First Isomorphism Theorem for Modules - Cohn Theorem 1.17

[Math Amateur]

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

In Chapter 1: Basics we find Theorem 1.17 (First Isomorphism Theorem for Modules) regarding module homomorphisms and quotient modules. I need help with some aspects of the proof.

Theorem 1.17 reads as follows:

View attachment 3260

Now since the proof of Theorem 1.17 refers to the Factor Theorem (Corollary 1.16) I will provide the text of that result, as follows:

View attachment 3261

Now the first line of the proof of Theorem 1.17 (First Isomorphism Theorem for Modules) reads as follows:

"By the factor theorem (Corollary 1.16) we have \(\displaystyle f = \nu g\) for some homomorphism \(\displaystyle g: M/ker f \to N\), and since the image of \(\displaystyle g\) is clearly I am \(\displaystyle f\), we can write \(\displaystyle g = f' \lambda\) for some \(\displaystyle f'\) ... ... ... "

Can someone please explain to me why the image of \(\displaystyle g\) is I am \(\displaystyle f\) and further, why then we can write \(\displaystyle g = f' \lambda\) for some \(\displaystyle f'\)?

Peter

 

[Math Amateur]

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

In Chapter 1: Basics we find Theorem 1.17 (First Isomorphism Theorem for Modules) regarding module homomorphisms and quotient modules. I need help with some aspects of the proof.

Theorem 1.17 reads as follows:

View attachment 3260

Now since the proof of Theorem 1.17 refers to the Factor Theorem (Corollary 1.16) I will provide the text of that result, as follows:

View attachment 3261

Now the first line of the proof of Theorem 1.17 (First Isomorphism Theorem for Modules) reads as follows:

"By the factor theorem (Corollary 1.16) we have \(\displaystyle f = \nu g\) for some homomorphism \(\displaystyle g: M/ker f \to N\), and since the image of \(\displaystyle g\) is clearly I am \(\displaystyle f\), we can write \(\displaystyle g = f' \lambda\) for some \(\displaystyle f'\) ... ... ... "

Can someone please explain to me why the image of \(\displaystyle g\) is I am \(\displaystyle f\) and further, why then we can write \(\displaystyle g = f' \lambda\) for some \(\displaystyle f'\)?

Peter

Just reflecting on my own question:

"... ... why the image of \(\displaystyle g\) is I am \(\displaystyle f\) ... ... "

Presumably this is because \(\displaystyle f = \nu g\) - for this equation to hold, (I think) it must be the case that I am \(\displaystyle g\) = I am \(\displaystyle f\) ... ... hmmm ... but why (exactly) then can we write: \(\displaystyle g = f' \lambda\) for some \(\displaystyle f'\)? ... ...

Can someone confirm that my thinking and analysis is correct ... as far as it goes ...

Peter***EDIT***

Just a further note to say that I am somewhat puzzled as to why Cohn bothers to introduce the inclusion mapping \(\displaystyle \lambda\) into the proof ... why is it necessary ... what is gained ...

Further, although it seems extremely plausible, why exactly can we claim that \(\displaystyle f'\) is a homomorphism - I suppose that since \(\displaystyle \lambda\) essentially 'does nothing' and \(\displaystyle g\) is a homomorphism, then informally we can argue that since \(\displaystyle g = f' \lambda\) then \(\displaystyle f\) is also, like \(\displaystyle g\), a homomorphism ... ...

Hope someone can clarify these further issues ... ...

 

[Deveno]

Let's look at Corollary 1.16, first.

Basically it says any homomorphism $f: M \to N$ (yes, ANY homomorphism) factors through the quotient morphism induced by any submodule $M'$ contained in the kernel of $f$.

One way to see this clearer, is to note that the kernel of $f$ induces a partition on $M$, into cosets of the kernel.

Now if $M' \subseteq \text{ker }f$, I claim that the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$. In other words, every coset just gets "chopped up" into smaller cosets, we do not have any elements from DIFFERENT cosets of $\text{ker }f$ winding up in the SAME coset of $M'$.

In other words, I claim that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$. This is immediate from the fact that: $M' \subseteq \text{ker }f$.

IN other words, if a mapping respects the partition induced by $\text{ker }f$ (that is to say, it is constant on any coset of $\text{ker }f$) then it CERTAINLY also respects the partition induced by $M'$ (in fact it is constant on "clusters of cosets", not just individual cosets of $M'$).

