MHB Explicit idempotents - Berrick and Keating - Exercise 2.1.5

[Math Amateur]

Exercise 2.1.5 in Berrick and Keating: An Introduction to Rings and Modules reads as follows:

Let $$M$$ be an abelian group with $$Mc = 0$$ for some positive integer $$c$$, and put $$c = ab$$ for coprime integers $$a,b$$.

Write $$1 = ar + bs$$, and define endomorphisms $$\alpha$$ and $$\beta$$ of $$M$$ by:

$$\alpha (m) = arm $$

and

$$\beta (m) = bsm$$.

Verify that $$\{ \alpha , \beta \}$$ is a set of projections for the direct sum decomposition $$M = Ma \oplus Mb$$ of $$M$$.

Hence, find the full set of orthogonal idempotents of End$$(M)$$ corresponding to the decomposition $$M = M_1 \oplus M_2 \ ... \ ... \ \oplus M_k$$.

Can someone please help me get started on this problem?

Help would be appreciated.

Peter

 

[Euge]

Exercise 2.1.5 in Berrick and Keating: An Introduction to Rings and Modules reads as follows:

Let $$M$$ be an abelian group with $$Mc = 0$$ for some positive integer $$c$$, and put $$c = ab$$ for coprime integers $$a,b$$.

Write $$1 = ar + bs$$, and define endomorphisms $$\alpha$$ and $$\beta$$ of $$M$$ by:

$$\alpha (m) = arm $$

and

$$\beta (m) = bsm$$.

Verify that $$\{ \alpha , \beta \}$$ is a set of projections for the direct sum decomposition $$M = Ma \oplus Mb$$ of $$M$$.

Hence, find the full set of orthogonal idempotents of End$$(M)$$ corresponding to the decomposition $$M = M_1 \oplus M_2 \ ... \ ... \ \oplus M_k$$.

Can someone please help me get started on this problem?

Help would be appreciated.

Peter

Hi Peter,

To answer the first part, first note that $\alpha : M \to Ma$ and $\beta : M \to Mb$. What needs to be shown is that $\alpha$ and $\beta$ are idempotent, i.e., $\alpha^2 = \alpha$ and $\beta^2 = \beta$. Given $m\in M$,

$\alpha(\alpha(m)) = \alpha(arm) = (ar)^2 m$.

Since $1 = ar + bs$, and $c = ab$, multiplication by $ar$ yields $ar = (ar)^2 + src$ or $(ar)^2 = ar - src$. Since $Mc = 0$, we find

$\displaystyle (ar)^2 m = arm - srcm = arm - 0 = arm = \alpha(m)$.

Hence, $\alpha(\alpha(m)) = \alpha(m)$. A similar argument gives $\beta(\beta(m)) = \beta(m)$. Since $m$ was arbitrary, $\alpha$ and $\beta$ are idempotent.

Even further, $\alpha$ and $\beta$ are orthogonal as

$\alpha(\beta(m)) = \alpha(bsm) = arbsm = rs(cm) = rs(0) = 0$

and similarly $\beta(\alpha(m)) = 0$ for all $m\in M$. So in fact, $\{\alpha, \beta\}$ is an orthogonal set of projections.

For the second part, are the $M$'s abelian groups?

 

[Deveno]

Here is an example to illustrate this is a generalization of stuff you learned some time ago:

Let $M = \Bbb Z_{24}$ so that $c = 24$.

We have $24 = 3\cdot 8$ with $\text{gcd}(3,8) = 1$.

And we find (for example) that $2\cdot 8 + (-5)\cdot 3 = 1$, so that $r = 2,s = -5$.

Hence $\alpha([k]_{24}) = [16k]_{24}$ and $\beta([k]_{24}) = [-15k]_{24} = [9k]_{24}$.

Let's find $M[3]_{24}$ and $M[8]_{24}$ explicitly.

$M[3]_{24} = \{[0]_{24},[3]_{24},[6]_{24},[9]_{24},[12]_{24},[15]_{24},[18]_{24},[21]_{24}\}$

$M[8]_{24} = \{[0]_{24},[8]_{24},[16]_{24}\}$

As you can see we can write:

$[k]_{24} = [1]_{24}\cdot k = [9 + 16]_{24}\cdot k = [3]_{24}\cdot 3k + [8]_{24}\cdot 2k$

So $M = Ma + Mb$, and $Ma \cap Mb = \{[0]_{24}\} = \{0_M\}$, so this sum is direct.

