Quotient Modules and Module Homomorphisms - Cohn - Corollary 1.16

[Math Amateur]

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

In Chapter 1: Basics we find Corollary 1.16 on module homomorphisms and quotient modules. I need help with some aspects of the proof.

Corollary 1.16 reads as follows:

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In the above text, the first line of the Proof reads as follows:

"If such a mapping \(\displaystyle f'\) exists, it must satisfy

\(\displaystyle (x + M')f' = xf\) ... ... ... (1.17)

and this shows that there can be at most one such mapping. ... ... "

Can someone please explain why \(\displaystyle f'\) must satisfy 1.17 and, further, why there can be at most one such mapping?

Further, the next sentence of the proof reads:

"Since \(\displaystyle M' \subseteq ker f, xf\) is independent of its choice in the coset ... ... "

Can someone please explain why \(\displaystyle xf\) being independent of its choice in the coset, depends on \(\displaystyle M' \subseteq ker f\)?

Help would be appreciated.

Peter

 

[Euge]

Can someone please explain why \(\displaystyle f'\) must satisfy 1.17 and, further, why there can be at most one such mapping?

If $f'$ exists, then $f = \nu f'$; hence, $(x + M')f' = (x\nu)f' = x(\nu f') = xf$. In particular, this shows that $f'$ is completely determined by $f$. So there can only be one such mapping. In case you're not satisfied, suppose there was another function $g' : M/M' \to N$ such that $f = \nu g'$. Then $(x + M')g' = x(\nu g') = xf = (x + M')f'$ for all cosets $x + M' \in M/M'$. Hence, $g' = f'$.

Can someone please explain why \(\displaystyle xf\) being independent of its choice in the coset, depends on \(\displaystyle M' \subseteq ker f\)?

You've done this argument before, but I'll explain it here. The hint to showing this is in Cohn's own words, "so $xf = x'f$ whenever $x + M' = x' + M'$ and this means that the mapping $f'$ given by (1.17) is well defined". If $x + M' = x' + M$, then $x - x' \in M'$ (since $M'$ is a submodule of $M$) and thus $(x - x')f = 0$ (as $M'\subseteq M$). Since $f$ is a homomorphism, $(x - x')f = xf - x'f$. Therefore, $xf - x'f = 0$, i.e., $xf = x'f$.

 

[Math Amateur]

If $f'$ exists, then $f = \nu f'$; hence, $(x + M')f' = (x\nu)f' = x(\nu f') = xf$. In particular, this shows that $f'$ is completely determined by $f$. So there can only be one such mapping. In case you're not satisfied, suppose there was another function $g' : M/M' \to N$ such that $f = \nu g'$. Then $(x + M')g' = x(\nu g') = xf = (x + M')f'$ for all cosets $x + M' \in M/M'$. Hence, $g' = f'$.
You've done this argument before, but I'll explain it here. The hint to showing this is in Cohn's own words, "so $xf = x'f$ whenever $x + M' = x' + M'$ and this means that the mapping $f'$ given by (1.17) is well defined". If $x + M' = x' + M$, then $x - x' \in M'$ (since $M'$ is a submodule of $M$) and thus $(x - x')f = 0$ (as $M'\subseteq M$). Since $f$ is a homomorphism, $(x - x')f = xf - x'f$. Therefore, $xf - x'f = 0$, i.e., $xf = x'f$.