This means we get a well-defined mapping: $f':M/M' \to N$ defined by $(x + M')f = xf$. If $\nu: M \to M/M'$ is the quotient homomorphism:

$x\nu = x + M'$

It is evident that $x(\nu f') = (x\nu)f' = (x + M')f = xf$ for any $x \in M$, that is: $f = \nu f'$.

(I am attempting to keep my mappings on the right, which feels odd to me).

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

Now, we can consider the special case $M' = \text{ker }f$.

This gives us $f = \nu f'$ where $f': M/\text{ker }f \to N$. The trouble is, here, that $N$ might be "much larger" than $M$ (in other words, $f$ is not surjective).

If $f$ is not surjective, then $f'$ cannot be, either, since $f'$ only takes on values that $f$ does (because of how its defined).

Lets' prove $\text{im }f = \text{im }f'$.

Suppose $n \in \text{im }f$. Then $n = xf$ for some $x \in M$, whence $n = (x + \text{ker }f)f'$, so that $n \in \text{im }f'$. Thus $\text{im }f \subseteq \text{im }f'$.

On the other hand, suppose $n \in \text{im }f'$. This means that $n = (x + \text{ker }f)f'$ for some coset $x +\text{ker }f \in M/\text{ker }f$.

Pick $y \in x + \text{ker }f$. Then $yf = y(\nu f') = (y + \text{ker }f)f' = (x + \text{ker }f)f'$ (since $y\nu = x\nu$, because $y \in x + \text{ker }f$)

$=n$, which shows that $n \in \text{im }f$, so that $\text{im }f' \subseteq \text{im }f$.

This means that we can define $f': M/\text{ker }f \to \text{im }f$ (the whole point of this is to make sure we get the proper co-domain).

Yes, you are correct the inclusion $\lambda: \text{im }f \to N$ given by $(xf)\lambda = xf$ "does nothing". But $f$ MAY NOT BE SURJECTIVE, and to get an isomorphism, we need a BIJECTIVE map.

In many texts, the way they get around this, is to consider instead only surjective homomorphisms:

$f: M \to (M)f$. The trouble with this in practice, is that we may have the modules $M,N$ given before-hand. So we "modify" $f$ to be the composition of a surjective map and an inclusion. The we use the factor theorem on the "modified $f$". Since the factor theorem gives us an injective homomorphism (when $M' = \text{ker }f$), guaranteeing that the original map is surjective then pops out an isomorphism.

 

[Math Amateur]

Let's look at Corollary 1.16, first.

Basically it says any homomorphism $f: M \to N$ (yes, ANY homomorphism) factors through the quotient morphism induced by any submodule $M'$ contained in the kernel of $f$.

One way to see this clearer, is to note that the kernel of $f$ induces a partition on $M$, into cosets of the kernel.

Now if $M' \subseteq \text{ker }f$, I claim that the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$. In other words, every coset just gets "chopped up" into smaller cosets, we do not have any elements from DIFFERENT cosets of $\text{ker }f$ winding up in the SAME coset of $M'$.

In other words, I claim that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$. This is immediate from the fact that: $M' \subseteq \text{ker }f$.

IN other words, if a mapping respects the partition induced by $\text{ker }f$ (that is to say, it is constant on any coset of $\text{ker }f$) then it CERTAINLY also respects the partition induced by $M'$ (in fact it is constant on "clusters of cosets", not just individual cosets of $M'$).

This means we get a well-defined mapping: $f':M/M' \to N$ defined by $(x + M')f = xf$. If $\nu: M \to M/M'$ is the quotient homomorphism:

$x\nu = x + M'$

It is evident that $x(\nu f') = (x\nu)f' = (x + M')f = xf$ for any $x \in M$, that is: $f = \nu f'$.

(I am attempting to keep my mappings on the right, which feels odd to me).

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

Now, we can consider the special case $M' = \text{ker }f$.

This gives us $f = \nu f'$ where $f': M/\text{ker }f \to N$. The trouble is, here, that $N$ might be "much larger" than $M$ (in other words, $f$ is not surjective).

If $f$ is not surjective, then $f'$ cannot be, either, since $f'$ only takes on values that $f$ does (because of how its defined).

