Modules - Decomposibiity of abelian groups

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with understanding Example 2.1.3 (ii) (page 39) which concerns $$L$$ as a submodule of the quotient module $$ \mathbb{Z}/p^r \mathbb{Z}$$ ... ...

Example 2.1.3 (ii) (page 39) reads as follows:View attachment 2972I need help to demonstrate rigorously that the inverse image of L under the canonical homomorphism from $$\mathbb{Z}$$ to $$ \mathbb{Z} / p^r \mathbb{Z} $$ must be an ideal $$a \mathbb{Z}$$ of $$\mathbb{Z}$$ containing $$ p^r \mathbb{Z}$$ ... ...

Hope someone can help ...

My thinking so far is as follows:The canonical homomorphism $$\pi \ : \ \mathbb{Z} \to \mathbb{Z} / p^r \mathbb{Z}$$ is as follows:

$$ \pi (n) = \overline{n} = n + p^r \mathbb{Z} $$

where

$$ p^r \mathbb{Z} = \{ ... \ -3 p^r \mathbb{Z}, -2 p^r \mathbb{Z}, - p^r \mathbb{Z}, \ 0, \ p^r \mathbb{Z}, \ 2 p^r \mathbb{Z}, \ 3 p^r \mathbb{Z}, ... \ \} $$ Now we need to show that the inverse image of $$L$$ under the canonical homomorphism from $$\mathbb{Z}$$ to $$ \mathbb{Z} / p^r \mathbb{Z} $$ must be an ideal $$a \mathbb{Z}$$ of $$\mathbb{Z}$$ ... ...

If we denote the inverse image of $$L$$ as $$\pi^* (L)$$ then we have

$$ \pi^* (L) = \{ n \in \mathbb{Z} \ | \ \pi (n) \in L \} $$

thus

$$ \pi^* (L) = \{ n \in \mathbb{Z} \ | \ \overline{n} \in L \} $$

Now if $$\pi^* (L)$$ is an ideal in $$\mathbb{Z}$$ then it has the form $$ a \mathbb{Z}$$ and we must have that $$0 \in \pi^* (L)$$ and, further, we must have additive and multiplicative closure ...

(1) We must have $$ 0 \in \pi^* (L) $$

Since L is a submodule of $$ \mathbb{Z} / p^r \mathbb{Z} $$ we have $$ \overline{0} \in L $$ and so $$0 \in \pi^* (L)$$(2) Additive closure : if $$x,y \in \pi^* (L)$$ then we must have $$x + y \in \pi^* (L) $$

$$x,y \in \pi^* (L) $$

$$\Longrightarrow \overline{x}, \overline{y} \in L $$

$$\Longrightarrow \overline{x}+\overline{y} \in L$$ since L is a submodule

$$\Longrightarrow \overline{x + y} \in L$$

$$\Longrightarrow x + y \in \pi^* (L)$$

so we have additive closure ...


(3) Multiplicative closure : if $$n \in \mathbb{Z}$$ and $$x \in \pi^* (L)$$ then we must have $$xn \in \pi^* (L) $$

$$ x \in \pi^* (L) $$

$$ \Longrightarrow \overline{x} \in L $$

$$\Longrightarrow \overline{x}n \in L$$

$$\Longrightarrow \overline{xn} \in L
$$

$$\Longrightarrow xn \in \pi^* (L)
$$

Given (1), (2) and (3) above $$\pi^* (L)$$ is an ideal of $$\mathbb{Z}$$ and hence is an ideal of the form$$ a \mathbb{Z}$$ (since all ideals of $$\mathbb{Z}$$ are of this form ...
Question 1: Could someone please critique the above analysis ...or at least let me know if it is basically OK ...

In particular are the logical steps of (3) above OK ... as I am particularly unsure of the following particular steps ...

" ...

$$\Longrightarrow \overline{x} \in L
$$

$$\Longrightarrow \overline{x}n \in L
$$

$$\Longrightarrow \overline{xn} \in L
$$

... ... "

Question 2:

Can someone please help me to show that, as stated in the example, that the ideal $$ \pi^* (L) $$ actually contains $$ p^r \mathbb{Z} $$

Question 3:

Can someone please help me to show that any submodule must have the form $$p^i \mathbb{Z} / p^r \mathbb{Z}$$, and further show that this leads to the fact that no submodules can complement each other ...

Hope someone can help ...

Peter

 

[Euge]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with understanding Example 2.1.3 (ii) (page 39) which concerns $$L$$ as a submodule of the quotient module $$ \mathbb{Z}/p^r \mathbb{Z}$$ ... ...

