Modules in Cohn's book on ring theory - simple notational issue

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 1: Basics, on Page 33 we find a definition of a module homomorphism (or R-linear mapping) and a definition of Hom.

I need help to interpret one of Cohn's expressions when he deals with the R-module structure of Hom(M,N) … ...

The relevant text on page 33 is as follows:

View attachment 3189
[ Note that Cohn writes mappings on the right, expressing the usual \(\displaystyle y = f(x)\) as \(\displaystyle y = xf\) ]

In the above text Cohn writes:

" … … When R is commutative, Hom(M,N) also has an R-module structure given by \(\displaystyle x(fr) = (xr)f = (xf)r\) … … "

My problem is how to interpret the term \(\displaystyle x(fr)\) and in particular the term \(\displaystyle fr\) - what can this term mean when we are writing, as Cohn does, mappings on the right?

Hoping someone can help.

Just in case MHB members need to see how Cohn defines a module I am providing the relevant text as follows:View attachment 3190Would welcome help,

Peter

 

[Euge]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 1: Basics, on Page 33 we find a definition of a module homomorphism (or R-linear mapping) and a definition of Hom.

I need help to interpret one of Cohn's expressions when he deals with the R-module structure of Hom(M,N) … ...

The relevant text on page 33 is as follows:

View attachment 3189
[ Note that Cohn writes mappings on the right, expressing the usual \(\displaystyle y = f(x)\) as \(\displaystyle y = xf\) ]

In the above text Cohn writes:

" … … When R is commutative, Hom(M,N) also has an R-module structure given by \(\displaystyle x(fr) = (xr)f = (xf)r\) … … "

My problem is how to interpret the term \(\displaystyle x(fr)\) and in particular the term \(\displaystyle fr\) - what can this term mean when we are writing, as Cohn does, mappings on the right?

Hoping someone can help.

Just in case MHB members need to see how Cohn defines a module I am providing the relevant text as follows:View attachment 3190Would welcome help,

Peter

The notation $x(fr)$ means $rf(x)$ in "our" notation. So the statement

\(\displaystyle x(fr) = (xr)f = (xf)r\)

is equivalent to the statement

\(\displaystyle rf(x) = f(rx) = f(x)r\).

 

[Math Amateur]

The notation $x(fr)$ means $rf(x)$ in "our" notation. So the statement

\(\displaystyle x(fr) = (xr)f = (xf)r\)

is equivalent to the statement

\(\displaystyle rf(x) = f(rx) = f(x)r\).

Thanks for the help, Euge

I must say that I find interpreting x(fr) as rf(x) as most UN-intuitive indeed … but if that is what the notation means then … … that is what the notation means!

I would never have guessed it!

Indeed I would have thought rf(x) would be expressed r(xf) under Cohn's notation!Thanks again,

Peter

 

[Euge]

Thanks for the help, Euge

I must say that I find interpreting x(fr) as rf(x) as most UN-intuitive indeed … but if that is what the notation means then … … that is what the notation means!

I would never have guessed it!

Indeed I would have thought rf(x) would be expressed r(xf) under Cohn's notation!Thanks again,

Peter

Sorry Peter, please disregard the last interpretation; I misread part of Cohn's discussion. Let's try this again...

Given $R$ is commutative, Cohn wants to define an $R$-action on $\text{Hom}_R(M,N)$ to make $\text{Hom}_R(M,N)$ and $R$-module. Given $r\in R$ and $f\in \text{Hom}_R(M,N)$, $fr$ is a function from $M$ to $N$ sending $x$ to $(xr)f$. In other words, in "our notation", $x(fr)$ means $f(xr)$. The actual statement $x(fr) = (xr)f = (xf)r$ defines $fr$. You can verify that $fr$ is an element of $\text{Hom}_R(M,N)$. You can verify that this multiplication defines an $R$-module structure of $\text{Hom}_R(M,N)$.

There is a more general setting where commutativity is not assumed. Let ${}_SL_R$, ${}_RM_T$, and ${}_SN_T$ be bimodules. Then $\text{Hom}_S(L,N)$ is a bimodule over $R$ (to the left) and $T$ (to the right). A left $R$-module structure is defined by $x(rf) := (xr)f$, and a right $T$-module structure is defined by $x(ft) := (xf)t$, for $x\in L$, $r\in R$, $t\in T$, and $f\in \text{Hom}_S(M,N)$.

 

[Deveno]

If we write the mapping on the left (as is customary in analysis), but want to define a RIGHT action, we have:

$(f\cdot r)(x) \stackrel{\text{def}}{=} (f(x))\cdot r$.

That is, if $f: x\mapsto y$, then $fr: x \mapsto yr$

This "looks weird", and is remedied in one of three ways:

1. Studying left actions instead (but sometimes we wind up with "anti-homomorphisms" because $fg$ is usually "first $g$, then $f$").

2. Writing $(x)f$ instead of $f(x)$.

3. Considering "opposite rings" , where the order of multiplication is reversed.

Option 3 is the best, in my opinion, as it let's us "choose handed-ness" as suits us.

 

[Math Amateur]

Sorry Peter, please disregard the last interpretation; I misread part of Cohn's discussion. Let's try this again...

Given $R$ is commutative, Cohn wants to define an $R$-action on $\text{Hom}_R(M,N)$ to make $\text{Hom}_R(M,N)$ and $R$-module. Given $r\in R$ and $f\in \text{Hom}_R(M,N)$, $fr$ is a function from $M$ to $N$ sending $x$ to $xr$. In other words, in "our notation", $x(fr)$ means $f(xr)$. The actual statement $x(fr) = (xr)f = (xf)r$ defines $fr$. You can verify that $fr$ is an element of $\text{Hom}_R(M,N)$. You can verify that this multiplication defines an $R$-module structure of $\text{Hom}_R(M,N)$.

There is a more general setting where commutativity is not assumed. Let ${}_SL_R$, ${}_RM_T$, and ${}_SN_T$ be bimodules. Then $\text{Hom}_S(L,N)$ is a bimodule over $R$ (to the left) and $T$ (to the right). A left $R$-module structure is defined by $x(rf) := (xr)f$, and a right $T$-module structure is defined by $x(ft) := (xf)t$, for $x\in L$, $r\in R$, $t\in T$, and $f\in \text{Hom}_S(M,N)$.

Thanks Euge … just working through this ...