Thanks Euge ... Really appreciate your help,

Peter

 

[Math Amateur]

If $f'$ exists, then $f = \nu f'$; hence, $(x + M')f' = (x\nu)f' = x(\nu f') = xf$. In particular, this shows that $f'$ is completely determined by $f$. So there can only be one such mapping. In case you're not satisfied, suppose there was another function $g' : M/M' \to N$ such that $f = \nu g'$. Then $(x + M')g' = x(\nu g') = xf = (x + M')f'$ for all cosets $x + M' \in M/M'$. Hence, $g' = f'$.
You've done this argument before, but I'll explain it here. The hint to showing this is in Cohn's own words, "so $xf = x'f$ whenever $x + M' = x' + M'$ and this means that the mapping $f'$ given by (1.17) is well defined". If $x + M' = x' + M$, then $x - x' \in M'$ (since $M'$ is a submodule of $M$) and thus $(x - x')f = 0$ (as $M'\subseteq M$). Since $f$ is a homomorphism, $(x - x')f = xf - x'f$. Therefore, $xf - x'f = 0$, i.e., $xf = x'f$.

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

You write:

"If $f'$ exists, then $f = \nu f'$; hence, $(x + M')f' = (x\nu)f' = x(\nu f') = xf$. ... ... "

Why, exactly, does it follow that \(\displaystyle (x\nu)f' = x(\nu f')\)?

Peter

 

[Euge]

By definition of function composition.

 

[Math Amateur]

By definition of function composition.

oh! ... Yes, of course ... Thanks

Peter

 

[Deveno]

Let's back up a little bit, and just talk about sets.

A PARTITION of a set $A$ is a collection of non-empty subsets of $A, \mathscr{P}$, such that:

(1) $X,Y \in \mathscr{P} \implies X = Y \text{ or } X \cap Y = \emptyset$

(2) $\displaystyle A = \bigcup_{X \in \mathscr{P}} X$

For example:

$\{\{1,2\},\{3\},\{4,5\}\}$ is a partition of $\{1,2,3,4,5\}$.

A basic result on partitions (very important!) is:

Every equivalence $E$ on $A$ induces a partition of $A$, into the equivalence classes of $E$. This is the "set version" of algebraic "quotient objects". Note that since we don't have any operations to worry about, there is just a quotient FUNCTION $q: A \to A/E$:

$a \mapsto [a]$ (in other words $q(a)$ is the equivalence class $a$ resides in). This function $q$ is surjective (by (2) above, (1) ensures $q$ is "well-defined").

Now, for any group $G$ (and remember that Abelian groups, rings, fields, modules, vector spaces and algebras are ALL, first and foremost, groups), we can define an equivalence relation (called "modulo $H$") via ANY subgroup $H$, by:

$g_1 \sim_H g_2 \iff g_1g_2^{-1} \in H$. Equivalence classes are called (right) cosets of $H$.

Since $H$ is itself a group we can do the same thing with any subgroup $K$ of $H$. But $K$ is also a subgroup of $G$, so we have:

(1) The cosets of $H$ in $G$
(2) The cosets of $K$ in $H$
(3) The cosets of $K$ in $G$

What relationships exist among these three things?

Basically the partition (3) is a REFINEMENT of partition (1), that preserves partition (2).

For example, suppose $G$ is the group of integers under addition, with $H = 2\Bbb Z$ and $K = 4\Bbb Z$.

We have the partition induced by $H$:

$\Bbb Z = \{\text{even integers}\} \cup \{\text{odd integers}\}$

The partition induced on the even integers ($H$) by $K$:

$H = \{\text{odd multiples of 2}\} \cup \{\text{even multiples of 2}\}$

The partition induced on $\Bbb Z$ by $K$:

$\Bbb Z = \{4k\} \cup \{4k+1\} \cup \{4k+2\} \cup\{4k+3\}$.

For "abelian thingies", this kind of partition is always an (additive) congruence, it respects addition (for non-abelian groups, we need to have normality, which makes life more complicated). This means that the "partition map":

$x \mapsto [x]_H$

for a sub-abelian thingie $H$ of an abelian thingie $A$ is typically a "abelian thingie homomorphism" (one has to check that the "additional structure" is "reasonably (well-) defined").

This is a basic, primitive thing: this kind of "splitting" into sub-thingie and quotient thingie cuts deep across many kinds of structures, and has its roots in replacing EQUALITY by EQUIVALENCE (the very basis of abstraction).