Example 2.1.3 (ii) (page 39) reads as follows:View attachment 2972I need help to demonstrate rigorously that the inverse image of L under the canonical homomorphism from $$\mathbb{Z}$$ to $$ \mathbb{Z} / p^r \mathbb{Z} $$ must be an ideal $$a \mathbb{Z}$$ of $$\mathbb{Z}$$ containing $$ p^r \mathbb{Z}$$ ... ...

Hope someone can help ...

My thinking so far is as follows:The canonical homomorphism $$\pi \ : \ \mathbb{Z} \to \mathbb{Z} / p^r \mathbb{Z}$$ is as follows:

$$ \pi (n) = \overline{n} = n + p^r \mathbb{Z} $$

where

$$ p^r \mathbb{Z} = \{ ... \ -3 p^r \mathbb{Z}, -2 p^r \mathbb{Z}, - p^r \mathbb{Z}, \ 0, \ p^r \mathbb{Z}, \ 2 p^r \mathbb{Z}, \ 3 p^r \mathbb{Z}, ... \ \} $$ Now we need to show that the inverse image of $$L$$ under the canonical homomorphism from $$\mathbb{Z}$$ to $$ \mathbb{Z} / p^r \mathbb{Z} $$ must be an ideal $$a \mathbb{Z}$$ of $$\mathbb{Z}$$ ... ...

If we denote the inverse image of $$L$$ as $$\pi^* (L)$$ then we have

$$ \pi^* (L) = \{ n \in \mathbb{Z} \ | \ \pi (n) \in L \} $$

thus

$$ \pi^* (L) = \{ n \in \mathbb{Z} \ | \ \overline{n} \in L \} $$

Now if $$\pi^* (L)$$ is an ideal in $$\mathbb{Z}$$ then it has the form $$ a \mathbb{Z}$$ and we must have that $$0 \in \pi^* (L)$$ and, further, we must have additive and multiplicative closure ...

(1) We must have $$ 0 \in \pi^* (L) $$

Since L is a submodule of $$ \mathbb{Z} / p^r \mathbb{Z} $$ we have $$ \overline{0} \in L $$ and so $$0 \in \pi^* (L)$$(2) Additive closure : if $$x,y \in \pi^* (L)$$ then we must have $$x + y \in \pi^* (L) $$

$$x,y \in \pi^* (L) $$

$$\Longrightarrow \overline{x}, \overline{y} \in L $$

$$\Longrightarrow \overline{x}+\overline{y} \in L$$ since L is a submodule

$$\Longrightarrow \overline{x + y} \in L$$

$$\Longrightarrow x + y \in \pi^* (L)$$

so we have additive closure ...


(3) Multiplicative closure : if $$n \in \mathbb{Z}$$ and $$x \in \pi^* (L)$$ then we must have $$xn \in \pi^* (L) $$

$$ x \in \pi^* (L) $$

$$ \Longrightarrow \overline{x} \in L $$

$$\Longrightarrow \overline{x}n \in L$$

$$\Longrightarrow \overline{xn} \in L
$$

$$\Longrightarrow xn \in \pi^* (L)
$$

Given (1), (2) and (3) above $$\pi^* (L)$$ is an ideal of $$\mathbb{Z}$$ and hence is an ideal of the form$$ a \mathbb{Z}$$ (since all ideals of $$\mathbb{Z}$$ are of this form ...
Question 1: Could someone please critique the above analysis ...or at least let me know if it is basically OK ...

In particular are the logical steps of (3) above OK ... as I am particularly unsure of the following particular steps ...

" ...

$$\Longrightarrow \overline{x} \in L
$$

$$\Longrightarrow \overline{x}n \in L
$$

$$\Longrightarrow \overline{xn} \in L
$$

... ... "

Question 2:

Can someone please help me to show that, as stated in the example, that the ideal $$ \pi^* (L) $$ actually contains $$ p^r \mathbb{Z} $$

Question 3:

Can someone please help me to show that any submodule must have the form $$p^i \mathbb{Z} / p^r \mathbb{Z}$$, and further show that this leads to the fact that no submodules can complement each other ...

Hope someone can help ...

Peter

Hi Peter,

Your analysis is just fine, but the short answer for Questions 1 and 2 is the lattice isomorphism theorem for modules. Recall that this theorem gives a one-to-one correspondence between the submodules of a quotient module $A/J$ and the submodules of $A$ containing $B$. You can find this theorem (or something similar) in Dummit and Foote's book and Beachy's rings and modules book.