You write:

" … … Cohn wants to define an $R$-action on $\text{Hom}_R(M,N)$ to make $\text{Hom}_R(M,N)$ an $R$-module. … "So to make $\text{Hom}_R(M,N)$ into an R-Module we define a right action of R on $\text{Hom}_R(M,N)$ as follows:

… … for \(\displaystyle r \in R\) and \(\displaystyle f \in \text{Hom}_R(M,N)\) we define a right action

\(\displaystyle \text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)\)

denoted by \(\displaystyle fr \in \text{Hom}_R(M,N)\)

where \(\displaystyle f \in \text{Hom}_R(M,N)\) and \(\displaystyle r \in R \)
Now we further define \(\displaystyle fr\) by (using \(\displaystyle g\) instead of Cohn's \(\displaystyle x\))

\(\displaystyle g(fr) = (gr)f = (gf)r\) where f\(\displaystyle ,g \in M, r \in R\)But how do we verify that \(\displaystyle fr\) is an element of \(\displaystyle \text{Hom}_R(M,N)\)?Further, how do we show that Cohn's M.1 to M.4 hold with the right action defined as above?Specifically we have to show that:

M.1 \(\displaystyle (f + g) r = fr + gr\) for \(\displaystyle r \in R, f,g \in\text{Hom}_R(M,N)\)

M.2 \(\displaystyle f(r + s) = fr + fs\) for \(\displaystyle r, s \in R, f \in\text{Hom}_R(M,N)\)

M.3 \(\displaystyle f(rs) = (fr)s\) for \(\displaystyle r, s \in R, f \in\text{Hom}_R(M,N)\)

M.4 \(\displaystyle f.1 = f\) for \(\displaystyle f \in\text{Hom}_R(M,N)\)
So … how do we show that M.1 to M.4 hold for the right action as defined above?
It may be important to recall Cohn's definition of a module homomorphism as a mapping \(\displaystyle f \ : \ M \to N\) such that

\(\displaystyle (x + y)f = xf + yf\) for \(\displaystyle x, y \in M\) … … (1)

and

\(\displaystyle (xf)r = (xr)f\) for \(\displaystyle r \in R, x \in M\) … … (2)

Note that (2) resembles the equation used in defining the right action …
Can someone please indicate whether my thoughts are on the right track, and also indicate how we verify that \(\displaystyle fr\) is an element of \(\displaystyle \text{Hom}_R(M,N)\) and further, show that M.1 to M.4 hold given our definition of the right action \(\displaystyle fr\)?

Hope someone can help.

Peter
***EDIT***1. Cohn writes in respect to the above:

"… … When \(\displaystyle R\) is commutative, Hom\(\displaystyle (M,N)\) also has an \(\displaystyle R\)-Module structure … … "

BUT … where exactly does the commutativity of R enter the analysis above … …?2. Perhaps some of my issues in the analysis above are answered by considering Deveno's statement in his most recent post in this thread:

" … … If we write the mapping on the left (as is customary in analysis), but want to define a RIGHT action, we have:

$(f\cdot r)(x) \stackrel{\text{def}}{=} (f(x))\cdot r$.

That is, if $f: x\mapsto y$, then $fr: x \mapsto yr$ … … "

Still reflecting on this … hoping someone can clarify … ...

 

[Math Amateur]

If we write the mapping on the left (as is customary in analysis), but want to define a RIGHT action, we have:

$(f\cdot r)(x) \stackrel{\text{def}}{=} (f(x))\cdot r$.

That is, if $f: x\mapsto y$, then $fr: x \mapsto yr$

This "looks weird", and is remedied in one of three ways:

1. Studying left actions instead (but sometimes we wind up with "anti-homomorphisms" because $fg$ is usually "first $g$, then $f$").

2. Writing $(x)f$ instead of $f(x)$.

3. Considering "opposite rings" , where the order of multiplication is reversed.

Option 3 is the best, in my opinion, as it let's us "choose handed-ness" as suits us.

Hi Deveno … thanks for your post … BUT, just a clarification …

You write:

" … … f we write the mapping on the left (as is customary in analysis), but want to define a RIGHT action, we have:

$(f\cdot r)(x) \stackrel{\text{def}}{=} (f(x))\cdot r$.

That is, if $f: x\mapsto y$, then $fr: x \mapsto yr$ … … "Now, I am uncertain of the context of your definition … are you in fact dealing with an action of \(\displaystyle R\) on \(\displaystyle \text{Hom}_R(M,N)\)?

In that case what is the nature of \(\displaystyle x\)? Does \(\displaystyle x \in M\)?

Hope you can help.

Peter

 

[Euge]

Thanks Euge … just working through this ...

You write:

" … … Cohn wants to define an $R$-action on $\text{Hom}_R(M,N)$ to make $\text{Hom}_R(M,N)$ an $R$-module. … "So to make $\text{Hom}_R(M,N)$ into an R-Module we define a right action of R on $\text{Hom}_R(M,N)$ as follows:

… … for \(\displaystyle r \in R\) and \(\displaystyle f \in \text{Hom}_R(M,N)\) we define a right action

\(\displaystyle \text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)\)

denoted by \(\displaystyle fr \in \text{Hom}_R(M,N)\)

where \(\displaystyle f \in \text{Hom}_R(M,N)\) and \(\displaystyle r \in R \)
Now we further define \(\displaystyle fr\) by (using \(\displaystyle g\) instead of Cohn's \(\displaystyle x\))

\(\displaystyle g(fr) = (gr)f = (gf)r\) where f\(\displaystyle ,g \in M, r \in R\)But how do we verify that \(\displaystyle fr\) is an element of \(\displaystyle \text{Hom}_R(M,N)\)?Further, how do we show that Cohn's M.1 to M.4 hold with the right action defined as above?Specifically we have to show that:

M.1 \(\displaystyle (f + g) r = fr + gr\) for \(\displaystyle r \in R, f,g \in\text{Hom}_R(M,N)\)

M.2 \(\displaystyle f(r + s) = fr + fs\) for \(\displaystyle r, s \in R, f \in\text{Hom}_R(M,N)\)

M.3 \(\displaystyle f(rs) = (fr)s\) for \(\displaystyle r, s \in R, f \in\text{Hom}_R(M,N)\)

M.4 \(\displaystyle f.1 = f\) for \(\displaystyle f \in\text{Hom}_R(M,N)\)
So … how do we show that M.1 to M.4 hold for the right action as defined above?
It may be important to recall Cohn's definition of a module homomorphism as a mapping \(\displaystyle f \ : \ M \to N\) such that

\(\displaystyle (x + y)f = xf + yf\) for \(\displaystyle x, y \in M\) … … (1)

and

\(\displaystyle (xf)r = (xr)f\) for \(\displaystyle r \in R, x \in M\) … … (2)

Note that (2) resembles the equation used in defining the right action …
Can someone please indicate whether my thoughts are on the right track, and also indicate how we verify that \(\displaystyle fr\) is an element of \(\displaystyle \text{Hom}_R(M,N)\) and further, show that M.1 to M.4 hold given our definition of the right action \(\displaystyle fr\)?

Hope someone can help.

Peter
***EDIT***1. Cohn writes in respect to the above:

"… … When \(\displaystyle R\) is commutative, Hom\(\displaystyle (M,N)\) also has an \(\displaystyle R\)-Module structure … … "

BUT … where exactly does the commutativity of R enter the analysis above … …?2. Perhaps some of my issues in the analysis above are answered by considering Deveno's statement in his most recent post in this thread:

" … … If we write the mapping on the left (as is customary in analysis), but want to define a RIGHT action, we have:

$(f\cdot r)(x) \stackrel{\text{def}}{=} (f(x))\cdot r$.

That is, if $f: x\mapsto y$, then $fr: x \mapsto yr$ … … "

Still reflecting on this … hoping someone can clarify … ...