Lets' prove $\text{im }f = \text{im }f'$.

Suppose $n \in \text{im }f$. Then $n = xf$ for some $x \in M$, whence $n = (x + \text{ker }f)f'$, so that $n \in \text{im }f'$. Thus $\text{im }f \subseteq \text{im }f'$.

On the other hand, suppose $n \in \text{im }f'$. This means that $n = (x + \text{ker }f)f'$ for some coset $x +\text{ker }f \in M/\text{ker }f$.

Pick $y \in x + \text{ker }f$. Then $yf = y(\nu f') = (y + \text{ker }f)f' = (x + \text{ker }f)f'$ (since $y\nu = x\nu$, because $y \in x + \text{ker }f$)

$=n$, which shows that $n \in \text{im }f$, so that $\text{im }f' \subseteq \text{im }f$.

This means that we can define $f': M/\text{ker }f \to \text{im }f$ (the whole point of this is to make sure we get the proper co-domain).

Yes, you are correct the inclusion $\lambda: \text{im }f \to N$ given by $(xf)\lambda = xf$ "does nothing". But $f$ MAY NOT BE SURJECTIVE, and to get an isomorphism, we need a BIJECTIVE map.

In many texts, the way they get around this, is to consider instead only surjective homomorphisms:

$f: M \to (M)f$. The trouble with this in practice, is that we may have the modules $M,N$ given before-hand. So we "modify" $f$ to be the composition of a surjective map and an inclusion. The we use the factor theorem on the "modified $f$". Since the factor theorem gives us an injective homomorphism (when $M' = \text{ker }f$), guaranteeing that the original map is surjective then pops out an isomorphism.

Thanks for the extensive help Deveno ... just working through your post in detail now ...

Thanks again,Peter

 

[Math Amateur]

Let's look at Corollary 1.16, first.

Basically it says any homomorphism $f: M \to N$ (yes, ANY homomorphism) factors through the quotient morphism induced by any submodule $M'$ contained in the kernel of $f$.

One way to see this clearer, is to note that the kernel of $f$ induces a partition on $M$, into cosets of the kernel.

Now if $M' \subseteq \text{ker }f$, I claim that the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$. In other words, every coset just gets "chopped up" into smaller cosets, we do not have any elements from DIFFERENT cosets of $\text{ker }f$ winding up in the SAME coset of $M'$.

In other words, I claim that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$. This is immediate from the fact that: $M' \subseteq \text{ker }f$.

IN other words, if a mapping respects the partition induced by $\text{ker }f$ (that is to say, it is constant on any coset of $\text{ker }f$) then it CERTAINLY also respects the partition induced by $M'$ (in fact it is constant on "clusters of cosets", not just individual cosets of $M'$).

This means we get a well-defined mapping: $f':M/M' \to N$ defined by $(x + M')f = xf$. If $\nu: M \to M/M'$ is the quotient homomorphism:

$x\nu = x + M'$

It is evident that $x(\nu f') = (x\nu)f' = (x + M')f = xf$ for any $x \in M$, that is: $f = \nu f'$.

(I am attempting to keep my mappings on the right, which feels odd to me).

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

Now, we can consider the special case $M' = \text{ker }f$.

This gives us $f = \nu f'$ where $f': M/\text{ker }f \to N$. The trouble is, here, that $N$ might be "much larger" than $M$ (in other words, $f$ is not surjective).

If $f$ is not surjective, then $f'$ cannot be, either, since $f'$ only takes on values that $f$ does (because of how its defined).

Lets' prove $\text{im }f = \text{im }f'$.

Suppose $n \in \text{im }f$. Then $n = xf$ for some $x \in M$, whence $n = (x + \text{ker }f)f'$, so that $n \in \text{im }f'$. Thus $\text{im }f \subseteq \text{im }f'$.

On the other hand, suppose $n \in \text{im }f'$. This means that $n = (x + \text{ker }f)f'$ for some coset $x +\text{ker }f \in M/\text{ker }f$.

Pick $y \in x + \text{ker }f$. Then $yf = y(\nu f') = (y + \text{ker }f)f' = (x + \text{ker }f)f'$ (since $y\nu = x\nu$, because $y \in x + \text{ker }f$)

$=n$, which shows that $n \in \text{im }f$, so that $\text{im }f' \subseteq \text{im }f$.

