these are called permutations of $X$ These are precisely the set-automorphisms of $X$.).

The symmetric group $S_n$ has a natural action on any set with $n$ elements, since:

$S_n \cong \text{Sym}(\{1,2,\dots,n\})$ (so $S_n$ has a representation as permutations which are precisely the elements of $S_n$ itself...in group-action theory, $S_n$ is an "invisible group", much like the standard basis for $\Bbb R^n$ is an "invisible basis").

Now, with an abelian group, not all endomorphisms are automorphisms. But we can form a similar notion of "monoid action" (which is actually a better analogy, since the multiplicative monoid of a ring with unity is typically NOT a group), which is a monoid-homomorphism:

$\phi: M \to X^X$ (the set $X^X$ forms what is called the monoid of transformations of $X$, that is: ALL functions $f:X \to X$).

This is pretty basic: what monoids essentially are, are "things we can do right after another". What groups essentially are, are "things we can do that can be un-done". The process of "undoing" is "inversion".

For example, with an object, say $a$, we can "add more $a$'s, to get things like:

$aa$
$aaaaaa$
$aaa$

We can "partially" undo this (subtracting $a$'s), as long as we have enough $a$'s to start with. But if we have:

$aa$

it is not immediately clear how to "take away 3 $a$'s".

What I've done above should remind you of something: free objects. Free monoids, for example, have their humble beginnings in something very basic: natural numbers. In a similar vein, free groups owe THEIR origins to something similar, integers.

I don't think I can stress enough, that these basic structures are essential for more abstract ones. If you have a doubt about modules, for example, one way to test your ideas is to see if they hold for square matrices with integer entries. While these form rather "special" modules, it is often easy to find simple counter-examples there. If you need a non-commutative ring $R$, these also make a good "test case" for $R$. A lot of sophisticated mathematics eventually boils down to some involved kind of arithmetic.

Thanks again to Deveno and Euge ...

Now to show \(\displaystyle (a) \Longrightarrow (b)\) ...

Assume \(\displaystyle M_R\) is an \(\displaystyle S-R\) bimodule (so we have \(\displaystyle r(ms) = (rm)s\) for all \(\displaystyle r \in R, s \in S\), and \(\displaystyle m \in M)\) ... ...

Now define \(\displaystyle \eta\) by \(\displaystyle \eta (s) = s \cdot m \) is a ring homomorphism (where \(\displaystyle s \in S, m \in M\))

We claim that \(\displaystyle \eta\) is a ring homomorphism \(\displaystyle \eta \ : \ S \to \text{ End} (M_R) \) ... ... so we have to show:

\(\displaystyle \eta (s + s') = \eta (s) + \eta (s') \)

and

\(\displaystyle \eta (ss') = \eta (s) \eta (s')
\)

Now we have ... ...

\(\displaystyle \eta (s + s') = (s + s') \cdot m \)

\(\displaystyle = s \cdot m + s' \cdot m \) by distributivity in M as a left S module

\(\displaystyle = \eta (s) + \eta (s')
\)and also we have

\(\displaystyle \eta (ss') = (ss') \cdot m
\)

\(\displaystyle = s \cdot (s' \cdot m)
\) (Is this step correct?)

\(\displaystyle = s \cdot ( \eta (s) )
\)

= ... ... ?

BUT ... ... how do we proceed from here ...

Hope someone can help?

Peter

 

[Euge]

Assume \(\displaystyle M_R\) is an \(\displaystyle S-R\) bimodule (so we have \(\displaystyle r(ms) = (rm)s\) for all \(\displaystyle r \in R, s \in S\), and \(\displaystyle m \in M)\) ... ...

You've reversed the actions of $S$ and $R$. It should be $s(mr) = (sm)r$ for all $s\in S$, $m\in M$, and $r\in R$.

Now define \(\displaystyle \eta\) by \(\displaystyle \eta (s) = s \cdot m \) is a ring homomorphism (where \(\displaystyle s \in S, m \in M\))

We claim that \(\displaystyle \eta\) is a ring homomorphism \(\displaystyle \eta \ : \ S \to \text{ End} (M_R) \) ... ... so we have to show:

\(\displaystyle \eta (s + s') = \eta (s) + \eta (s') \)

and

\(\displaystyle \eta (ss') = \eta (s) \eta (s')
\)

Now we have ... ...

\(\displaystyle \eta (s + s') = (s + s') \cdot m \)

\(\displaystyle = s \cdot m + s' \cdot m \) by distributivity in M as a left S module

\(\displaystyle = \eta (s) + \eta (s')
\)and also we have

\(\displaystyle \eta (ss') = (ss') \cdot m
\)

\(\displaystyle = s \cdot (s' \cdot m)
\) (Is this step correct?)