Hi Peter,

Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $(xr)f$; thus $x(fr) := (xr)f$. Alternatively, $x(fr) := (xf)r$ since, by the homomorphism property of $f$, $(xr)f = (xf)r$. You need commutativity in $R$ to ensure that $fr$ is an element of $\text{Hom}_R(M,N)$. Given $x\in M$ and $r' \in R$,

\(\displaystyle (x + y)(fr) = [(x + y)r]f = (xr + yr)f = (xr)f + (yr)f = x(fr) + y(fr)\)

\(\displaystyle (xr')(fr) = [(xr')r]f = [x(r'r)]f = [x(rr')]f = [(xr)r']f = [(xr)f]r' = [x(fr)]r'\).

So, $fr\in \text{Hom}_R(M,N)$. Note that the commutativity in R was used in the third equality of the second line.

Properties M1 - M4 are proven straight from the definitions. For M1,

$\displaystyle x[(f + g)r] = (xr)(f + g) = (xr)f + (xr)g = x(fr) + x(gr) = x(fr + gr)$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. Therefore M1 holds. For M2,

$\displaystyle x[f(r + s)] = [x(r + s)]f = (xr + xs)f = (xr)f + (xs)f = x(fr) + x(fs) = x(fr + fs)$

for $x \in M$, $r, s\in R$, and $f\in \text{Hom}_R(M,N)$. Thus M2 holds. For M3,

$\displaystyle x[f(rs)] = [x(rs)]f = [(xr)s]f = [(xr)f]s = [x(fr)]s = x[(fr)s]$

for all $x\in M$, $r,s\in R$, and $f\in \text{Hom}_R(M,N)$. So M3 holds. Finally, $x(f1) = (x1)f = xf$ for all $x\in M$ and $f\in \text{Hom}_R(M,N)$; this means that M4 holds. Thus, we have shown that the multiplication $fr$ (for $f \in \text{Hom}_R(M,N)$ and $r\in R$) gives an $R$-module structure on $\text{Hom}_R(M,N)$.

 

[Deveno]

Yes, $f \in \text{Hom}_R(M,N)$ and $x \in M$.

In order to turn $\text{Hom}_R(M,N)$ into a right $R$-module, we need an addition.

To this end, we define $f+g$ to be the mapping that takes $x \in M$ to $xf + xg \in N$.

Of course, this alone doesn't guarantee $f+g \in \text{Hom}_R(M,N)$.

Let $x,y \in M$.

Then $(x+y)(f+g) = (x+y)f + (x+y)g$ (by definition)

$= xf + yf + xg + yg$ (because $f,g \in \text{Hom}_R(M,N)$

$= xf + xg +yf + yg$ (because addition in $N$ is commutative)

$= x(f+g) + y(f+g)$ (again, by definition).

Now let $r \in R$. We need to show that $(xr)(f+g) = (x(f+g))r$.

$(xr)(f+g) = (xr)f + (xr)g = (xf)r + (xg)r = (xf + xg)r = (x(f+g))r$.

This shows that this indeed defines an addition on $\text{Hom}_R(M,N)$.

It is straight-forward to show that this addition is associative, commutative, with additive identity the 0-map, and with additive inverses:

$x(-f) = -(xf)$, which turns $\text{Hom}_R(M,N)$ into an abelian group.

My previous post gives a right $R$-action (see below), which we then need to show obeys the module axioms. So, here we go:

For any $x \in M, r \in R$ and $f,g \in \text{Hom}_R(M,N)$, we have:

$x((f+g)r) = (x(f+g))r = (xf+xg)r = (xf)r + (xg)r = x(fr) + x(gr) = x(fr + gr)$

which establishes that $(f+g)r = fr + gr$.

Moreover, if $r' \in R$, then:

$x(fr + fr') = x(fr) + x(fr') = (xf)r + (xf)r' = (xf)(r+r') = x(f(r+r'))$, so that:

$fr + fr' = f(r+r')$.

If $R$ is unital, then:

$x(f1_R) = (xf)1_R = xf$, so that $f1_R = f$, and:

$x((fr)r') = (x(fr))r' = ((xf)r)r' = (xf)(rr') = x(f(rr'))$, so that $(fr)r' = f(rr')$.

As Euge points out, to establish that $fr$ is in $\text{Hom}_R(M,N)$ we need to have commutativity of the multiplication of $R$:

$(xr')(fr) = ((xr')f)r = ((xf)r')r = (xf)(r'r)$ and if $R$ is commutative:

$= (xf)(rr') = ((xf)r)r' = (x(fr))r'$.

 

[Math Amateur]

Hi Peter,

Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. Alternatively, $x(fr) := (xf)r$ since, by the homomorphism property of $f$, $(xr)f = (xf)r$. You need commutativity in $R$ to ensure that $fr$ is an element of $\text{Hom}_R(M,N)$. Given $x\in M$ and $r' \in R$,

\(\displaystyle (x + y)(fr) = [(x + y)r]f = (xr + yr)f = (xr)f + (yr)f = x(fr) + y(fr)\)

\(\displaystyle (xr')(fr) = [(xr')r]f = [x(r'r)]f = [x(rr')]f = [(xr)r']f = [(xr)f]r' = [x(fr)]r'\).

So, $fr\in \text{Hom}_R(M,N)$. Note that the commutativity in R was used in the third equality of the second line.

Properties M1 - M4 are proven straight from the definitions. For M1,

$\displaystyle x(f + g)r = (xr)(f + g) = (xr)f + (xr)g = (xf)r + (xg)r$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. Therefore M1 holds. For M2,

$\displaystyle x[f(r + s)] = [x(r + s)]f = (xr + xs)f = (xr)f + (xs)f = x(fr) + x(fs) = x(fr + fs)$

for $x \in M$, $r, s\in R$, and $f\in \text{Hom}_R(M,N)$. Thus M2 holds. For M3,

$\displaystyle x[f(rs)] = [x(rs)]f = [(xr)s]f = [(xr)f]s = [x(fr)]s = x[(fr)s]$

for all $x\in M$, $r,s\in R$, and $f\in \text{Hom}_R(M,N)$. So M3 holds. Finally, $x(f1) = (x1)f = xf$ for all $x\in M$ and $f\in \text{Hom}_R(M,N)$; this means that M4 holds. Thus, we have shown that the multiplication $fr$ (for $f \in \text{Hom}_R(M,N)$ and $r\in R$) gives an $R$-module structure on $\text{Hom}_R(M,N)$.

Thanks Euge … …

BUT … …

You write:

" … … Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. … … "

However in order to show that \(\displaystyle \text{Hom}_R(M,N)\) is an R-module, don't we need an action (function) \(\displaystyle fr \ : \ \text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)\) and not a function $fr : M \to N$?

Peter

 

[Euge]

Thanks Euge … …

BUT … …

You write:

" … … Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. … … "

However in order to show that \(\displaystyle \text{Hom}_R(M,N)\) is an R-module, don't we need an action (function) \(\displaystyle fr \ : \ \text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)\) and not a function $fr : M \to N$?

Peter

You're misunderstanding -- the action is $(f,r)\to fr$, not $fr$. Certainly, you can't show something is an action without having a multiplication. Part of the proof was showing that $fr$ is an element in $\text{Hom}_R(M,N)$ for every $f\in \text{Hom}_R(M,N)$ and $r\in R$. Then you can claim that the multiplication is well-defined. After that, you can verify M1 - M4, showing that the map $\text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)$ sending $(f,r)$ to $fr$ is an $R$-action on $\text{Hom}_R(M,N)$.