This means that we can define $f': M/\text{ker }f \to \text{im }f$ (the whole point of this is to make sure we get the proper co-domain).

Yes, you are correct the inclusion $\lambda: \text{im }f \to N$ given by $(xf)\lambda = xf$ "does nothing". But $f$ MAY NOT BE SURJECTIVE, and to get an isomorphism, we need a BIJECTIVE map.

In many texts, the way they get around this, is to consider instead only surjective homomorphisms:

$f: M \to (M)f$. The trouble with this in practice, is that we may have the modules $M,N$ given before-hand. So we "modify" $f$ to be the composition of a surjective map and an inclusion. The we use the factor theorem on the "modified $f$". Since the factor theorem gives us an injective homomorphism (when $M' = \text{ker }f$), guaranteeing that the original map is surjective then pops out an isomorphism.

Thanks again for the help Deveno ... but just a minor clarification ... ...

You write:

" ... ... In other words, I claim that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$. This is immediate from the fact that: $M' \subseteq \text{ker }f$. ... ... "

Your claim, I must say seems intuitive ... but how do we explicitly and formally prove that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$?

Can you help?

I wish to follow your statement

" ... ... Now if $M' \subseteq \text{ker }f$, I claim that the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$. In other words, every coset just gets "chopped up" into smaller cosets, we do not have any elements from DIFFERENT cosets of $\text{ker }f$ winding up in the SAME coset of $M'$."

However I am having trouble seeing why this is true ... but suspect that I am having trouble because I do not completely follow your claim that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$

Hoping you can clarify.

Peter

***EDIT***

Reflecting further, I am even puzzled as to why the claim "if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$" justifies the statement "if $M' \subseteq \text{ker }f$, I claim that the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$."

Can you help my confusion and puzzlement with these issues?

 

[Math Amateur]

Thanks again for the help Deveno ... but just a minor clarification ... ...

You write:

" ... ... In other words, I claim that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$. This is immediate from the fact that: $M' \subseteq \text{ker }f$. ... ... "

Your claim, I must say seems intuitive ... but how do we explicitly and formally prove that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$?

Can you help?

I wish to follow your statement

" ... ... Now if $M' \subseteq \text{ker }f$, I claim that the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$. In other words, every coset just gets "chopped up" into smaller cosets, we do not have any elements from DIFFERENT cosets of $\text{ker }f$ winding up in the SAME coset of $M'$."

However I am having trouble seeing why this is true ... but suspect that I am having trouble because I do not completely follow your claim that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$

Hoping you can clarify.

Peter

***EDIT***

Reflecting further, I am even puzzled as to why the claim "if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$" justifies the statement "if $M' \subseteq \text{ker }f$, I claim that the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$."

Can you help my confusion and puzzlement with these issues?

A little reading and reflection has, I think, enabled me to answer one of my own questions ...I asked above:

"how do we explicitly and formally prove that if $x +M' = y + M'$, then $x + \text{ker }f = y + \text{ker }f$?"

The mechanics of an explicit proof of this is (it now seems to me) quite simple and is as follows:

\(\displaystyle x + M' = y + M'\)

\(\displaystyle \Longrightarrow \ \ x - y \in M'\)

\(\displaystyle \Longrightarrow \ \ x - y \in \text{ ker } f \ \ \ \text{ since } M' \subseteq \text{ ker } f\)

\(\displaystyle \Longrightarrow \ \ x + \text{ ker } f = y + \text{ ker } f
\)So that part of my question is OK but the explicit proof of your contention "if $M' \subseteq \text{ker }f$, then the partition of $M$ by $M'$ (into cosets of $M'$) is a REFINEMENT of the partition induced by $\text{ker }f$." still escapes me ... ...

Peter

 

[Deveno]

Before we continue, do you know what a refinement of a partition is?

 

[Math Amateur]

Before we continue, do you know what a refinement of a partition is?

I am not sure of the precise formal definition, but essentially a refinement of a partition is a partition which keeps or respects the boundaries of the sets in the partition but divides some of the sets into smaller sets ...