\(\displaystyle = s \cdot ( \eta (s) )
\)

= ... ... ?

BUT ... ... how do we proceed from here ...

Hope someone can help?

Peter

The definition of $\eta$ does not make sense: For each $s\in S$, $\eta(s)$ should be a function, not an element of $M_R$. I think this is why you're getting stuck. Instead you should have, for each $s\in S$, $\eta(s) : m \mapsto sm$, or $\eta(s)(m) = sm$. So to prove $\eta(s + s') = \eta(s) + \eta(s')$ and $\eta(ss') = \eta(s)\eta(s')$, you need to show $\eta(s + s')(m) = \eta(s)(m) + \eta(s')(m)$ and $\eta(ss')(m) = \eta(s)[\eta(s')(m)]$ for all $m\in M$.

I don't know if you've already done this, but you need to show that for each $s\in S$, the map from $M$ to itself sending $m$ to $sm$ is an endomorphism of $M_R$. Otherwise, the $\eta$ map is not well-defined.

 

[Math Amateur]

You've reversed the actions of $S$ and $R$. It should be $s(mr) = (sm)r$ for all $s\in S$, $m\in M$, and $r\in R$.
The definition of $\eta$ does not make sense: For each $s\in S$, $\eta(s)$ should be a function, not an element of $M_R$. I think this is why you're getting stuck. Instead you should have, for each $s\in S$, $\eta(s) : m \mapsto sm$, or $\eta(s)(m) = sm$. So to prove $\eta(s + s') = \eta(s) + \eta(s')$ and $\eta(ss') = \eta(s)\eta(s')$, you need to show $\eta(s + s')(m) = \eta(s)(m) + \eta(s')(m)$ and $\eta(ss')(m) = \eta(s)[\eta(s')(m)]$ for all $m\in M$.

I don't know if you've already done this, but you need to show that for each $s\in S$, the map from $M$ to itself sending $m$ to $sm$ is an endomorphism of $M_R$. Otherwise, the $\eta$ map is not well-defined.

Thanks Euge ...

Well you make two VERY helpful clarifications ...

You write:

" ... The definition of $\eta$ does not make sense: For each $s\in S$, $\eta(s)$ should be a function, not an element of $M_R$. ..."

Yes indeed ...

and

" ... ... you need to show that for each $s\in S$, the map from $M$ to itself sending $m$ to $sm$ is an endomorphism of $M_R$. Otherwise, the $\eta$ map is not well-defined. ..."

Yes, that is now obvious to me ... thanks ...

Your post helped a lot ... thanks!

Peter

 

[Deveno]

I just want to amplify a couple of things:

If we have a mapping $\eta: S \to \text{End}(M_r)$

Then $\eta(s)$ for $s \in S$ is a FUNCTION.

To specify this function, we say what $\eta(s)$ does to a given $m \in M_R$.

Now, the notation:

$\eta(s)(m)$ is sort of "awkward" so many texts will use a subscript:

$s \mapsto \eta_s$ that is: $\eta(s) = \eta_s$, which looks "more natural", as then we can write:

$\eta_s(m)$ instead of $\eta(s)(m)$.

To amplify what I said about "currying", I'll use a simple example:

Suppose we have: $f:\Bbb R^2 \to \Bbb R$ given by $f(x,y) = x+y$ (so $f$ is the ordinary binary operation of real addition).

We could, instead of considering this as a binary operation, consider it as a mapping of real numbers to real functions:

$L: \Bbb R \to \Bbb R^{\Bbb R}$, where $L(x) = L_x$, and $L_x: \Bbb R \to \Bbb R$ is given by: $L_x(y) = x+y$. Note we get a DIFFERENT function $L_x$ for every $x$.

The point being, every time you have a mapping of some sort, you have to pay attention to "what kind of thing the domain elements are", and "what kind of thing the co-domain elements are".

 

[Math Amateur]

I just want to amplify a couple of things:

If we have a mapping $\eta: S \to \text{End}(M_r)$

Then $\eta(s)$ for $s \in S$ is a FUNCTION.

To specify this function, we say what $\eta(s)$ does to a given $m \in M_R$.

Now, the notation:

$\eta(s)(m)$ is sort of "awkward" so many texts will use a subscript:

$s \mapsto \eta_s$ that is: $\eta(s) = \eta_s$, which looks "more natural", as then we can write:

$\eta_s(m)$ instead of $\eta(s)(m)$.