 

[Math Amateur]

You're misunderstanding -- the action is $(f,r)\to fr$, not $fr$. Certainly, you can't show something is an action without having a multiplication. Part of the proof was showing that $fr$ is an element in $\text{Hom}_R(M,N)$ for every $f\in \text{Hom}_R(M,N)$ and $r\in R$. Then you can claim that the multiplication is well-defined. After that, you can verify M1 - M4, showing that the map $\text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)$ sending $(f,r)$ to $fr$ is an $R$-action on $\text{Hom}_R(M,N)$.

Thanks for that clarification Euge ...

So to make sure I am understanding you, you write:

" … … You're misunderstanding -- the action is $(f,r)\to fr$, not $fr$. Certainly, you can't show something is an action without having a multiplication. … "

So if I am understanding you correctly, defining an action involves giving an explicit form to the nature of the multiplication … i.e. to define an action we need to define multiplication …

Am I interpreting you correctly?

Peter

 

[Euge]

Yes, that's right.

 

[Deveno]

You can view an $R$-action on a module $M$ "two ways":

1. As a function: $M \times R \to M$

2. As a family of functions (one for every $r \in R$): $M \to M$.

The first way gives us $(m,r) \mapsto mr$.

The second way gives us $m \to mr$.

With way #1, we often say we want the "scalar multiplication" to be "compatible" by the operation + in $M$. This means:

$(m_1 + m_2)r = m_1r + m_2r$, for the relevant entities.

With way #2, we can express this more succinctly: for each $r$, we want the map "multiply by $r$" to be an abelian group homomorphism.

In a similar vein, we want the scalar multiplication to be "compatible" with the sum of $R$ (way #1), which is to say we want the point-wise sum of the maps multiplication by $r_1$ and multiplication by $r_2$ to be the map multiplication by $r_1 + r_2$ (way #2). This is equivalent to saying we have an abelian group homomorphism:

$R \to \text{End}_{\Bbb Z}(M)$.

Note that way #1 focuses on ELEMENTS of $R$ and $M$, while way #2 focuses on MAPPINGS. The homomorphism properties of these mappings in effect "tell us what happens to elements" without a lot of tedious computation.

The $R$-module properties of $\text{Hom}_R(M,N)$ are, in a sense, "inherited" from $N$. To see why this is so, try to figure out which module properties get lost if $M$ is merely a set.

The endomorphism ring of an abelian group is, in a sense, a "self-referential structure". It is something similar to the "automorphism group of a set". Module actions are "the same type of thing" as "group actions": we use one structure (groups/rings) to "enrich" an object with "lesser structure" (sets/abelian groups) using a "canonical way" of using the "lesser structure" to get the "greater structure" (permutations of a set/endomorphisms of an abelian group).

This continues with linear algebra: we have fields, and they act on abelian groups in a "compatible way". The endomorphism ring gives rise to "matrices", which in turn ALSO have a vector space structure. Thus we get an ALGEBRA. This is "very fortunate", as it allows us to use various techniques from fields, vector spaces and ring theory ALL AT ONCE, giving us a diverse "tool-kit". Moreover, often the algebra itself has SPATIAL structure, perhaps even a NORM, which allows us to use analytic tools to "estimate" things.

All of these techniques can be brought to bear on studying the behavior of FUNCTIONS, particularly when these functions are "algebraic", in real $n$-dimensional space. We can look at ideals, or geometry, or bounds, or connectedness, and often have several different ways to "compute" something.

I mention this to bring into focus an important fact: things we want to study have STRUCTURE, and structures have FORM. After a while, statements like: "...and the rest follows by linearity" become second nature, the "gritty details" always have the same SHAPE.

 

[Math Amateur]

You can view an $R$-action on a module $M$ "two ways":

1. As a function: $M \times R \to M$

2. As a family of functions (one for every $r \in R$): $M \to M$.

The first way gives us $(m,r) \mapsto mr$.

The second way gives us $m \to mr$.

With way #1, we often say we want the "scalar multiplication" to be "compatible" by the operation + in $M$. This means:

$(m_1 + m_2)r = m_1r + m_2r$, for the relevant entities.

With way #2, we can express this more succinctly: for each $r$, we want the map "multiply by $r$" to be an abelian group homomorphism.

In a similar vein, we want the scalar multiplication to be "compatible" with the sum of $R$ (way #1), which is to say we want the point-wise sum of the maps multiplication by $r_1$ and multiplication by $r_2$ to be the map multiplication by $r_1 + r_2$ (way #2). This is equivalent to saying we have an abelian group homomorphism:
$R \to \text{End}_{\Bbb Z}(M)$.

Note that way #1 focuses on ELEMENTS of $R$ and $M$, while way #2 focuses on MAPPINGS. The homomorphism properties of these mappings in effect "tell us what happens to elements" without a lot of tedious computation.

The $R$-module properties of $\text{Hom}_R(M,N)$ are, in a sense, "inherited" from $N$. To see why this is so, try to figure out which module properties get lost if $M$ is merely a set.

The endomorphism ring of an abelian group is, in a sense, a "self-referential structure". It is something similar to the "automorphism group of a set". Module actions are "the same type of thing" as "group actions": we use one structure (groups/rings) to "enrich" an object with "lesser structure" (sets/abelian groups) using a "canonical way" of using the "lesser structure" to get the "greater structure" (permutations of a set/endomorphisms of an abelian group).

This continues with linear algebra: we have fields, and they act on abelian groups in a "compatible way". The endomorphism ring gives rise to "matrices", which in turn ALSO have a vector space structure. Thus we get an ALGEBRA. This is "very fortunate", as it allows us to use various techniques from fields, vector spaces and ring theory ALL AT ONCE, giving us a diverse "tool-kit". Moreover, often the algebra itself has SPATIAL structure, perhaps even a NORM, which allows us to use analytic tools to "estimate" things.

All of these techniques can be brought to bear on studying the behavior of FUNCTIONS, particularly when these functions are "algebraic", in real $n$-dimensional space. We can look at ideals, or geometry, or bounds, or connectedness, and often have several different ways to "compute" something.

I mention this to bring into focus an important fact: things we want to study have STRUCTURE, and structures have FORM. After a while, statements like: "...and the rest follows by linearity" become second nature, the "gritty details" always have the same SHAPE.

Thanks Deveno … this post is really helpful and informative in ways that the various textbooks largely omit … …

Peter

 

[Math Amateur]

Hi Peter,

Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. Alternatively, $x(fr) := (xf)r$ since, by the homomorphism property of $f$, $(xr)f = (xf)r$. You need commutativity in $R$ to ensure that $fr$ is an element of $\text{Hom}_R(M,N)$. Given $x\in M$ and $r' \in R$,

\(\displaystyle (x + y)(fr) = [(x + y)r]f = (xr + yr)f = (xr)f + (yr)f = x(fr) + y(fr)\)

\(\displaystyle (xr')(fr) = [(xr')r]f = [x(r'r)]f = [x(rr')]f = [(xr)r']f = [(xr)f]r' = [x(fr)]r'\).

So, $fr\in \text{Hom}_R(M,N)$. Note that the commutativity in R was used in the third equality of the second line.