Now we calculate $\alpha$ and $\beta$, explicitly. Starting with $[0]_{24}$ and then $[1]_{24}$, etc. the images of $\alpha$ are:

$[0]_{24},[9]_{24},[18]_{24},[3]_{24},[12]_{24},[21]_{24},[6]_{24},[15]_{24}$

and then we repeat: $\alpha([k]_{24}) = \alpha([k+8]_{24})$

So we see that $\alpha(M) = Ma$.

Similarly, the images of $\beta$ are:

$[0]_{24},[16]_{24},[8]_{24}$ and then we repeat since $\beta([k]_{24}) = \beta([k+3]_{24})$.

We can verify explicitly that $\alpha^2 = \alpha$ and $\beta^2 = \beta$:

$\alpha^2([k]_{24}) = \alpha([9k]_{24}) = [81k]_{24} = [9k]_{24} = \alpha([k]_{24})$

since 81 = 9 (mod 24), and

$\beta^2([k]_{24}) = \beta([16k]_{24}) = [256k]_{24} = [16k]_{24} = \beta([k]_{24})$

since 256 = 16 (mod 24).

So we see the internal direct sum decomposition gives us a way to make an explicit isomorphism of $\Bbb Z_3 \times \Bbb Z_8$ with $\Bbb Z_{24}$:

We have:

$([a]_3,_8) \mapsto [16a + 9b]_{24}$ which clearly has the inverse map:

$[k]_{24} \mapsto ([k]_3,[k]_8)$.

In other words, we recover the Chinese Remainder Theorem (for $n = 24$).

Naively, one would think that the "natural" projection of $\Bbb Z_{24}$ onto $\langle [3]_{24}\rangle$ would be:

$[k]_{24} \mapsto [3k]_{24}$

but this turns out to be "the wrong generator" (we want $[9]_{24}$).

 

[Math Amateur]

Here is an example to illustrate this is a generalization of stuff you learned some time ago:

Let $M = \Bbb Z_{24}$ so that $c = 24$.

We have $24 = 3\cdot 8$ with $\text{gcd}(3,8) = 1$.

And we find (for example) that $2\cdot 8 + (-5)\cdot 3 = 1$, so that $r = 2,s = -5$.

Hence $\alpha([k]_{24}) = [16k]_{24}$ and $\beta([k]_{24}) = [-15k]_{24} = [9k]_{24}$.

Let's find $M[3]_{24}$ and $M[8]_{24}$ explicitly.

$M[3]_{24} = \{[0]_{24},[3]_{24},[6]_{24},[9]_{24},[12]_{24},[15]_{24},[18]_{24},[21]_{24}\}$

$M[8]_{24} = \{[0]_{24},[8]_{24},[16]_{24}\}$

As you can see we can write:

$[k]_{24} = [1]_{24}\cdot k = [9 + 16]_{24}\cdot k = [3]_{24}\cdot 3k + [8]_{24}\cdot 2k$

So $M = Ma + Mb$, and $Ma \cap Mb = \{[0]_{24}\} = \{0_M\}$, so this sum is direct.

Now we calculate $\alpha$ and $\beta$, explicitly. Starting with $[0]_{24}$ and then $[1]_{24}$, etc. the images of $\alpha$ are:

$[0]_{24},[9]_{24},[18]_{24},[3]_{24},[12]_{24},[21]_{24},[6]_{24},[15]_{24}$

and then we repeat: $\alpha([k]_{24}) = \alpha([k+8]_{24})$

So we see that $\alpha(M) = Ma$.

Similarly, the images of $\beta$ are:

$[0]_{24},[16]_{24},[8]_{24}$ and then we repeat since $\beta([k]_{24}) = \beta([k+3]_{24})$.

We can verify explicitly that $\alpha^2 = \alpha$ and $\beta^2 = \beta$:

$\alpha^2([k]_{24}) = \alpha([9k]_{24}) = [81k]_{24} = [9k]_{24} = \alpha([k]_{24})$

since 81 = 9 (mod 24), and

$\beta^2([k]_{24}) = \beta([16k]_{24}) = [256k]_{24} = [16k]_{24} = \beta([k]_{24})$

since 256 = 16 (mod 24).