To amplify what I said about "currying", I'll use a simple example:

Suppose we have: $f:\Bbb R^2 \to \Bbb R$ given by $f(x,y) = x+y$ (so $f$ is the ordinary binary operation of real addition).

We could, instead of considering this as a binary operation, consider it as a mapping of real numbers to real functions:

$L: \Bbb R \to \Bbb R^{\Bbb R}$, where $L(x) = L_x$, and $L_x: \Bbb R \to \Bbb R$ is given by: $L_x(y) = x+y$. Note we get a DIFFERENT function $L_x$ for every $x$.

The point being, every time you have a mapping of some sort, you have to pay attention to "what kind of thing the domain elements are", and "what kind of thing the co-domain elements are".

Thanks Deveno and Euge for considerable help and clarification with this exercise ...

Sorry to be slow and careful ... but just checking that I am understanding your various messages on this problem ...

So to sum up what I think you are saying to me ...

We are assuming that \(\displaystyle M_R\) is an \(\displaystyle S-R-\)bimodule.

We need to show that there exists a ring homomorphism \(\displaystyle S \to \text{ End } (M_R) \).

So we proceed as follows:

For each \(\displaystyle s \in S\) define \(\displaystyle \eta_s \ : \ M_R \to M_R \) as follows:

\(\displaystyle \eta_s(m) = s \cdot m\)

Now we are claiming (and have to show) that \(\displaystyle \eta_s\) is an endomorphism on \(\displaystyle M_R\).

Then we go on to claim that there exists a ring homomorphism \(\displaystyle \alpha \ : \ S \to \text{ End }(M_R)\) where \(\displaystyle \alpha(s) = \eta_s\) ... ...

and, specifically, we have to show that:

\(\displaystyle \alpha (s + s') = \alpha (s) + \alpha (s')\)

and

\(\displaystyle \alpha (ss') = \alpha (s) \alpha (s')\)

Please let me know if I have interpreted your advice/guidance correctly.

Peter

 

[Euge]

You've made just one error:

$\displaystyle \alpha (s + s') = \alpha (s) \alpha (s')$

should be

$\displaystyle \alpha (ss') = \alpha (s) \alpha (s')$

Other than that, you've got it! Great job summarizing the concepts!

 

[Math Amateur]

You've made just one error:

$\displaystyle \alpha (s + s') = \alpha (s) \alpha (s')$

should be

$\displaystyle \alpha (ss') = \alpha (s) \alpha (s')$

Other than that, you've got it! Great job summarizing the concepts!

oh! ... of course you are right ... basically a result of copying and pasting too quickly and not amending the paste ... will edit it now ...

Thanks for helping me understand the exercise ...

Peter

 

[Math Amateur]

I just want to amplify a couple of things:

If we have a mapping $\eta: S \to \text{End}(M_r)$

Then $\eta(s)$ for $s \in S$ is a FUNCTION.

To specify this function, we say what $\eta(s)$ does to a given $m \in M_R$.

Now, the notation:

$\eta(s)(m)$ is sort of "awkward" so many texts will use a subscript:

$s \mapsto \eta_s$ that is: $\eta(s) = \eta_s$, which looks "more natural", as then we can write:

$\eta_s(m)$ instead of $\eta(s)(m)$.

To amplify what I said about "currying", I'll use a simple example:

Suppose we have: $f:\Bbb R^2 \to \Bbb R$ given by $f(x,y) = x+y$ (so $f$ is the ordinary binary operation of real addition).

We could, instead of considering this as a binary operation, consider it as a mapping of real numbers to real functions:

$L: \Bbb R \to \Bbb R^{\Bbb R}$, where $L(x) = L_x$, and $L_x: \Bbb R \to \Bbb R$ is given by: $L_x(y) = x+y$. Note we get a DIFFERENT function $L_x$ for every $x$.

The point being, every time you have a mapping of some sort, you have to pay attention to "what kind of thing the domain elements are", and "what kind of thing the co-domain elements are".

Thanks for this post Deveno ... it was critical to me finally understanding the structure of things in this exercise ...

I have learned a lot from you and Euge regarding the concepts involved in this exercise ... ...

Peter

 

[Math Amateur]

Thanks Deveno and Euge for considerable help and clarification with this exercise ...

Sorry to be slow and careful ... but just checking that I am understanding your various messages on this problem ...

So to sum up what I think you are saying to me ...

We are assuming that \(\displaystyle M_R\) is an \(\displaystyle S-R-\)bimodule.