Properties M1 - M4 are proven straight from the definitions. For M1,

$\displaystyle x(f + g)r = (xr)(f + g) = (xr)f + (xr)g = (xf)r + (xg)r$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. Therefore M1 holds. For M2,

$\displaystyle x[f(r + s)] = [x(r + s)]f = (xr + xs)f = (xr)f + (xs)f = x(fr) + x(fs) = x(fr + fs)$

for $x \in M$, $r, s\in R$, and $f\in \text{Hom}_R(M,N)$. Thus M2 holds. For M3,

$\displaystyle x[f(rs)] = [x(rs)]f = [(xr)s]f = [(xr)f]s = [x(fr)]s = x[(fr)s]$

for all $x\in M$, $r,s\in R$, and $f\in \text{Hom}_R(M,N)$. So M3 holds. Finally, $x(f1) = (x1)f = xf$ for all $x\in M$ and $f\in \text{Hom}_R(M,N)$; this means that M4 holds. Thus, we have shown that the multiplication $fr$ (for $f \in \text{Hom}_R(M,N)$ and $r\in R$) gives an $R$-module structure on $\text{Hom}_R(M,N)$.

Hi Euge,

I am still reflecting on the helpful series of posts sent by you and Deveno.

In the post which begins as follows:

"Hi Peter,

Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. Alternatively, $x(fr) := (xf)r$ since, by the homomorphism property of $f$, $(xr)f = (xf)r$. You need commutativity in $R$ to ensure that $fr$ is an element of $\text{Hom}_R(M,N)$. Given $x\in M$ and $r' \in R$, … … … etc etc … "

You write:

"Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. … … "

If \(\displaystyle fr\) is the function sending \(\displaystyle x \to xr\) then \(\displaystyle xr\) is the image of \(\displaystyle fr\) acting on \(\displaystyle r\) - that is, it is the image of \(\displaystyle x\) under \(\displaystyle fr\) - so shouldn't we write \(\displaystyle x.fr = xr\)

Can you explain why you write \(\displaystyle x(fr) = (xr)f\)?

Surely it is \(\displaystyle x(fr) = xr\)?

Can you please clarify?

Peter

 

[Math Amateur]

Hi Peter,

Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. Alternatively, $x(fr) := (xf)r$ since, by the homomorphism property of $f$, $(xr)f = (xf)r$. You need commutativity in $R$ to ensure that $fr$ is an element of $\text{Hom}_R(M,N)$. Given $x\in M$ and $r' \in R$,

\(\displaystyle (x + y)(fr) = [(x + y)r]f = (xr + yr)f = (xr)f + (yr)f = x(fr) + y(fr)\)

\(\displaystyle (xr')(fr) = [(xr')r]f = [x(r'r)]f = [x(rr')]f = [(xr)r']f = [(xr)f]r' = [x(fr)]r'\).

So, $fr\in \text{Hom}_R(M,N)$. Note that the commutativity in R was used in the third equality of the second line.

Properties M1 - M4 are proven straight from the definitions. For M1,

$\displaystyle x(f + g)r = (xr)(f + g) = (xr)f + (xr)g = (xf)r + (xg)r$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. Therefore M1 holds. For M2,

$\displaystyle x[f(r + s)] = [x(r + s)]f = (xr + xs)f = (xr)f + (xs)f = x(fr) + x(fs) = x(fr + fs)$

for $x \in M$, $r, s\in R$, and $f\in \text{Hom}_R(M,N)$. Thus M2 holds. For M3,

$\displaystyle x[f(rs)] = [x(rs)]f = [(xr)s]f = [(xr)f]s = [x(fr)]s = x[(fr)s]$

for all $x\in M$, $r,s\in R$, and $f\in \text{Hom}_R(M,N)$. So M3 holds. Finally, $x(f1) = (x1)f = xf$ for all $x\in M$ and $f\in \text{Hom}_R(M,N)$; this means that M4 holds. Thus, we have shown that the multiplication $fr$ (for $f \in \text{Hom}_R(M,N)$ and $r\in R$) gives an $R$-module structure on $\text{Hom}_R(M,N)$.

Hi Euge,

In order to fully understand the structure of \(\displaystyle \text{Hom}_R(M,N)\), and also what you and Deveno are saying, I have gone back to the definition of a right module over \(\displaystyle R\) in both Dummit and Foote (page 337) and Cohn (page 31).

Following these authors, we define a right module as follows:

Let \(\displaystyle R\) be a ring.

A right \(\displaystyle R\)-module or a right module over R is a set M together with

(1) a binary operation + on \(\displaystyle M\) under which \(\displaystyle M\) is an abelian group

(2) an action of \(\displaystyle R\) on \(\displaystyle M\) (that is a map \(\displaystyle M \times R \to M\) )
denoted by \(\displaystyle mr\) for all \(\displaystyle r \in R\) and for all \(\displaystyle m \in M\) (Note that \(\displaystyle mr \in M\)) which satisfies:

M.1 \(\displaystyle \ \ (x + y)r = xr + yr\) for all \(\displaystyle r \in R, x,y \in M\)

M.2 \(\displaystyle \ \ x(r + s) = xr + xs\) for all \(\displaystyle r,s \in R, x, \in M\)

M.3 \(\displaystyle \ \ x(rs) = (xr)s \) for all \(\displaystyle r,s \in R, x, \in M\)

M.4 \(\displaystyle \ \ x.1 = x\) for all \(\displaystyle x, \in M\)
So … … for a situation where we are trying to construct a right \(\displaystyle R\)-module structure on \(\displaystyle \text{Hom}_R(M,N)\) we proceed as follows:

Firstly … we have that R is a ring.

We also have (or, better, need) a set of homomorphisms \(\displaystyle \text{Hom}_R(M,N)\) together with

(1) a binary operation + on \(\displaystyle \text{Hom}_R(M,N)\) under which \(\displaystyle \text{Hom}_R(M,N)\) is an abelian group

(2) an action of \(\displaystyle R\) on \(\displaystyle \text{Hom}_R(M,N)\) (that is a map \(\displaystyle \text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)\) )
denoted by \(\displaystyle fr\) for all \(\displaystyle r \in R\) and for all \(\displaystyle f \in \text{Hom}_R(M,N) \) (Note that we require \(\displaystyle fr \in \text{Hom}_R(M,N)\) ) which satisfies:

M.1 \(\displaystyle \ \ (f + g)r = fr + gr \) for all \(\displaystyle r \in R, f,g \in \text{Hom}_R(M,N)\)

M.2 \(\displaystyle \ \ f(r + s) = fr + fs\) for all \(\displaystyle r,s \in R, f \in \text{Hom}_R(M,N)\)

M.3 \(\displaystyle \ \ f(rs) = (fr)s\) for all \(\displaystyle r,s \in R, f \in \text{Hom}_R(M,N)\)

M.4 \(\displaystyle \ \ f.1 = f\) for all \(\displaystyle f \in \text{Hom}_R(M,N)\)

Now we proceed to define the effect of the action fr as follows:

Following your post we have

Define a function \(\displaystyle fr \ : \ M \to N\) by \(\displaystyle x \mapsto xr\)

BUT … …

in your post above where you prove property M.1 for the situation of \(\displaystyle \text{Hom}_R(M,N)\) as a right \(\displaystyle R\)-module, you write:

" … … For M1,

$\displaystyle x(f + g)r = (xr)(f + g) = (xr)f + (xr)g = (xf)r + (xg)r$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. … … "

But as far as I can see the relevant property is as follows:

M.1 \(\displaystyle (f + g)r = fr + gr\) for all \(\displaystyle r \in R, f,g \in \text{Hom}_R(M,N)\)

Can you please clarify?

