MHB 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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View attachment 3259

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

"If such a mapping $$f'$$ exists, it must satisfy

$$(x + M')f' = xf$$ ... ... ... (1.17)

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

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

Further, the next sentence of the proof reads:

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

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

Help would be appreciated.

Peter

 

[Euge]

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

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 $$(x\nu)f' = x(\nu f')$$?

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).