So we see the internal direct sum decomposition gives us a way to make an explicit isomorphism of $\Bbb Z_3 \times \Bbb Z_8$ with $\Bbb Z_{24}$:

We have:

$([a]_3,_8) \mapsto [16a + 9b]_{24}$ which clearly has the inverse map:

$[k]_{24} \mapsto ([k]_3,[k]_8)$.

In other words, we recover the Chinese Remainder Theorem (for $n = 24$).

Naively, one would think that the "natural" projection of $\Bbb Z_{24}$ onto $\langle [3]_{24}\rangle$ would be:

$[k]_{24} \mapsto [3k]_{24}$

but this turns out to be "the wrong generator" (we want $[9]_{24}$).

Thanks so much Euge and Deveno!

Will be working through your posts in detail shortly.

Peter

- - - Updated - - -

Here is an example to illustrate this is a generalization of stuff you learned some time ago:

Let $M = \Bbb Z_{24}$ so that $c = 24$.

We have $24 = 3\cdot 8$ with $\text{gcd}(3,8) = 1$.

And we find (for example) that $2\cdot 8 + (-5)\cdot 3 = 1$, so that $r = 2,s = -5$.

Hence $\alpha([k]_{24}) = [16k]_{24}$ and $\beta([k]_{24}) = [-15k]_{24} = [9k]_{24}$.

Let's find $M[3]_{24}$ and $M[8]_{24}$ explicitly.

$M[3]_{24} = \{[0]_{24},[3]_{24},[6]_{24},[9]_{24},[12]_{24},[15]_{24},[18]_{24},[21]_{24}\}$

$M[8]_{24} = \{[0]_{24},[8]_{24},[16]_{24}\}$

As you can see we can write:

$[k]_{24} = [1]_{24}\cdot k = [9 + 16]_{24}\cdot k = [3]_{24}\cdot 3k + [8]_{24}\cdot 2k$

So $M = Ma + Mb$, and $Ma \cap Mb = \{[0]_{24}\} = \{0_M\}$, so this sum is direct.

Now we calculate $\alpha$ and $\beta$, explicitly. Starting with $[0]_{24}$ and then $[1]_{24}$, etc. the images of $\alpha$ are:

$[0]_{24},[9]_{24},[18]_{24},[3]_{24},[12]_{24},[21]_{24},[6]_{24},[15]_{24}$

and then we repeat: $\alpha([k]_{24}) = \alpha([k+8]_{24})$

So we see that $\alpha(M) = Ma$.

Similarly, the images of $\beta$ are:

$[0]_{24},[16]_{24},[8]_{24}$ and then we repeat since $\beta([k]_{24}) = \beta([k+3]_{24})$.

We can verify explicitly that $\alpha^2 = \alpha$ and $\beta^2 = \beta$:

$\alpha^2([k]_{24}) = \alpha([9k]_{24}) = [81k]_{24} = [9k]_{24} = \alpha([k]_{24})$

since 81 = 9 (mod 24), and

$\beta^2([k]_{24}) = \beta([16k]_{24}) = [256k]_{24} = [16k]_{24} = \beta([k]_{24})$

since 256 = 16 (mod 24).

So we see the internal direct sum decomposition gives us a way to make an explicit isomorphism of $\Bbb Z_3 \times \Bbb Z_8$ with $\Bbb Z_{24}$:

We have:

$([a]_3,_8) \mapsto [16a + 9b]_{24}$ which clearly has the inverse map:

$[k]_{24} \mapsto ([k]_3,[k]_8)$.

In other words, we recover the Chinese Remainder Theorem (for $n = 24$).

Naively, one would think that the "natural" projection of $\Bbb Z_{24}$ onto $\langle [3]_{24}\rangle$ would be:

$[k]_{24} \mapsto [3k]_{24}$

but this turns out to be "the wrong generator" (we want $[9]_{24}$).

Thanks so much for an example ... I have to say I find your examples extremely helpful in building a good understanding of a topic ... they give such insights into a topic ... and give one a solid and concrete feeling/understanding of abstract topics that goes beyond the theorems and propositions ...

Peter

 

[Math Amateur]

Hi Peter,

To answer the first part, first note that $\alpha : M \to Ma$ and $\beta : M \to Mb$. What needs to be shown is that $\alpha$ and $\beta$ are idempotent, i.e., $\alpha^2 = \alpha$ and $\beta^2 = \beta$. Given $m\in M$,

$\alpha(\alpha(m)) = \alpha(arm) = (ar)^2 m$.