We need to show that there exists a ring homomorphism \(\displaystyle S \to \text{ End } (M_R) \).

So we proceed as follows:

For each \(\displaystyle s \in S\) define \(\displaystyle \eta_s \ : \ M_R \to M_R \) as follows:

\(\displaystyle \eta_s(m) = s \cdot m\)

Now we are claiming (and have to show) that \(\displaystyle \eta_s\) is an endomorphism on \(\displaystyle M_R\).

Then we go on to claim that there exists a ring homomorphism \(\displaystyle \alpha \ : \ S \to \text{ End }(M_R)\) where \(\displaystyle \alpha(s) = \eta_s\) ... ...

and, specifically, we have to show that:

\(\displaystyle \alpha (s + s') = \alpha (s) + \alpha (s')\)

and

\(\displaystyle \alpha (ss') = \alpha (s) \alpha (s')\)

Please let me know if I have interpreted your advice/guidance correctly.

Peter

So now ... solving the exercise ... but ... help still needed ...

I think that the solution to \(\displaystyle (a) \Longrightarrow (b) \) proceeds as follows:Assume that \(\displaystyle M_R\) is an \(\displaystyle S-R-\)bimodule.

We need to show that there exists a ring homomorphism \(\displaystyle S \to \text{ End } (M_R) \).

Proceed as follows:

For each \(\displaystyle s \in S\) define \(\displaystyle \eta_s \ : \ M_R \to M_R \) as follows:

\(\displaystyle \eta_s(m) = s \cdot m \)Now we are claiming (and have to show) that \(\displaystyle \eta_s\) is an endomorphism on \(\displaystyle M_R\).

Then we go on to claim that there exists a ring homomorphism \(\displaystyle \alpha \ : \ S \to \text{ End }(M_R)\) where \(\displaystyle \alpha(s) = \eta_s\)So ... ... first we need to establish that \(\displaystyle \eta_s \ : \ M_R \to M_R \) is an endomorphism on \(\displaystyle M_R\). ... ...

So we need to show that:

\(\displaystyle \eta_s (m +n) = \eta_s (m) + \eta_s (n) \) for \(\displaystyle m,n \in M, s \in S \)

and

\(\displaystyle \eta_s (mn) = \eta_s (m) \eta_s (n)\) for \(\displaystyle m,n \in M, s \in S \)Now to show this, proceed as follows:

\(\displaystyle \eta_s (m +n) = s \cdot (m + n) \)

= \(\displaystyle sm + sn\) by distributivity of \(\displaystyle s\) over \(\displaystyle (m +n)\) (definition of module)

\(\displaystyle = \eta_s (m) + \eta_s (n)\)

and

\(\displaystyle \eta_s (mn) = s \cdot (mn) \)

= ?

BUT ... how to proceed?

Can someone please help?

Peter

 

[Deveno]

So now ... solving the exercise ... but ... help still needed ...

I think that the solution to \(\displaystyle (a) \Longrightarrow (b) \) proceeds as follows:Assume that \(\displaystyle M_R\) is an \(\displaystyle S-R-\)bimodule.

We need to show that there exists a ring homomorphism \(\displaystyle S \to \text{ End } (M_R) \).

Proceed as follows:

For each \(\displaystyle s \in S\) define \(\displaystyle \eta_s \ : \ M_R \to M_R \) as follows:

\(\displaystyle \eta_s(m) = s \cdot m \)Now we are claiming (and have to show) that \(\displaystyle \eta_s\) is an endomorphism on \(\displaystyle M_R\).

Then we go on to claim that there exists a ring homomorphism \(\displaystyle \alpha \ : \ S \to \text{ End }(M_R)\) where \(\displaystyle \alpha(s) = \eta_s\)So ... ... first we need to establish that \(\displaystyle \eta_s \ : \ M_R \to M_R \) is an endomorphism on \(\displaystyle M_R\). ... ...

So we need to show that:

\(\displaystyle \eta_s (m +n) = \eta_s (m) + \eta_s (n) \) for \(\displaystyle m,n \in M, s \in S \)

and

\(\displaystyle \eta_s (mn) = \eta_s (m) \eta_s (n)\) for \(\displaystyle m,n \in M, s \in S \)Now to show this, proceed as follows:

\(\displaystyle \eta_s (m +n) = s \cdot (m + n) \)

= \(\displaystyle sm + sn\) by distributivity of \(\displaystyle s\) over \(\displaystyle (m +n)\) (definition of module)

\(\displaystyle = \eta_s (m) + \eta_s (n)\)

and

\(\displaystyle \eta_s (mn) = s \cdot (mn) \)

= ?