I cannot fully answer your third question yet since there are different meanings of complementation and you have not specified which one you mean. So I'll answer the first part of the question. Recall that if $m$ and $n$ are integers such that $m\Bbb Z \subseteq n\Bbb Z$, then $n|m$. Since $p^r \Bbb Z \subseteq L = a\Bbb Z$, we know that $a|p^r$ and thus $\pi^*(L) = p^j \Bbb Z$ for some nonnegative integer $j \le r$. Since $\pi$ is an onto map,

$\displaystyle L = \pi(\pi^*(L)) = \pi(p^j \Bbb Z) = p^j \Bbb Z/p^r \Bbb Z $.

Since $L$ was arbitrary, it follows that every submodule of $\Bbb Z/p^r \Bbb Z$ is of the form $p^j \Bbb Z/p^r \Bbb Z$.

 

[Deveno]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with understanding Example 2.1.3 (ii) (page 39) which concerns $$L$$ as a submodule of the quotient module $$ \mathbb{Z}/p^r \mathbb{Z}$$ ... ...

Example 2.1.3 (ii) (page 39) reads as follows:View attachment 2972I need help to demonstrate rigorously that the inverse image of L under the canonical homomorphism from $$\mathbb{Z}$$ to $$ \mathbb{Z} / p^r \mathbb{Z} $$ must be an ideal $$a \mathbb{Z}$$ of $$\mathbb{Z}$$ containing $$ p^r \mathbb{Z}$$ ... ...

Hope someone can help ...

My thinking so far is as follows:The canonical homomorphism $$\pi \ : \ \mathbb{Z} \to \mathbb{Z} / p^r \mathbb{Z}$$ is as follows:

$$ \pi (n) = \overline{n} = n + p^r \mathbb{Z} $$

where

$$ p^r \mathbb{Z} = \{ ... \ -3 p^r \mathbb{Z}, -2 p^r \mathbb{Z}, - p^r \mathbb{Z}, \ 0, \ p^r \mathbb{Z}, \ 2 p^r \mathbb{Z}, \ 3 p^r \mathbb{Z}, ... \ \} $$

These are sets. The elemensts of $p^r\Bbb Z$ are just integers.

Now we need to show that the inverse image of $$L$$ under the canonical homomorphism from $$\mathbb{Z}$$ to $$ \mathbb{Z} / p^r \mathbb{Z} $$ must be an ideal $$a \mathbb{Z}$$ of $$\mathbb{Z}$$ ... ...

Just a note, here. Presumably we are talking about abelian groups, that is $\Bbb Z$-modules. Recall, that any ring $R$ is an $R$-module over itself, and that an $R$-submodule of itself is just an ideal.

If we denote the inverse image of $$L$$ as $$\pi^* (L)$$ then we have

$$ \pi^* (L) = \{ n \in \mathbb{Z} \ | \ \pi (n) \in L \} $$

thus

$$ \pi^* (L) = \{ n \in \mathbb{Z} \ | \ \overline{n} \in L \} $$

Now if $$\pi^* (L)$$ is an ideal in $$\mathbb{Z}$$ then it has the form $$ a \mathbb{Z}$$ and we must have that $$0 \in \pi^* (L)$$ and, further, we must have additive and multiplicative closure ...

(1) We must have $$ 0 \in \pi^* (L) $$

Since L is a submodule of $$ \mathbb{Z} / p^r \mathbb{Z} $$ we have $$ \overline{0} \in L $$ and so $$0 \in \pi^* (L)$$(2) Additive closure : if $$x,y \in \pi^* (L)$$ then we must have $$x + y \in \pi^* (L) $$

$$x,y \in \pi^* (L) $$

$$\Longrightarrow \overline{x}, \overline{y} \in L $$

$$\Longrightarrow \overline{x}+\overline{y} \in L$$ since L is a submodule

$$\Longrightarrow \overline{x + y} \in L$$

$$\Longrightarrow x + y \in \pi^* (L)$$

so we have additive closure ...


(3) Multiplicative closure : if $$n \in \mathbb{Z}$$ and $$x \in \pi^* (L)$$ then we must have $$xn \in \pi^* (L) $$

$$ x \in \pi^* (L) $$

$$ \Longrightarrow \overline{x} \in L $$

$$\Longrightarrow \overline{x}n \in L$$

$$\Longrightarrow \overline{xn} \in L
$$

$$\Longrightarrow xn \in \pi^* (L)
$$

Given (1), (2) and (3) above $$\pi^* (L)$$ is an ideal of $$\mathbb{Z}$$ and hence is an ideal of the form$$ a \mathbb{Z}$$ (since all ideals of $$\mathbb{Z}$$ are of this form ...
Question 1: Could someone please critique the above analysis ...or at least let me know if it is basically OK ...