Since $1 = ar + bs$, and $c = ab$, multiplication by $ar$ yields $ar = (ar)^2 + src$ or $(ar)^2 = ar - src$. Since $Mc = 0$, we find

$\displaystyle (ar)^2 m = arm - srcm = arm - 0 = arm = \alpha(m)$.

Hence, $\alpha(\alpha(m)) = \alpha(m)$. A similar argument gives $\beta(\beta(m)) = \beta(m)$. Since $m$ was arbitrary, $\alpha$ and $\beta$ are idempotent.

Even further, $\alpha$ and $\beta$ are orthogonal as

$\alpha(\beta(m)) = \alpha(bsm) = arbsm = rs(cm) = rs(0) = 0$

and similarly $\beta(\alpha(m)) = 0$ for all $m\in M$. So in fact, $\{\alpha, \beta\}$ is an orthogonal set of projections.

For the second part, are the $M$'s abelian groups?

Thanks!

OK, I am following some of the above ... BUT where have we shown that $$\alpha$$ and $$\beta$$ are projections ...?

I thought the main objective was to show that $$\alpha$$ and $$\beta$$ were projections ?

Peter

 

[Deveno]

Thanks!

OK, I am following some of the above ... BUT where have we shown that $$\alpha$$ and $$\beta$$ are projections ...?

I thought the main objective was to show that $$\alpha$$ and $$\beta$$ were projections ?

Peter

What is your text's definition of projection?

The following facts (prove these) should be useful, in any case:

$\alpha(Ma) = Ma$
$\alpha(Mb) = \{0\}$
$\beta(Ma) = \{0\}$
$\beta(Mb) = Mb$.

The three possible definitions I can think of are as follows:

a) $\alpha,\beta$ are surjective
b) $\alpha,\beta$ satisfy the UMP for a direct product
c) $\alpha,\beta$ are idempotent

If it is (c), we have already proven this.

If it is (a), observe that since:

$ar + bs = 1$

$a = a^2r + abs = a^2r + cs$.

Thus $\alpha(am) = (ar)(am) = a^2rm = a^2rm + s0 = a^2rm + scm = (a^2r + cs)m = am$, so any element $am \in Ma$ is its own pre-image under $\alpha$.

A similar statement holds for $\beta$.

If it is (b), show that for any abelian group $G$, with homomorphisms:

$h_1:G \to Ma$
$h_2: G \to Mb$,

that the mapping $\phi:G \to M$ given by $\phi(g) = h_1(g) + h_2(g)$ is the unique group homomorphism with:

$\alpha\phi = h_1$
$\beta\phi = h_2$

the facts I listed above for you to prove may come in handy.

 

[Math Amateur]

What is your text's definition of projection?

The following facts (prove these) should be useful, in any case:

$\alpha(Ma) = Ma$
$\alpha(Mb) = \{0\}$
$\beta(Ma) = \{0\}$
$\beta(Mb) = Mb$.

The three possible definitions I can think of are as follows:

a) $\alpha,\beta$ are surjective
b) $\alpha,\beta$ satisfy the UMP for a direct product
c) $\alpha,\beta$ are idempotent

If it is (c), we have already proven this.

If it is (a), observe that since:

$ar + bs = 1$

$a = a^2r + abs = a^2r + cs$.

Thus $\alpha(am) = (ar)(am) = a^2rm = a^2rm + s0 = a^2rm + scm = (a^2r + cs)m = am$, so any element $am \in Ma$ is its own pre-image under $\alpha$.

A similar statement holds for $\beta$.

If it is (b), show that for any abelian group $G$, with homomorphisms:

$h_1:G \to Ma$
$h_2: G \to Mb$,

that the mapping $\phi:G \to M$ given by $\phi(g) = h_1(g) + h_2(g)$ is the unique group homomorphism with:

$\alpha\phi = h_1$
$\beta\phi = h_2$

the facts I listed above for you to prove may come in handy.

Hi Deveno ... thanks for the clarifying ... ... BUT I am still somewhat confused ...

Are you saying, for example, that if $$\alpha$$ is defined by

$$\alpha (m) = arm$$

and then we show, for example, that $$\alpha$$ is surjective (or idempotent, or satisfies the UMP for direct products) that $$\alpha$$ is a projection ...