BUT ... how to proceed?

Can someone please help?

Peter

One thing I do not know, because you have not specified, is whether $\text{End}(M_R)$ refers to abelian group endomorphisms, or right $R$-module endomorphisms. If it is the former, then additivity is all you need to show for each $\eta_s$.

If it is the latter (as I suspect is the case) then you need to show that:

$\eta_s(m\cdot r) = (\eta_s(m))\cdot r$ (I am putting the "dots" into make clear the SCOPE of the right-action).

This means that:

$\eta_s(m\cdot r) = s\cdot(m\cdot r) = (s\cdot m)\cdot r = (\eta_s(m))\cdot r$, because $M_R$ is an $S,R$-bimodule.

$\eta_s$ is NOT a ring-homomorphism, in general $M_R$ will NOT be a ring! Writing $mn$ for $m,n \in M_R$ doesn't even make sense.

$\eta$ is the ring-homomorphism, not $\eta_s$.

EDIT:

Let me give you an example of an $S,R$-bimodule that may help clear things up.

Let $M = \text{Mat}_{2\times 3}(\Bbb Z)$ with $R = \text{Mat}_{3\times 3}(\Bbb Z)$, and $S = \text{Mat}_{2\times 2}(\Bbb Z)$.

We can turn $M$ into a right $R$-module $M_R$ by defining, for $A \in M,\ T \in R$:

$A \cdot T = AT$ (the RHS is matrix multiplication).

We can define an $S$-action on $M_R$ for $B \in S,\ A \in M_R$, by:

$B \cdot A = BA$.

This is a bimodule, because $B(AT) = (BA)T$ (matrix multiplication is associative). Note we always get a 2x3 matrix as the result of either, or both of our actions being applied.

Note that for $A_1,A_2 \in M_R$, that the product $A_1A_2$ isn't even defined, we have a "type mismatch", as the number of rows doesn't equal the number of columns.

 

[Math Amateur]

One thing I do not know, because you have not specified, is whether $\text{End}(M_R)$ refers to abelian group endomorphisms, or right $R$-module endomorphisms. If it is the former, then additivity is all you need to show for each $\eta_s$.

If it is the latter (as I suspect is the case) then you need to show that:

$\eta_s(m\cdot r) = (\eta_s(m))\cdot r$ (I am putting the "dots" into make clear the SCOPE of the right-action).

This means that:

$\eta_s(m\cdot r) = s\cdot(m\cdot r) = (s\cdot m)\cdot r = (\eta_s(m))\cdot r$, because $M_R$ is an $S,R$-bimodule.

$\eta_s$ is NOT a ring-homomorphism, in general $M_R$ will NOT be a ring! Writing $mn$ for $m,n \in M_R$ doesn't even make sense.

$\eta$ is the ring-homomorphism, not $\eta_s$.

EDIT:

Let me give you an example of an $S,R$-bimodule that may help clear things up.

Let $M = \text{Mat}_{2\times 3}(\Bbb Z)$ with $R = \text{Mat}_{3\times 3}(\Bbb Z)$, and $S = \text{Mat}_{2\times 2}(\Bbb Z)$.

We can turn $M$ into a right $R$-module $M_R$ by defining, for $A \in M,\ T \in R$:

$A \cdot T = AT$ (the RHS is matrix multiplication).

We can define an $S$-action on $M_R$ for $B \in S,\ A \in M_R$, by:

$B \cdot A = BA$.

This is a bimodule, because $B(AT) = (BA)T$ (matrix multiplication is associative). Note we always get a 2x3 matrix as the result of either, or both of our actions being applied.

Note that for $A_1,A_2 \in M_R$, that the product $A_1A_2$ isn't even defined, we have a "type mismatch", as the number of rows doesn't equal the number of columns.

Thanks Deveno ...

You write: " ... ... $\eta_s$ is NOT a ring-homomorphism ... " indeed, how careless/silly of me ... had realized this but you beat me to it!

I am assuming that End(\(\displaystyle M_R\)) = \(\displaystyle Hom_R (M,M)\) is the set of all R-module homomorphisms from \(\displaystyle M_R \) to \(\displaystyle M_R\) ...

So we have to show that:

\(\displaystyle \eta_s (m + n) = \eta_s (m) + \eta_s (n)\)

and

\(\displaystyle \eta_s (mr) = \eta_s (m)r\)

which has been done ... ...

Thanks again to you and Euge for the significant help and guidance ... ...

Peter