As an example of a refinement of a partition consider the family of pairwise disjoint sets \(\displaystyle \mathscr{F}\) where:

\(\displaystyle \mathscr{F} = \Biggl\{ \{ a, b, c \}, \{ d, e, f, g \}, \{ h, i, j \} \Biggr\}
\)

Clearly \(\displaystyle \mathscr{F}\) is a partition of \(\displaystyle \Biggl\{ a, b, c, d, e, f, g, h, i, j, k \Biggr\}\)

One refinement of \(\displaystyle \mathscr{F}\) would be \(\displaystyle \mathscr{F_1}\) where:

\(\displaystyle \mathscr{F_1} = \Biggl\{ \{a, b \}, \{ c \} , \{ d, e \}, \{ f, g \}, \{ h, i, j \Biggr\}\)Peter

 

[Deveno]

I am not sure of the precise formal definition, but essentially a refinement of a partition is a partition which keeps or respects the boundaries of the sets in the partition but divides some of the sets into smaller sets ...

Peter

That will do for now. Let $M$ be a module, and $f: M \to N$ a module homomorphism, with $K = \text{ker }f$. Let $M'$ be any submodule of $K$.

So we have the partition of $M$ into cosets of $K$, and we have the partition of $M$ into cosets of $M'$.

Consider any coset $x + K$. I claim that $M'$ partitions $x + K$ into cosets of $M'$.

Clearly $M'$ partitions $K$ into cosets:

$\{k_j + M'\}$.

I claim that this gives us a partition $\{(x + k_j) + M'\}$ of $x + K$.

Let $m \in x + K$. Then $m = x + k$ for some $k \in K$. Now since the $k_j + M'$ partition $K$, we have:

$k = k_j + m'$, for some $k_j \in K$ and $m' \in M'$. Therefore:

$m = x + k = x + k_j + m' \in (x + k_j) + M'$, so every element of $x + K$ is in SOME coset $(x + k_j) + M'$.

This shows that the union of all these cosets is $x + K$.

Now suppose that $k_1 + M' \neq k_2 + M'$. We will show that $(x + k_1) + M'$ and $(x + k_2) + M'$ are disjoint sets.

For suppose not, and we have $x + k_1 + m_1' = x + k_2 + m_2'$, for some $m_1',m_2' \in M'$.

Then (adding $-x$ to both sides) we have:

$k_1 + m_1' = k_2 + m_2'$, in other words $k_1 - k_2 \in M'$ (since this equals $m_2' - m_1' \in M'$ by closure), and we have:

$k_1 + M' = k_2 + M'$, contradicting our choice of $k_1,k_2$.

You might want to think about why this suffices.

What "quotienting by $M'$" DOES, is turn $M$ into $M'$-sized "chunks", each of which we regard as "a single element". This is the same thing as "shrinking $M'$ down to 0", since the addition forces us to apply this shrinkage UNIFORMLY over $M$.

 

[Math Amateur]

That will do for now. Let $M$ be a module, and $f: M \to N$ a module homomorphism, with $K = \text{ker }f$. Let $M'$ be any submodule of $K$.

So we have the partition of $M$ into cosets of $K$, and we have the partition of $M$ into cosets of $M'$.

Consider any coset $x + K$. I claim that $M'$ partitions $x + K$ into cosets of $M'$.

Clearly $M'$ partitions $K$ into cosets:

$\{k_j + M'\}$.

I claim that this gives us a partition $\{(x + k_j) + M'\}$ of $x + K$.

Let $m \in x + K$. Then $m = x + k$ for some $k \in K$. Now since the $k_j + M'$ partition $K$, we have:

$k = k_j + m'$, for some $k_j \in K$ and $m' \in M'$. Therefore:

$m = x + k = x + k_j + m' \in (x + k_j) + M'$, so every element of $x + K$ is in SOME coset $(x + k_j) + M'$.

This shows that the union of all these cosets is $x + K$.

Now suppose that $k_1 + M' \neq k_2 + M'$. We will show that $(x + k_1) + M'$ and $(x + k_2) + M'$ are disjoint sets.

For suppose not, and we have $x + k_1 + m_1' = x + k_2 + m_2'$, for some $m_1',m_2' \in M'$.

Then (adding $-x$ to both sides) we have:

$k_1 + m_1' = k_2 + m_2'$, in other words $k_1 - k_2 \in M'$ (since this equals $m_2' - m_1' \in M'$ by closure), and we have:

$k_1 + M' = k_2 + M'$, contradicting our choice of $k_1,k_2$.

You might want to think about why this suffices.