Can you clarify?

Peter

 

[Deveno]

Hi Deveno ... thanks for the clarifying ... ... BUT I am still somewhat confused ...

Are you saying, for example, that if $$\alpha$$ is defined by

$$\alpha (m) = arm$$

and then we show, for example, that $$\alpha$$ is surjective (or idempotent, or satisfies the UMP for direct products) that $$\alpha$$ is a projection ...

Can you clarify?

Peter

What definition of projection is IN YOUR TEXT?

 

[Math Amateur]

What definition of projection is IN YOUR TEXT?

OK ... yes ... sorry, you did ask ...

It appears to me that the definition is that of a standard projection:

that is, where

$$M = L_1 \oplus L_2 \oplus L_3 \ ... \ ... \ L_k$$

then

$$\pi (l_1, l_2, \ ... \ ... \ l_k) = l_i$$

But I will scan a couple of pages of the relevant text and send shortly ...

Peter

 

[Math Amateur]

OK ... yes ... sorry, you did ask ...

It appears to me that the definition is that of a standard projection:

that is, where

$$M = L_1 \oplus L_2 \oplus L_3 \ ... \ ... \ L_k$$

then

$$\pi (l_1, l_2, \ ... \ ... \ l_k) = l_i$$

But I will scan a couple of pages of the relevant text and send shortly ...

Peter

Text pertaining to where Berrick and Keating introduce inclusions and projections - in Chapter 2: Direct Sums and Free Modules.

Text follows:

https://www.physicsforums.com/attachments/3080
View attachment 3081

***EDIT***

I just noted a brief comment in B&K (page 43) under Section 2.1.9 Idempotents that states the following:

" ... ... a full set of orthogonal idempotents of End($$M$$) gives rise to a full set of inclusions and projections for $$M$$: for each $$i$$, take $$L_i = e_iM$$, $$\pi_i$$ to be $$ \pi_i \ : \ \mapsto e_im$$ and $$\sigma_i $$ to be the evident inclusion map. ... ... "

Not quite sure how to interpret and/or apply this, however ...

Hope you can clarify ...

Peter

 

[Euge]

Text pertaining to where Berrick and Keating introduce inclusions and projections - in Chapter 2: Direct Sums and Free Modules.

Text follows:

https://www.physicsforums.com/attachments/3080
View attachment 3081

***EDIT***

I just noted a brief comment in B&K (page 43) under Section 2.1.9 Idempotents that states the following:

" ... ... a full set of orthogonal idempotents of End($$M$$) gives rise to a full set of inclusions and projections for $$M$$: for each $$i$$, take $$L_i = e_iM$$, $$\pi_i$$ to be $$ \pi_i \ : \ \mapsto e_im$$ and $$\sigma_i $$to be the evident inclusion map. ... ... "

Alright, let my try to explain it differently. We can identify a sum $x + y$ in $Ma \oplus Mb$ with the ordered pair $(x, y)$ in $Ma \times Mb$. So, depending on which makes you feel more comfortable, you prove $\alpha(x + y) = x$ for all $x\in Ma$ and $y\in Mb$ or, using the ordered pair notation, $\alpha(x, y) = (x,0)$ for all $x\in Ma$ and $y\in Mb$. You also have to prove something similar for $\beta$. The authors of your text seem to interchange the sum and ordered pair notations dealing with finite products, which might be the reason you're getting confused.

 

[Deveno]

OK, so the relevant conditions in your text are (SIP1) and (SIP2). So we need to show they hold.

Since we have an INTERNAL direct sum, we can take the identity function restricted to $Ma$ or $Mb$ as our inclusion functions.

So we need to show:

$\alpha\circ (1_M|_{Ma}) = 1_{Ma}$ (which is just showing that $\alpha(am) = am$, as we did above)

$\alpha\circ(1_M|_{Mb}) = 0_{\text{End}(M)}$ (which is the same as showing $\alpha(bm) = 0$)

$\beta\circ (1_M|_{Mb}) = 1_{Mb}$

$\beta \circ(1_M|_{Ma}) = 0_{\text{End}(M)}$

This will establish (SIP1).