Peter***EDIT***

I have just been studying and reflecting on one of Deveno's posts following the above one by you and am now thinking that you are showing that x mapped under (f+g)r is the same as the sum of x mapped under fr and x mapped under gr - is that correct?

 

[Euge]

Hi Euge,

I am still reflecting on the helpful series of posts sent by you and Deveno.

In the post which begins as follows:

"Hi Peter,

Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. Alternatively, $x(fr) := (xf)r$ since, by the homomorphism property of $f$, $(xr)f = (xf)r$. You need commutativity in $R$ to ensure that $fr$ is an element of $\text{Hom}_R(M,N)$. Given $x\in M$ and $r' \in R$, … … … etc etc … "

You write:

"Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $xr$; thus $x(fr) := (xr)f$. … … "

If \(\displaystyle fr\) is the function sending \(\displaystyle x \to xr\) then \(\displaystyle xr\) is the image of \(\displaystyle fr\) acting on \(\displaystyle r\) - that is, it is the image of \(\displaystyle x\) under \(\displaystyle fr\) - so shouldn't we write \(\displaystyle x.fr = xr\)

Can you explain why you write \(\displaystyle x(fr) = (xr)f\)?

Surely it is \(\displaystyle x(fr) = xr\)?

Can you please clarify?

Peter

It's a typo, I was missing an $f$. It should be that $fr$ sends $x$ to $(xr)f$, not $xr$. I made the correction for two of my posts.

 

[Euge]

Hi Euge,

In order to fully understand the structure of \(\displaystyle \text{Hom}_R(M,N)\), and also what you and Deveno are saying, I have gone back to the definition of a right module over \(\displaystyle R\) in both Dummit and Foote (page 337) and Cohn (page 31).

Following these authors, we define a right module as follows:

Let \(\displaystyle R\) be a ring.

A right \(\displaystyle R\)-module or a right module over R is a set M together with

(1) a binary operation + on \(\displaystyle M\) under which \(\displaystyle M\) is an abelian group

(2) an action of \(\displaystyle R\) on \(\displaystyle M\) (that is a map \(\displaystyle M \times R \to M\) )
denoted by \(\displaystyle mr\) for all \(\displaystyle r \in R\) and for all \(\displaystyle m \in M\) (Note that \(\displaystyle mr \in M\)) which satisfies:

M.1 \(\displaystyle \ \ (x + y)r = xr + yr\) for all \(\displaystyle r \in R, x,y \in M\)

M.2 \(\displaystyle \ \ x(r + s) = xr + xs\) for all \(\displaystyle r,s \in R, x, \in M\)

M.3 \(\displaystyle \ \ x(rs) = (xr)s \) for all \(\displaystyle r,s \in R, x, \in M\)

M.4 \(\displaystyle \ \ x.1 = x\) for all \(\displaystyle x, \in M\)
So … … for a situation where we are trying to construct a right \(\displaystyle R\)-module structure on \(\displaystyle \text{Hom}_R(M,N)\) we proceed as follows:

Firstly … we have that R is a ring.

We also have (or, better, need) a set of homomorphisms \(\displaystyle \text{Hom}_R(M,N)\) together with

(1) a binary operation + on \(\displaystyle \text{Hom}_R(M,N)\) under which \(\displaystyle \text{Hom}_R(M,N)\) is an abelian group

(2) an action of \(\displaystyle R\) on \(\displaystyle \text{Hom}_R(M,N)\) (that is a map \(\displaystyle \text{Hom}_R(M,N) \times R \to \text{Hom}_R(M,N)\) )
denoted by \(\displaystyle fr\) for all \(\displaystyle r \in R\) and for all \(\displaystyle f \in \text{Hom}_R(M,N) \) (Note that we require \(\displaystyle fr \in \text{Hom}_R(M,N)\) ) which satisfies:

M.1 \(\displaystyle \ \ (f + g)r = fr + gr \) for all \(\displaystyle r \in R, f,g \in \text{Hom}_R(M,N)\)

M.2 \(\displaystyle \ \ f(r + s) = fr + fs\) for all \(\displaystyle r,s \in R, f \in \text{Hom}_R(M,N)\)

M.3 \(\displaystyle \ \ f(rs) = (fr)s\) for all \(\displaystyle r,s \in R, f \in \text{Hom}_R(M,N)\)

M.4 \(\displaystyle \ \ f.1 = f\) for all \(\displaystyle f \in \text{Hom}_R(M,N)\)

Now we proceed to define the effect of the action fr as follows:

Following your post we have

Define a function \(\displaystyle fr \ : \ M \to N\) by \(\displaystyle x \mapsto xr\)

BUT … …

in your post above where you prove property M.1 for the situation of \(\displaystyle \text{Hom}_R(M,N)\) as a right \(\displaystyle R\)-module, you write:

" … … For M1,

$\displaystyle x(f + g)r = (xr)(f + g) = (xr)f + (xr)g = (xf)r + (xg)r$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. … … "

But as far as I can see the relevant property is as follows:

M.1 \(\displaystyle (f + g)r = fr + gr\) for all \(\displaystyle r \in R, f,g \in \text{Hom}_R(M,N)\)

Can you please clarify?

Peter***EDIT***

I have just been studying and reflecting on one of Deveno's posts following the above one by you and am now thinking that you are showing that x mapped under (f+g)r is the same as the sum of x mapped under fr and x mapped under gr - is that correct?

I've changed the ending of those series of equations just slightly so that it will be clearer. The line now reads

\(\displaystyle x[(f + g)r] = (xr)(f + g) = (xr)f + (xr)g = x(fr) + x(gr) = x(fr + gr)\)

This way, you can see the relevance to M1. Since the above equations hold true for $x\in M$, $f,g\in \text{Hom}_R(M,N)$, and $r\in R$, we have $(f + g)r = fr + gr$ for all $f, g \in \text{Hom}_R(M,N)$ and $r\in R$ showing M1 holds.

To show two functions, say, $F$ and $G$ (with the same domain) are equal, you must show that $xF = xG$ for all $x$ in the domain. That's why, for example, when dealing with M1, I verified that $x[(f + g)r] = x(fr + gr)$ for all $x$, $f$, $g$, and $r$ before claiming that $(f + g)r = fr + gr$ for all $f$, $g$, and $r$.

 

[Deveno]

Given a function:

$f:A \to B$

and another function $g: A \to B$, when is $f = g$?

Recall that $f$ is actually a subset of $A \times B$ such that:

$(a_1,b_1),(a_1,b_2) \in f \implies a_1 = a_2$, that is, for every $a \in A$ we get a unique $b \in B$ with $(a,b) \in f$.

This is NOT to say that if $b \in \text{im }f = f(A)$ that we have a unique $a$ (this only happens if $f$ is injective).