In particular are the logical steps of (3) above OK ... as I am particularly unsure of the following particular steps ...

" ...

$$\Longrightarrow \overline{x} \in L
$$

$$\Longrightarrow \overline{x}n \in L
$$

$$\Longrightarrow \overline{xn} \in L
$$

... ... "

This is all straight-forward: as $\pi$ is a $\Bbb Z$-module homomorphism (homomorphism of abelian groups).

Question 2:

Can someone please help me to show that, as stated in the example, that the ideal $$ \pi^* (L) $$ actually contains $$ p^r \mathbb{Z} $$

$\pi(p^r) = p^r + p^r\Bbb Z = 0 + p^r\Bbb Z = \overline{0} \in L$

Question 3:

Can someone please help me to show that any submodule must have the form $$p^i \mathbb{Z} / p^r \mathbb{Z}$$, and further show that this leads to the fact that no submodules can complement each other ...

Hope someone can help ...

Peter

What can a subgroup of $\Bbb Z$ containing $p^r$ possibly look like? We know it has to be of the form:

$k\Bbb Z = \{\dots,-3k,-2k,-k,0,k,2k,3k,\dots\}$

Since $p^r$ is one of these elements, $p^r = nk$. How many distinct primes can divide either $n$ or $k$?

Hopefully you can see we must have $k = p^i$ where $0 \leq i \leq r$.

(I'll state this another way to underscore its importance: the integers are an abelian group, in fact they are a CYCLIC abelian group. So when we consider them as a ring, we find that any ideal (which is a subgroup of a cyclic group, and hence cyclic) MUST BE PRINCIPAL. The integers are also a integral domain, so they form a principal ideal domain.

In a PID, ideals are all about "divisibility". So the reason the generator of an ideal, $k$, that is a pre-image under $\pi$ of some coset:

$n + p^r\Bbb Z$

is $p^i$, is because $p^i$ has to divide $p^r$.)

This means that under $\pi$ we have the image group ($\Bbb Z$-module):

$\pi(p^i\Bbb Z) = p^i\Bbb Z/p^r\Bbb Z$

Now here the thing: We have a cyclic group (module). It is generated by $\overline{1}$. Any subgroup (sub-module) is going to also be a cyclic group (module).

So let's form two subgroups: $L,L'$.

We have $L = \langle \overline{k}\rangle,\ L' = \langle \overline{k'}\rangle$

From the fore-going discussion, we know that:

$k = p^i,\ k' = p^{i'}$.

So if say, $i \leq i'$, we have $L' \subseteq L$, which shows that if:

$p^i\Bbb Z/p^r\Bbb Z = L \oplus L'$

that either $L \subseteq L'$ or $L' \subseteq L$, in which case $L \cap L' = \overline{0}$ implies one summand must be trivial.

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

I think the notation obscures the simplicity of what is going on, here. Let's take $p = 3, r = 4$. So we are talking about:

$\pi: \Bbb Z \to \Bbb Z_{81}$.

What are the divisors of 81? 1,3,9,27 and 81. The only subgroups of $\Bbb Z_{81}$ are:

$\langle \overline{0} \rangle = \{\overline{0}\}$
$\langle \overline{1} \rangle = \Bbb Z_{81}$
$\langle \overline{3} \rangle$ (this has 27 elements which I decline to list explicitly, I'm sure you can figure it out)
$\langle \overline{9} \rangle = \{\overline{0}, \overline{9},\dots,\overline{72}\}$
$\langle \overline{27} \rangle = \{\overline{0}, \overline{27},\overline{54}\}$

What subgroups of the integers map to these?

$\Bbb Z \to \Bbb Z_{81}$
$3\Bbb Z \to \langle \overline{3} \rangle$
$9\Bbb Z \to \langle \overline{9} \rangle$
$27\Bbb Z \to \langle \overline{27} \rangle$
$81\Bbb Z \to \langle \overline{0} \rangle$

 

[Math Amateur]

Hi Peter,

Your analysis is just fine, but the short answer for Questions 1 and 2 is the lattice isomorphism theorem for modules. Recall that this theorem gives a one-to-one correspondence between the submodules of a quotient module $A/J$ and the submodules of $A$ containing $B$. You can find this theorem (or something similar) in Dummit and Foote's book and Beachy's rings and modules book.