What "quotienting by $M'$" DOES, is turn $M$ into $M'$-sized "chunks", each of which we regard as "a single element". This is the same thing as "shrinking $M'$ down to 0", since the addition forces us to apply this shrinkage UNIFORMLY over $M$.

Thanks for the help, Deveno ... Much appreciated ... ...

Just working through your post carefully and in detail now ...

I would not not fully understand this material without your help ... So thanks again!

Peter

 

[Math Amateur]

That will do for now. Let $M$ be a module, and $f: M \to N$ a module homomorphism, with $K = \text{ker }f$. Let $M'$ be any submodule of $K$.

So we have the partition of $M$ into cosets of $K$, and we have the partition of $M$ into cosets of $M'$.

Consider any coset $x + K$. I claim that $M'$ partitions $x + K$ into cosets of $M'$.

Clearly $M'$ partitions $K$ into cosets:

$\{k_j + M'\}$.

I claim that this gives us a partition $\{(x + k_j) + M'\}$ of $x + K$.

Let $m \in x + K$. Then $m = x + k$ for some $k \in K$. Now since the $k_j + M'$ partition $K$, we have:

$k = k_j + m'$, for some $k_j \in K$ and $m' \in M'$. Therefore:

$m = x + k = x + k_j + m' \in (x + k_j) + M'$, so every element of $x + K$ is in SOME coset $(x + k_j) + M'$.

This shows that the union of all these cosets is $x + K$.

Now suppose that $k_1 + M' \neq k_2 + M'$. We will show that $(x + k_1) + M'$ and $(x + k_2) + M'$ are disjoint sets.

For suppose not, and we have $x + k_1 + m_1' = x + k_2 + m_2'$, for some $m_1',m_2' \in M'$.

Then (adding $-x$ to both sides) we have:

$k_1 + m_1' = k_2 + m_2'$, in other words $k_1 - k_2 \in M'$ (since this equals $m_2' - m_1' \in M'$ by closure), and we have:

$k_1 + M' = k_2 + M'$, contradicting our choice of $k_1,k_2$.

You might want to think about why this suffices.

What "quotienting by $M'$" DOES, is turn $M$ into $M'$-sized "chunks", each of which we regard as "a single element". This is the same thing as "shrinking $M'$ down to 0", since the addition forces us to apply this shrinkage UNIFORMLY over $M$.

Hi Deveno ... thanks again for the help ... but just a minor point ...

You write:

" ... ... I claim that this gives us a partition $\{(x + k_j) + M'\}$ of $x + K$.

Let $m \in x + K$. Then $m = x + k$ for some $k \in K$. Now since the $k_j + M'$ partition $K$, we have:

$k = k_j + m'$, for some $k_j \in K$ and $m' \in M'$. Therefore:

$m = x + k = x + k_j + m' \in (x + k_j) + M'$, so every element of $x + K$ is in SOME coset $(x + k_j) + M'$.

This shows that the union of all these cosets is $x + K$. ... ... "


We have shown that so every element of $x + K$ is in SOME coset \(\displaystyle (x + k_j) + M'\) - but what if every element is actually in one coset equal to \(\displaystyle x + K\) ... ... then we do not have a refinement (except in a trivial sense) ... don't we have to show that there is at least two distinct cosets?

Can you clarify?

Peter

 

[Math Amateur]

That will do for now. Let $M$ be a module, and $f: M \to N$ a module homomorphism, with $K = \text{ker }f$. Let $M'$ be any submodule of $K$.

So we have the partition of $M$ into cosets of $K$, and we have the partition of $M$ into cosets of $M'$.

Consider any coset $x + K$. I claim that $M'$ partitions $x + K$ into cosets of $M'$.

Clearly $M'$ partitions $K$ into cosets:

$\{k_j + M'\}$.

I claim that this gives us a partition $\{(x + k_j) + M'\}$ of $x + K$.

Let $m \in x + K$. Then $m = x + k$ for some $k \in K$. Now since the $k_j + M'$ partition $K$, we have:

$k = k_j + m'$, for some $k_j \in K$ and $m' \in M'$. Therefore:

$m = x + k = x + k_j + m' \in (x + k_j) + M'$, so every element of $x + K$ is in SOME coset $(x + k_j) + M'$.

This shows that the union of all these cosets is $x + K$.