For SIP2, note that $m = 1m = (ar + bs)m = arm + bsm = \alpha(m) + \beta(m)$

and since $\alpha(m) \in Ma$, and $\beta(m) \in Mb$, we have:

$1_M(m) = 1_M(\alpha(m) + \beta(m)) = 1_M\alpha(m) + 1_M\beta(m)$

$= [(1_M|_{Ma})\circ\alpha](m) + [(1_M|_{Mb})\circ\beta](m)$

so that $1_M = [(1_M|_{Ma})\circ\alpha] + [(1_M|_{Mb})\circ\beta]$, which is (SIP2).

In this case, the characterization (Idp1) is actually easier to establish, it is clear we have:

$1_M = \alpha + \beta$

 

[Math Amateur]

Alright, let my try to explain it differently. We can identify a sum $x + y$ in $Ma \oplus Mb$ with the ordered pair $(x, y)$ in $Ma \times Mb$. So, depending on which makes you feel more comfortable, you prove $\alpha(x + y) = x$ for all $x\in Ma$ and $y\in Mb$ or, using the ordered pair notation, $\alpha(x, y) = (x,0)$ for all $x\in Ma$ and $y\in Mb$. You also have to prove something similar for $\beta$. The authors of your text seem to interchange the sum and ordered pair notations dealing with finite products, which might be the reason you're getting confused.

Thanks Euge ... that helped ... but I am also battling with the link between projections and idempotents & orthogonality ...

Peter

 

[Math Amateur]

OK, so the relevant conditions in your text are (SIP1) and (SIP2). So we need to show they hold.

Since we have an INTERNAL direct sum, we can take the identity function restricted to $Ma$ or $Mb$ as our inclusion functions.

So we need to show:

$\alpha\circ (1_M|_{Ma}) = 1_{Ma}$ (which is just showing that $\alpha(am) = am$, as we did above)

$\alpha\circ(1_M|_{Mb}) = 0_{\text{End}(M)}$ (which is the same as showing $\alpha(bm) = 0$)

$\beta\circ (1_M|_{Mb}) = 1_{Mb}$

$\beta \circ(1_M|_{Ma}) = 0_{\text{End}(M)}$

This will establish (SIP1).

For SIP2, note that $m = 1m = (ar + bs)m = arm + bsm = \alpha(m) + \beta(m)$

and since $\alpha(m) \in Ma$, and $\beta(m) \in Mb$, we have:

$1_M(m) = 1_M(\alpha(m) + \beta(m)) = 1_M\alpha(m) + 1_M\beta(m)$

$= [(1_M|_{Ma})\circ\alpha](m) + [(1_M|_{Mb})\circ\beta](m)$

so that $1_M = [(1_M|_{Ma})\circ\alpha] + [(1_M|_{Mb})\circ\beta]$, which is (SIP2).

In this case, the characterization (Idp1) is actually easier to establish, it is clear we have:

$1_M = \alpha + \beta$

Thanks for the help Deveno ... will be working through your post in detail shortly ... BUT ... a minor preliminary clarification about our objective ...

We have been asked to show that $$\alpha$$ and $$\beta$$ as defined in the exercise are projections ...

BUT ... ...in the B&K text, B&K define standard inclusions and projections, $$\sigma_i$$ and $$\pi_i$$ respectively, and then write:

"It is easy to verify that the maps $$\sigma_i$$ and $$\pi_i$$ are homomorphisms of right R-modules, and the following identities hold."

B&K then stat SIP1 and SIP2 ...

In other words if $$\sigma_i$$ and $$\pi_i$$ are standard inclusions and projections then SIP1 and SIP2 apply ...BUT

We are trying to show that $$\alpha$$ and $$\beta$$ as defined in the exercise are projections ...?

Can you clarify?

Peter

(maybe if I work through your post in detail i will understand your strategy ...)

 

[Deveno]

Thanks for the help Deveno ... will be working through your post in detail shortly ... BUT ... a minor preliminary clarification about our objective ...

We have been asked to show that $$\alpha$$ and $$\beta$$ as defined in the exercise are projections ...

BUT ... ...in the B&K text, B&K define standard inclusions and projections, $$\sigma_i$$ and $$\pi_i$$ respectively, and then write:

"It is easy to verify that the maps $$\sigma_i$$ and $$\pi_i$$ are homomorphisms of right R-modules, and the following identities hold."

B&K then stat SIP1 and SIP2 ...

In other words if $$\sigma_i$$ and $$\pi_i$$ are standard inclusions and projections then SIP1 and SIP2 apply ...BUT

We are trying to show that $$\alpha$$ and $$\beta$$ as defined in the exercise are projections ...?