I cannot fully answer your third question yet since there are different meanings of complementation and you have not specified which one you mean. So I'll answer the first part of the question. Recall that if $m$ and $n$ are integers such that $m\Bbb Z \subseteq n\Bbb Z$, then $n|m$. Since $p^r \Bbb Z \subseteq L = a\Bbb Z$, we know that $a|p^r$ and thus $\pi^*(L) = p^j \Bbb Z$ for some nonnegative integer $j \le r$. Since $\pi$ is an onto map,

$\displaystyle L = \pi(\pi^*(L)) = \pi(p^j \Bbb Z) = p^j \Bbb Z/p^r \Bbb Z $.

Since $L$ was arbitrary, it follows that every submodule of $\Bbb Z/p^r \Bbb Z$ is of the form $p^j \Bbb Z/p^r \Bbb Z$.

Thanks for the help Euge ... will review the Lattice Isomorphism Theorem for Modules ... still dealing with your answer to the third question ...

By the way, Berrick and Keating's definition of a complement in the context of a direct sum is as follows:View attachment 2980Thanks again,

Peter

 

[Math Amateur]

These are sets. The elemensts of $p^r\Bbb Z$ are just integers.

Just a note, here. Presumably we are talking about abelian groups, that is $\Bbb Z$-modules. Recall, that any ring $R$ is an $R$-module over itself, and that an $R$-submodule of itself is just an ideal.
This is all straight-forward: as $\pi$ is a $\Bbb Z$-module homomorphism (homomorphism of abelian groups).
$\pi(p^r) = p^r + p^r\Bbb Z = 0 + p^r\Bbb Z = \overline{0} \in L$
What can a subgroup of $\Bbb Z$ containing $p^r$ possibly look like? We know it has to be of the form:

$k\Bbb Z = \{\dots,-3k,-2k,-k,0,k,2k,3k,\dots\}$

Since $p^r$ is one of these elements, $p^r = nk$. How many distinct primes can divide either $n$ or $k$?

Hopefully you can see we must have $k = p^i$ where $0 \leq i \leq r$.

(I'll state this another way to underscore its importance: the integers are an abelian group, in fact they are a CYCLIC abelian group. So when we consider them as a ring, we find that any ideal (which is a subgroup of a cyclic group, and hence cyclic) MUST BE PRINCIPAL. The integers are also a integral domain, so they form a principal ideal domain.

In a PID, ideals are all about "divisibility". So the reason the generator of an ideal, $k$, that is a pre-image under $\pi$ of some coset:

$n + p^r\Bbb Z$

is $p^i$, is because $p^i$ has to divide $p^r$.)

This means that under $\pi$ we have the image group ($\Bbb Z$-module):

$\pi(p^i\Bbb Z) = p^i\Bbb Z/p^r\Bbb Z$

Now here the thing: We have a cyclic group (module). It is generated by $\overline{1}$. Any subgroup (sub-module) is going to also be a cyclic group (module).

So let's form two subgroups: $L,L'$.

We have $L = \langle \overline{k}\rangle,\ L' = \langle \overline{k'}\rangle$

From the fore-going discussion, we know that:

$k = p^i,\ k' = p^{i'}$.

So if say, $i \leq i'$, we have $L' \subseteq L$, which shows that if:

$p^i\Bbb Z/p^r\Bbb Z = L \oplus L'$

that either $L \subseteq L'$ or $L' \subseteq L$, in which case $L \cap L' = \overline{0}$ implies one summand must be trivial.

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

I think the notation obscures the simplicity of what is going on, here. Let's take $p = 3, r = 4$. So we are talking about:

$\pi: \Bbb Z \to \Bbb Z_{81}$.

What are the divisors of 81? 1,3,9,27 and 81. The only subgroups of $\Bbb Z_{81}$ are:

$\langle \overline{0} \rangle = \{\overline{0}\}$
$\langle \overline{1} \rangle = \Bbb Z_{81}$
$\langle \overline{3} \rangle$ (this has 27 elements which I decline to list explicitly, I'm sure you can figure it out)
$\langle \overline{9} \rangle = \{\overline{0}, \overline{9},\dots,\overline{72}\}$
$\langle \overline{27} \rangle = \{\overline{0}, \overline{27},\overline{54}\}$

What subgroups of the integers map to these?

$\Bbb Z \to \Bbb Z_{81}$
$3\Bbb Z \to \langle \overline{3} \rangle$
$9\Bbb Z \to \langle \overline{9} \rangle$
$27\Bbb Z \to \langle \overline{27} \rangle$
$81\Bbb Z \to \langle \overline{0} \rangle$

Thanks for your help, Deveno ... working through your post now ...

BUT immediately, I note, you write:

" ... ... These are sets. The elements of $p^r\Bbb Z$ are just integers. ... ... "

Oh, indeed ... how careless and silly of me ... thanks for pointing that out!

Peter