Now suppose that $k_1 + M' \neq k_2 + M'$. We will show that $(x + k_1) + M'$ and $(x + k_2) + M'$ are disjoint sets.

For suppose not, and we have $x + k_1 + m_1' = x + k_2 + m_2'$, for some $m_1',m_2' \in M'$.

Then (adding $-x$ to both sides) we have:

$k_1 + m_1' = k_2 + m_2'$, in other words $k_1 - k_2 \in M'$ (since this equals $m_2' - m_1' \in M'$ by closure), and we have:

$k_1 + M' = k_2 + M'$, contradicting our choice of $k_1,k_2$.

You might want to think about why this suffices.

What "quotienting by $M'$" DOES, is turn $M$ into $M'$-sized "chunks", each of which we regard as "a single element". This is the same thing as "shrinking $M'$ down to 0", since the addition forces us to apply this shrinkage UNIFORMLY over $M$.

Just a further very minor clarification, Deveno ...

You write:

" ... ... $k_1 + m_1' = k_2 + m_2'$, in other words $k_1 - k_2 \in M'$ (since this equals $m_2' - m_1' \in M'$ by closure) ... ... "

Now it seems that:

\(\displaystyle k_1 + m_1' = k_2 + m_2' \Longrightarrow k_1 - k_2 \in M' \)

because adding \(\displaystyle -m_1' - k_2\) to each side of \(\displaystyle k_1 + m_1' = k_2 + m_2'\)

leads to the equation \(\displaystyle k_1 - k_2 = m_2' - m_1' \in M' \)
... ... BUT I am puzzled by why you attribute this result to 'closure' ... presumably the argument does not work without closure of \(\displaystyle M'\)?

Can you clarify why the argument depends on 'closure'?

Peter

***EDIT***

Just been reflecting on my own question - in the equation \(\displaystyle k_1 + m_1' = k_2 + m_2\)' we have that \(\displaystyle k_1\) and \(\displaystyle k_2\) belong to \(\displaystyle K\) and \(\displaystyle m_1'\) and \(\displaystyle m_2'\) belong to \(\displaystyle M'\) so presumably, in order to add \(\displaystyle -m_1' - k_2\) to each side, we need \(\displaystyle -k_2 \in K \)and \(\displaystyle -m_1 \in M'\) ... ? ... ... is that the reason?

 

[Deveno]

Hi Deveno ... thanks again for the help ... but just a minor point ...

You write:

" ... ... I claim that this gives us a partition $\{(x + k_j) + M'\}$ of $x + K$.

Let $m \in x + K$. Then $m = x + k$ for some $k \in K$. Now since the $k_j + M'$ partition $K$, we have:

$k = k_j + m'$, for some $k_j \in K$ and $m' \in M'$. Therefore:

$m = x + k = x + k_j + m' \in (x + k_j) + M'$, so every element of $x + K$ is in SOME coset $(x + k_j) + M'$.

This shows that the union of all these cosets is $x + K$. ... ... "


We have shown that so every element of $x + K$ is in SOME coset \(\displaystyle (x + k_j) + M'\) - but what if every element is actually in one coset equal to \(\displaystyle x + K\) ... ... then we do not have a refinement (except in a trivial sense) ... don't we have to show that there is at least two distinct cosets?

Can you clarify?

Peter

Well, actually, that CAN happen...if $M' = K$.

It might also be the case that $K = M$ (that is, $f$ is the 0-map). Here, the partition of $M$ is trivial, everything is in the one coset $M$.

But, if $K \neq M$, then we have at least one additional coset, choose $x \not \in K$, then $x + K \neq K$.

(If $x + k \in K$, then so is $x = x + (k - k) = (x + k) - k$, which is the difference of two elements of $K$).

If $M' \neq K$, then we can choose $k' \not\in M'$, in which case for our $x \not \in K$:

$x + M' \neq M'$ (since $x$ cannot be in $M'$ since $M' \subseteq K$), and thus:

$x + k' + M' \neq x + M'$.

That is, if there is more than one coset of $M'$ in $K$, we have more than one coset of $M'$ in $x + K$.

One of the facts underlying this, is that the mapping:

$L_a(x) = a + x$ is a bijection $M \to M$.

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

To answer your second question:

We need closure of $M'$ to ensure that $m_2' - m_1' \in M'$.