The unique $b$ is called the IMAGE of $a$ under $f$, and usually denoted $f(a)$ (except for Mr. Cohn, who would write $af$, I believe Herstein follows the same convention).

So to decide if $f$ and $g$ are the same subsets of $A \times B$, we need to compare:

$(a,f(a))$ and $(a,g(a))$ for every $a \in A$. If the second coordinates are equal, for every $a \in A$, then the first ones are too, since for every $a \in A$, we have $a = a$.

(recall that in a cartesian product set: $(x_1,y_1) = (x_2,y_2) \iff x_1 = x_2 \text{ and } y_1 = y_2$).

So for two functions with the same domain, it suffices to compare images of those functions at every point of the domain, to show equality of the two functions.

 

[Math Amateur]

Given a function:

$f:A \to B$

and another function $g: A \to B$, when is $f = g$?

Recall that $f$ is actually a subset of $A \times B$ such that:

$(a_1,b_1),(a_1,b_2) \in f \implies a_1 = a_2$, that is, for every $a \in A$ we get a unique $b \in B$ with $(a,b) \in f$.

This is NOT to say that if $b \in \text{im }f = f(A)$ that we have a unique $a$ (this only happens if $f$ is injective).

The unique $b$ is called the IMAGE of $a$ under $f$, and usually denoted $f(a)$ (except for Mr. Cohn, who would write $af$, I believe Herstein follows the same convention).

So to decide if $f$ and $g$ are the same subsets of $A \times B$, we need to compare:

$(a,f(a))$ and $(a,g(a))$ for every $a \in A$. If the second coordinates are equal, for every $a \in A$, then the first ones are too, since for every $a \in A$, we have $a = a$.

(recall that in a cartesian product set: $(x_1,y_1) = (x_2,y_2) \iff x_1 = x_2 \text{ and } y_1 = y_2$).

So for two functions with the same domain, it suffices to compare images of those functions at every point of the domain, to show equality of the two functions.

Thanks Deveno and Euge … will work through the posts in the morning (Melbourne time i.e. Eastern Australian time - same as Hobart time)

So grateful for the help!

Peter

 

[Math Amateur]

Hi Peter,

Recall that, given $f\in \text{Hom}_R(M,N)$ and $r\in R$, we define a function $fr : M \to N$ sending $x$ to $(xr)f$; thus $x(fr) := (xr)f$. Alternatively, $x(fr) := (xf)r$ since, by the homomorphism property of $f$, $(xr)f = (xf)r$. You need commutativity in $R$ to ensure that $fr$ is an element of $\text{Hom}_R(M,N)$. Given $x\in M$ and $r' \in R$,

\(\displaystyle (x + y)(fr) = [(x + y)r]f = (xr + yr)f = (xr)f + (yr)f = x(fr) + y(fr)\)

\(\displaystyle (xr')(fr) = [(xr')r]f = [x(r'r)]f = [x(rr')]f = [(xr)r']f = [(xr)f]r' = [x(fr)]r'\).

So, $fr\in \text{Hom}_R(M,N)$. Note that the commutativity in R was used in the third equality of the second line.

Properties M1 - M4 are proven straight from the definitions. For M1,

$\displaystyle x[(f + g)r] = (xr)(f + g) = (xr)f + (xr)g = x(fr) + x(gr) = x(fr + gr)$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. Therefore M1 holds. For M2,

$\displaystyle x[f(r + s)] = [x(r + s)]f = (xr + xs)f = (xr)f + (xs)f = x(fr) + x(fs) = x(fr + fs)$

for $x \in M$, $r, s\in R$, and $f\in \text{Hom}_R(M,N)$. Thus M2 holds. For M3,

$\displaystyle x[f(rs)] = [x(rs)]f = [(xr)s]f = [(xr)f]s = [x(fr)]s = x[(fr)s]$

for all $x\in M$, $r,s\in R$, and $f\in \text{Hom}_R(M,N)$. So M3 holds. Finally, $x(f1) = (x1)f = xf$ for all $x\in M$ and $f\in \text{Hom}_R(M,N)$; this means that M4 holds. Thus, we have shown that the multiplication $fr$ (for $f \in \text{Hom}_R(M,N)$ and $r\in R$) gives an $R$-module structure on $\text{Hom}_R(M,N)$.

Thanks to Euge and Deveno, I think I now understand the statement by Cohn:

" … … When \(\displaystyle R\) is commutative, \(\displaystyle \text{Hom}_R(M,N\) has an \(\displaystyle R\)-module structure given by \(\displaystyle x(fr) = (xr)f = (xf)\) … … "

In the above post, Euge writes:

" … … For M1,

$\displaystyle x[(f + g)r] = (xr)(f + g) = (xr)f + (xr)g = x(fr) + x(gr) = x(fr + gr)$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. Therefore M1 holds. … … "Can someone please confirm that my explicit reasons/justifications for each of the steps is correct?

\(\displaystyle x[(f + g)r] = (xr)(f + g)\) … … … definition of fr

\(\displaystyle (xr)(f + g) = (xr)f + (xr)g\) … … … definition of the sum of two functions

\(\displaystyle (xr)f + (xr)g = x(fr) + x(gr)\) … … … because f and g are module homomorphisms

\(\displaystyle x(fr) + x(gr) = x(fr + gr) \) … … … slightly uncertain, but think that this follows by definition of the sum of two functions

Again, can someone please confirm that my explicit reasons/justifications for each of the steps is correct?

Now for M.2, Euge writes:

" … … For M2,

$\displaystyle x[f(r + s)] = [x(r + s)]f = (xr + xs)f = (xr)f + (xs)f = x(fr) + x(fs) = x(fr + fs)$

for $x \in M$, $r, s\in R$, and $f\in \text{Hom}_R(M,N)$. Thus M2 holds. … …"Can someone please confirm that my explicit reasons/justifications for each of the steps is correct?x[f(r + s)] = [x(r + s)]f … … … Definition of fr

[x(r + s)]f = (xr + xs)f … … … Because M is an R-module

(xr + xs)f = (xr)f + (xs)f … … … (?? need help) Definition of a function … … but unsure

(xr)f + (xs)f = x(fr) + x(fs) … … … Definition of fr

x(fr) + x(fs) = x(fr + fs) … … … uncertain … but I think it is the definition of the sum of two functionsAgain, can someone please confirm that my explicit reasons/justifications for each of the steps is correct?
Peter

[Thanks again to Euge and Deveno for extensive and generous help on this topic!]

 

[Deveno]

Thanks to Euge and Deveno, I think I now understand the statement by Cohn:

" … … When \(\displaystyle R\) is commutative, \(\displaystyle \text{Hom}_R(M,N\) has an \(\displaystyle R\)-module structure given by \(\displaystyle x(fr) = (xr)f = (xf)\) … … "

In the above post, Euge writes:

" … … For M1,

$\displaystyle x[(f + g)r] = (xr)(f + g) = (xr)f + (xr)g = x(fr) + x(gr) = x(fr + gr)$

for all $x\in M$, $r\in R$, and $f,g\in \text{Hom}_R(M,N)$. Therefore M1 holds. … … "Can someone please confirm that my explicit reasons/justifications for each of the steps is correct?