Can you clarify?

Peter

(maybe if I work through your post in detail i will understand your strategy ...)

I took your text to mean if we have homomorphisms satisfying (SIP1) and (SIP2), we have a full set of inclusions and projections.

But let's, for the moment, adopt your point of view:

$M = Ma \oplus Mb$ so that:

$m = am_1 + bm_2$ for some unique $m_1,m_2$.

So we define $\pi_1(m) = am_1$ as "the standard projection". We want to show $\pi_1 = \alpha$.

$\alpha(m) = \alpha(am_1) + \alpha(bm_2) = ar(am_1) + ar(bm_2)$

$= a^2r(m_1) + rcm_2 = a^2r(m_1) + 0 = (a - abs)m_1 = am_1 - csm_1 = am_1 - 0 = am_1 = \pi_1(m)$.

A similar demonstration holds to show $\beta = \pi_2$, defined analogously.

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

(SIP1) and (SIP2) are simple consequences of the UMP for a direct product (= direct sum for a finite number of factors).

The projection maps that "come with" a direct product onto its factors uniquely determine the inclusion maps, and in the FINITE factors case, the inclusion maps of a direct sum uniquely determine the projection maps.

I suspect what this problem is leading to, is a characterization of ANY finite abelian group as a direct sum of abelian groups of prime-power exponent. These, in turn, are direct sums of cyclic groups of order $p^k$ (we can wind up with various $k$'s).

In summary: any finite abelian group is a direct sum of a finite number of cyclic groups.

 

[Math Amateur]

Thanks for the help Deveno ... will be working through your post in detail shortly ... BUT ... a minor preliminary clarification about our objective ...

We have been asked to show that $$\alpha$$ and $$\beta$$ as defined in the exercise are projections ...

BUT ... ...in the B&K text, B&K define standard inclusions and projections, $$\sigma_i$$ and $$\pi_i$$ respectively, and then write:

"It is easy to verify that the maps $$\sigma_i$$ and $$\pi_i$$ are homomorphisms of right R-modules, and the following identities hold."

B&K then stat SIP1 and SIP2 ...

In other words if $$\sigma_i$$ and $$\pi_i$$ are standard inclusions and projections then SIP1 and SIP2 apply ...BUT

We are trying to show that $$\alpha$$ and $$\beta$$ as defined in the exercise are projections ...?

Can you clarify?

Peter

(maybe if I work through your post in detail i will understand your strategy ...)

Although I have had significant help from Deveno and Euge on this problem ... indeed I am still reflecting on what they have written ... I do need someone to confirm/clarify my understanding of the basics of this exercise ...

... ... so ... ...

M is given as an abelian group and so can be regarded as a $$\mathbb{Z}$$-module ... ... and thus $$Ma$$ and $$Mb$$ would be sub-modules (actually sub-$$\mathbb{Z}$$-modules) where:

$$Ma = \{ ma \ | \ m \in M, a \in \mathbb{Z} \text{ and } ma = m + m + \ ...\ ... \ m \ (a \text{ terms }) \}
$$
and

$$Mb = \{ mb \ | \ m \in M, a \in \mathbb{Z} \text{ and } mb = m + m + \ ...\ ... \ m \ (b \text{ terms }) \}$$

Can someone please confirm that this understanding is a correct interpretation of the problem?
... BUT ... one further problem ...

I am still trying to figure how B&K can assume that $$M = Ma \oplus Mb$$ ?

Hope someone can help and clarify.

Now ... back to working on the posts from Deveno and Euge ... again a big thankyou to them!

Peter

 

[Math Amateur]

I took your text to mean if we have homomorphisms satisfying (SIP1) and (SIP2), we have a full set of inclusions and projections.

But let's, for the moment, adopt your point of view:

$M = Ma \oplus Mb$ so that:

$m = am_1 + bm_2$ for some unique $m_1,m_2$.

So we define $\pi_1(m) = am_1$ as "the standard projection". We want to show $\pi_1 = \alpha$.

$\alpha(m) = \alpha(am_1) + \alpha(bm_2) = ar(am_1) + ar(bm_2)$

$= a^2r(m_1) + rcm_2 = a^2r(m_1) + 0 = (a - abs)m_1 = am_1 - csm_1 = am_1 - 0 = am_1 = \pi_1(m)$.

