MHB 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 $$f = \nu g$$ for some homomorphism $$g: M/ker f \to N$$, and since the image of $$g$$ is clearly I am $$f$$, we can write $$g = f' \lambda$$ for some $$f'$$ ... ... ... "

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

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 $$f = \nu g$$ for some homomorphism $$g: M/ker f \to N$$, and since the image of $$g$$ is clearly I am $$f$$, we can write $$g = f' \lambda$$ for some $$f'$$ ... ... ... "

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

Peter

Just reflecting on my own question:

"... ... why the image of $$g$$ is I am $$f$$ ... ... "

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

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 $$\lambda$$ into the proof ... why is it necessary ... what is gained ...

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

$$x + M' = y + M'$$

$$\Longrightarrow \ \ x - y \in M'$$

$$\Longrightarrow \ \ x - y \in \text{ ker } f \ \ \ \text{ since } M' \subseteq \text{ ker } f$$

$$\Longrightarrow \ \ x + \text{ ker } f = y + \text{ ker } f
$$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 $$\mathscr{F}$$ where:

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

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

One refinement of $$\mathscr{F}$$ would be $$\mathscr{F_1}$$ where:

$$\mathscr{F_1} = \Biggl\{ \{a, b \}, \{ c \} , \{ d, e \}, \{ f, g \}, \{ h, i, j \Biggr\}$$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 $$(x + k_j) + M'$$ - but what if every element is actually in one coset equal to $$x + K$$ ... ... 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:

$$k_1 + m_1' = k_2 + m_2' \Longrightarrow k_1 - k_2 \in M' $$

because adding $$-m_1' - k_2$$ to each side of $$k_1 + m_1' = k_2 + m_2'$$

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

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

Peter

***EDIT***

Just been reflecting on my own question - in the equation $$k_1 + m_1' = k_2 + m_2$$' we have that $$k_1$$ and $$k_2$$ belong to $$K$$ and $$m_1'$$ and $$m_2'$$ belong to $$M'$$ so presumably, in order to add $$-m_1' - k_2$$ to each side, we need $$-k_2 \in K $$and $$-m_1 \in M'$$ ... ? ... ... 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 $$(x + k_j) + M'$$ - but what if every element is actually in one coset equal to $$x + K$$ ... ... 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 $m_2' - m_1' \in M'$.