\(\displaystyle x[(f + g)r] = (xr)(f + g)\) … … … definition of fr

Personally, I would use the definition $x(fr) = (xf)r$ rather than $x(fr) = (xr)f$. The second definition works, though, because $f$ is $R$-linear.

\(\displaystyle (xr)(f + g) = (xr)f + (xr)g\) … … … definition of the sum of two functions

\(\displaystyle (xr)f + (xr)g = x(fr) + x(gr)\) … … … because f and g are module homomorphisms

There's (small) step missing here-$f,g$ being $R$-module homomorphisms tells us $(xr)f = (xf)r$, and thus (by your initial definition of $fr$) is equal to $x(fr)$, and similarly for $g$.

\(\displaystyle x(fr) + x(gr) = x(fr + gr) \) … … … slightly uncertain, but think that this follows by definition of the sum of two functions

Again, can someone please confirm that my explicit reasons/justifications for each of the steps is correct?

Yes that is how the addition in $\text{Hom}_R(M,N)$ is defined.

Now for M.2, Euge writes:

" … … For M2,

$\displaystyle x[f(r + s)] = [x(r + s)]f = (xr + xs)f = (xr)f + (xs)f = x(fr) + x(fs) = x(fr + fs)$

for $x \in M$, $r, s\in R$, and $f\in \text{Hom}_R(M,N)$. Thus M2 holds. … …"Can someone please confirm that my explicit reasons/justifications for each of the steps is correct?x[f(r + s)] = [x(r + s)]f … … … Definition of fr

[x(r + s)]f = (xr + xs)f … … … Because M is an R-module

(xr + xs)f = (xr)f + (xs)f … … … (?? need help) Definition of a function … … but unsure

This is because $f$ is an $R$-module homomorphism, and thus preserves sums.

(xr)f + (xs)f = x(fr) + x(fs) … … … Definition of fr

x(fr) + x(fs) = x(fr + fs) … … … uncertain … but I think it is the definition of the sum of two functionsAgain, can someone please confirm that my explicit reasons/justifications for each of the steps is correct?Peter

[Thanks again to Euge and Deveno for extensive and generous help on this topic!]

Let's look at a particular $\Bbb Z$-module of this nature. Let $M = N = \Bbb Z_6$ (the integers mod 6).

Now if $f \in \text{Hom}_{\Bbb Z}(\Bbb Z_6,\Bbb Z_6)$, then we have:

$(k)f = (1 + 1 +\cdots + 1)f = (1)f + (1)f + \cdots + (1)f = ((1)f)k$.

So any such $f$ is completely determined by its action on 1.

Suppose $(1)f= a$. I claim then, that $f$ is the mapping $k \mapsto ak$. You should verify this yourself. For example, if $a = 3$, then f does this:

$(1)f = 3$
$(2)f = 0$
$(3)f = 3$
$(4)f = 0$
$(5)f = 3$
$(0)f = 0$.

Suppose $g(1) = 5$. I claim (and you should also verify this) that $g$ is the inversion homomorphism (this is only a homomorphism for abelian groups): $k \mapsto -k$. So $f+g$ is the mapping that does this:

$(1)(f+g) = (1)f + (1)g = 3 + 5 = 2$
$(2)(f+g) = (2)f + (2)g = 0 + 4 = 4$
$(3)(f+g) = (3)f + (3)g = 3 + 3 = 0$
$(4)(f+g) = (4)f + (4)g = 0 + 2 = 2$
$(5)(f+g) = (5)f + (5)g = 3 + 1 = 4$
$(0)(f+g) = (0)f + (0)g = 0 + 0 = 0$.

If you work with abelian groups, vector spaces, and rings for any length of time at all, you get the feeling that all these structures have "something in common". The idea of an $R$-module, is the umbrella we put all this similarity under:

An abelian group is a module over the ring $\Bbb Z$.
A vector space is a module over a ring that is a field (fields are commutative rings, so modules over commutative rings are "more like vector spaces" than modules over non-commutative rings).
A ring is a module over itself (equivalently, the free $R$-module over a single element set, which we identify with $1_R$).

So anything we prove for modules has immediate application to all 3 of these "other" structures. For example, the result on composition series we discussed in another thread becomes the Structure Theorem for finitely-generated abelian groups.

There is a good chance that any time you are dealing with things you can "add somehow" (usually indicated with the much-overloaded + sign), what you actually have is a module of some sort. They are the proto-typical "abelian thingies".

 

[Math Amateur]

Personally, I would use the definition $x(fr) = (xf)r$ rather than $x(fr) = (xr)f$. The second definition works, though, because $f$ is $R$-linear.
There's (small) step missing here-$f,g$ being $R$-module homomorphisms tells us $(xr)f = (xf)r$, and thus (by your initial definition of $fr$) is equal to $x(fr)$, and similarly for $g$.
Yes that is how the addition in $\text{Hom}_R(M,N)$ is defined.
This is because $f$ is an $R$-module homomorphism, and thus preserves sums.
Let's look at a particular $\Bbb Z$-module of this nature. Let $M = N = \Bbb Z_6$ (the integers mod 6).

Now if $f \in \text{Hom}_{\Bbb Z}(\Bbb Z_6,\Bbb Z_6)$, then we have:

$(k)f = (1 + 1 +\cdots + 1)f = (1)f + (1)f + \cdots + (1)f = ((1)f)k$.

So any such $f$ is completely determined by its action on 1.

Suppose $(1)f= a$. I claim then, that $f$ is the mapping $k \mapsto ak$. You should verify this yourself. For example, if $a = 3$, then f does this:

$(1)f = 3$
$(2)f = 0$
$(3)f = 3$
$(4)f = 0$
$(5)f = 3$
$(0)f = 0$.

Suppose $g(1) = 5$. I claim (and you should also verify this) that $g$ is the inversion homomorphism (this is only a homomorphism for abelian groups): $k \mapsto -k$. So $f+g$ is the mapping that does this:

$(1)(f+g) = (1)f + (1)g = 3 + 5 = 2$
$(2)(f+g) = (2)f + (2)g = 0 + 4 = 4$
$(3)(f+g) = (3)f + (3)g = 3 + 3 = 0$
$(4)(f+g) = (4)f + (4)g = 0 + 2 = 2$
$(5)(f+g) = (5)f + (5)g = 3 + 1 = 4$
$(0)(f+g) = (0)f + (0)g = 0 + 0 = 0$.

If you work with abelian groups, vector spaces, and rings for any length of time at all, you get the feeling that all these structures have "something in common". The idea of an $R$-module, is the umbrella we put all this similarity under:

An abelian group is a module over the ring $\Bbb Z$.
A vector space is a module over a ring that is a field (fields are commutative rings, so modules over commutative rings are "more like vector spaces" than modules over non-commutative rings).
A ring is a module over itself (equivalently, the free $R$-module over a single element set, which we identify with $1_R$).

So anything we prove for modules has immediate application to all 3 of these "other" structures. For example, the result on composition series we discussed in another thread becomes the Structure Theorem for finitely-generated abelian groups.

There is a good chance that any time you are dealing with things you can "add somehow" (usually indicated with the much-overloaded + sign), what you actually have is a module of some sort. They are the proto-typical "abelian thingies".

Thanks for the help, and in particular, the example … … just reflecting on what you have written …

Peter