MHB Quotient Modules and Homomorphisms

[Math Amateur]

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

In Chapter 1: Basics we find Theorem 1.15 on module homomorphisms and quotient modules. I need help with some aspects of the proof.

Theorem 1.15 reads as follows:

View attachment 3247

In the proof of the converse (see above text) Cohn states the following:

"If $$x$$ is replaced by $$x + m$$ where $$m \in M'$$,

then $$x + m + M' = x + M'$$ and $$mr \in M'$$,

hence $$(x + m)r + M' = xr + mr + M' = xr +M'$$,

therefore the action of $$R$$ on the cosets is well defined … … "

I am uncertain about what is going on here … … can someone please explain to me why/how the above text does indeed show that the action of $$R$$ on the cosets is well defined?

Help will be appreciated.

Peter

***EDIT*** (SOLVED?)

I think I should have reflected longer on this matter!

Now we take an element $$x \in M$$ and $$r \in R$$ and define the right action $$(x + M')r = xr + m$$.

I now think that we have to show that if we take another element from the coset $$x + M'$$ - say $$x + m$$ where $$m \in M'$$, that the effect of the action is the same - and this is what Cohn is doing ...

Can someone please confirm that this is correct?

 

[Euge]

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

In Chapter 1: Basics we find Theorem 1.15 on module homomorphisms and quotient modules. I need help with some aspects of the proof.

Theorem 1.15 reads as follows:

View attachment 3247

In the proof of the converse (see above text) Cohn states the following:

"If $$x$$ is replaced by $$x + m$$ where $$m \in M'$$,

then $$x + m + M' = x + M'$$ and $$mr \in M'$$,

hence $$(x + m)r + M' = xr + mr + M' = xr +M'$$,

therefore the action of $$R$$ on the cosets is well defined … … "

I am uncertain about what is going on here … … can someone please explain to me why/how the above text does indeed show that the action of $$R$$ on the cosets is well defined?

Help will be appreciated.

Peter

***EDIT*** (SOLVED?)

I think I should have reflected longer on this matter!

Now we take an element $$x \in M$$ and $$r \in R$$ and define the right action $$(x + M')r = xr + m$$.

I now think that we have to show that if we take another element from the coset $$x + M'$$ - say $$x + m$$ where $$m \in M'$$, that the effect of the action is the same - and this is what Cohn is doing ...

Can someone please confirm that this is correct?

To show that the action is well defined, we have to show that $x + M' = y + M'$ implies $(x + M')r = (y + M')r$. What Cohn did is equivalent to this, but follow what I'm doing here. If $x + M' = y + M'$, then $x - y \in M'$ and thus $x = y + m$ for some $m\in M'$. By right-distributivity of scalar multiplication in $M'$ over addition, $xr = (y + m)r = yr + mr$. Since $mr \in M'$, we have $xr + M' = yr + M'$, i.e., $(x + M')r = (y + M')r$.

 

[Math Amateur]

To show that the action is well defined, we have to show that $x + M' = y + M'$ implies $(x + M')r = (y + M')r$. What Cohn did is equivalent to this, but follow what I'm doing here. If $x + M' = y + M'$, then $x - y \in M'$ and thus $x = y + m$ for some $m\in M'$. By right-distributivity of scalar multiplication in $M'$ over addition, $xr = (y + m)r = yr + mr$. Since $mr \in M'$, we have $xr + M' = yr + M'$, i.e., $(x + M')r = (y + M')r$.

Thanks for clarifying that Euge … still reflecting on what you have said ...

Peter

 

[Deveno]

Here's the thing:

we can "define":

$(x + M')r = xr + M'$

but this definition uses $x$ (an element of $M'$), and $x$ does not uniquely determine $x + M'$; for example, if $y = x + m'$, with $m' \in M'$, then $y + M' = (x + m') + M' = (x + M') + (m' + M') = (x + M') + (0 + M') = x + M'$.

So we need to be sure that the assignment:

$x + M' \mapsto xr + M'$ is a function that only depends on $x + M'$, and not on $x$.

And what THAT means is, that if we have $y \neq x$ with $y + M' = x + M'$, we must get the same value $xr + M'$ no matter which $y$ we pick.