A similar demonstration holds to show $\beta = \pi_2$, defined analogously.

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

(SIP1) and (SIP2) are simple consequences of the UMP for a direct product (= direct sum for a finite number of factors).

The projection maps that "come with" a direct product onto its factors uniquely determine the inclusion maps, and in the FINITE factors case, the inclusion maps of a direct sum uniquely determine the projection maps.

I suspect what this problem is leading to, is a characterization of ANY finite abelian group as a direct sum of abelian groups of prime-power exponent. These, in turn, are direct sums of cyclic groups of order $p^k$ (we can wind up with various $k$'s).

In summary: any finite abelian group is a direct sum of a finite number of cyclic groups.

Thanks Deveno ...

You write:

" ... ... I took your text to mean if we have homomorphisms satisfying (SIP1) and (SIP2), we have a full set of inclusions and projections. ... .. "oh! OK ... yes, you may well be (as usual) correct on that ... since ... running back to my text ... I read, underneath the SIP1, SIP2 conditions:

"In general, given a module $$M$$, a collection

$$\{ L_1, \ ... \ ... \ L_k; \sigma_1, \ ... \ ... \ \sigma_k; \pi_1, \ ... \ ... \ , \pi_k \}$$

of modules and homomorphisms satisfying the above conditions is termed a full set of inclusions and projections for $$M$$.

Not quite sure what "is termed" means exactly, but as I said above, you are likely correct!

Thanks.

Now ... back to your post ...

Peter

 

[Deveno]

We are given, at the outset:

$c = ab$ with $\text{gcd}(a,b) = 1$ and that $c$ annihilates $M$.

This means that $M$ is what is called a "torsion group", as every element is of finite additive order (of $c$ or less).

Since $\text{gcd}(a,b) = 1$ there exist (by the Euclidean algorithm) integers $r,s$ with:

$ar + bs = 1$.

So for ANY $m \in M$:

$m = 1m = (ar + bs)m = a(rm) + b(sm)$.

Since $M$ is a (left, or right, makes no difference, here) $\Bbb Z$-module, clearly $rm, sm \in M$.

Thus $a(rm) \in Ma$ and $b(sm) \in Mb$.

So we have $M \subseteq Ma + Mb = \{x + y: x \in Ma,y \in Mb\}$ and it is clear that $Ma + Mb \subseteq M$ by closure of addition in $M$.

So the "join"(sum) of the $\Bbb Z$-submodules $Ma,Mb$ is $M$. In other words $Ma,Mb$ GENERATE $M$.

To show this sum is DIRECT, we must show the intersection $Ma \cap Mb$ is trivial.

Suppose that $m = am_1 = bm_2$.

Then $am = abm_2 = cm_2 = 0$, and $bm = bam_1 = cm_1 = 0$, so:

$m = 1m = (ar + bs)m = (ar)m + (bs)m = r(am) + s(bm) = r0 + s0 = 0 + 0 = 0$.

This means that an abelian group that $c = ab$ annihilates, "splits cleanly" into a (direct) sum of an abelian group that $a$ annihilates, and one that $b$ annihilates. Examine in detail how this works for my example of $\Bbb Z_{24}$.

For example, what is the direct sum decomposition of $[13]_{24}$?

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

As an aside, this is how "algebra" gets applied to "number theory". Note that when we take $Ma$, we are actually applying the ideal $(a)$ of $\Bbb Z$ to $M$. In this case, saying $\text{gcd}(a,b) = 1$ is equivalent to the IDEAL condition:

$(a) + (b) = \Bbb Z$, that is, that $(a),(b)$ are co-prime ideals.

 

[Euge]

You can prove $M = Ma \oplus Mb$ as follows. Since $1 = ar + bs$, for each $m\in M$,

$\displaystyle m = (mr)a + (ms)b \in Ma + Mb$.

So $M = Ma + Mb$. In addition, $Ma \cap Mb = 0$. For if $x\in Ma \cap Mb$, then $x = ma = m'b$ for some $m,\, m'\in M$. Since $Mc = 0$, we have $xa = m'c = 0$ and $xb = mc = 0$. Hence, the additive order of $x$ divides $a$ and $b$. Since $a$ and $b$ are coprime, it follows that $x$ has additive order one. Therefore, $x = 0$. This proves $M = Ma \oplus Mb$.