In other words, we must show: $x + M' = y + M' \implies xr + M' = yr + M'$.

Now $x + M' = y + M'$ if and only if $x - y \in M'$. From this, we must deduce that $xr - yr \in M'$.

BECAUSE $M$' IS A SUBMODULE, it is closed under the right $R$-action.

This means that $(x - y)r$ is in $M'$ since $x - y$ is.

Since $M$ is a module, we have $(x - y)r = xr - yr$, and thus we have shown what we set out to.

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

This is an example of a quite general phenomenon:

For any SET $A$, and any FUNCTION $f:A \to B$, and any PARTITION of $A$, say $E$, we can form the set of partition subsets $A/E$.

We say $f$ RESPECTS $E$ if and only if we have a function $[f]: A/E \to B$ such that $[f][a]_E = f(a)$ for all $a \in A$.

(Here, $[a]_E$ is the element of the partition $E$ that contains $a$. Since $E$ is a partition, each $a \in A$ occurs in precisely one such partition element).

In other words: $f$ is constant on each element of the partition.

One way to partition $A$ that always works is to set $[a] = f^{-1}(a)$. This is called the partition induced by $f$, which automatically respects this partition.

So, what Cohn is saying is that scalar multiplication (by any $r \in R$) respects the partition of $M$ induced by the mapping:

$x \mapsto x + M'$

As a bonus, we get that this mapping is an $R$-module homomorphism, which is kind of the whole point.

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

Recall that as $\Bbb Z$-modules we have $\Bbb Z_6 \cong \Bbb Z_2 \oplus \Bbb Z_3$.

Now we have (fairly obviously):

$\Bbb Z_3 \cong (\Bbb Z_2 \oplus \Bbb Z_3)/(\Bbb Z_2 \oplus \{[0]_3\})$

Since this is finite, rather than invoking the Chinese Remainder theorem to find the "inverse isomorphism", let's just compute the "forward" one using:

$[k]_6 \mapsto ([k]_2,[k]_3)$

So:

$0 \mapsto (0,0)$ (I will omit the brackets and subscripts from here on).
$1\mapsto (1,1)$
$2 \mapsto (0,2)$
$3 \mapsto (1,0)$
$4 \mapsto (0,1)$
$5 \mapsto (1,2)$

If we call this mapping $\phi: \Bbb Z_6 \to \Bbb Z_2 \oplus \Bbb Z_3$, let's examine:

$\phi^{-1}(\Bbb Z_2 \oplus \{0\}) = \langle 3\rangle$.

So $\Bbb Z_6$ will be our "$M$", and $\langle 3\rangle$ will be our "$M'$".

In $M$ we define the $\Bbb Z$-action by:

$[k]_6 \cdot a = [ka]_6$.

In $M/M'$, we define: $([k]_6 + M')\cdot a = [ka]_6 + M'$. Let's look at this explicitly.

We have three cosets:

$0 + M' = \{0,3\}$
$1 + M' = \{1,4\}$
$2 + M' = \{2,5\}$

What we want to do is verify that for a coset $N$, that $Na$ (multiplying every element in $\Bbb Z_6$ that lies in $N$ by $a$) gives the same set as just multiplying a representative (say $k$) by $a$ and then adding $M'$. Let's do this for $k = 2$, and $a = 9$.

Now $2 + M' = \{2,5\}$ so $(2 + M')\cdot 9 = \{18,45\} = \{0,3\} = M'$ (since we are working mod 6).

On the other hand: $2 \cdot 9 = 18 = 0$, so we would expect that $(2 + M')\cdot 9 = 0 + M' = M'$.

Of course, we get the same multiples for $a+6$ as we do for $a$, so we can just compute explictly the action of $M/M'$ by looking at the action of $\Bbb Z_6$ on $M$:

$(k + M')0 = M'$
$(k + M')1 = k + M'$
$(k + M')2 = 2k + M'$ etc.

It is not hard to see that if $a$ is a multiple of $3$, we get $M'$, so we really only need to look at $a = 0,1,2$.

This, of course should NOT be surprising, after all, our quotient module is congruent to $\Bbb Z/3\Bbb Z$.