Free module

amfulger

Is $M=\mathbb{Z}^{\mathbb{N}}$ a free $\mathbb{Z}$ module?
----------------------------------------------------------------------------------
To explain a bit more:

$\mathbb{Z}^{\mathbb{N}}$ is $\mathbb{Z} \times \mathbb{Z} \times \ldots \times \mathbb{Z}$ a countable number of times.

The elements of M are $(x_n)_{n \in \mathbb{N}}$ , with $x_n \in \mathbb{Z}$ .
M has a group structure:
$(x_n)_{n\in\mathbb{N}}+(y_n)_{n\in\mathbb{N}}=(x_n+y_n)_{n\in\mathbb{N}}$ .
$(x_n=0)_{n\in\mathbb{N}}$ is the neutral element of M.

We define $n\cdot x=x+\ldots +x$ (n times) for n in N and (-1)x=-x for all x in M.
Thus we can define zx for all $z \in \mathbb{Z}$ and $x \in M$ . This means that M is a $\mathbb{Z}$ module

M is a free $\mathbb{Z}$ module, if it has a basis. That is if there exists B a subset of M such as:
1. for all x in M there exists a natural n, integers $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$ in B such as $x=\sum_{k=1}^n a_k\cdot b_k$ .
2. for all n, integers $a_1,\ldots,a_n$ and distinct elements of B, $b_1,\ldots,b_n$ , the condition $\sum_{k=1}^n a_k\cdot b_k=0$ , implies $a_k=0$ for all $k=\overline{1,n}$ .

**Posted:** Sat Mar 27, 2004 11:52 pm

amfulger

The following proof is due to "Kaplanski - Infinite Abelian Groups".
M is not free.

Definition: Let R be a ring and I be a set.
$R^I=\{(x_i)_{i\in I}| x_i\in R\ \forall i\in I\}$ is called the direct product (or complete direct sum) of R idexed by I.
$R^{(I)}=\{(x_i)_{i\in I}| x_i\in R\ \forall i\in I$ and only finitely many $x_i$ 's aren't 0} is called the direct sum of R indexed by I.

Suppose M is free. Let B be a basis for M.
Suppose B is countable. Then it is easy to see that the $\mathbb{Z}$ module M generated by B is countable. But $|M|=|\mathbb{Z}|^{|\mathbb{N}|}=\aleph_{0}^{\aleph_0}=\aleph_c$ . So M is uncountable. We get a contradiction.
So B is uncountable.

Let p be a prime number in $\mathbb{N}$ .
Let $v:\mathbb{Z}\to\mathbb{N}\cup \{\infty\}$ defined by $p^v(n)|n$ and $p^{v(n)+1}\nmid n$ for all $n\in \mathbb{Z}^*$ . Also, $v(0)=\infty$ .

Let $\displaystyle S=\{(x_n)_{n\in\mathbb{N}}\in M| \lim_{n\to\infty}v(x_n)=\infty \}$ .

Let $f:M\to M$ defined by $f((x_n)_{n\in \mathbb{N}})=(p^n\cdot x_n)_{n\in \mathbb{N}}$ . f is an injective morphism of M into S. The image of M by f must be an uncountable generated (cannot be countable generated) free submodule of S, so S cannot be countable generated. From this we should infer that S/pS cannot be countable generated as an $\mathbb{Z}/p\mathbb{Z}$ module.

On the other hand, we prove that $S/pS\simeq \mathbb{Z}^{(\mathbb{N})}$ , which is free and countable generated.
By the definition of S, for all $(x_n)_{n\in\mathbb{N}}\in S$ there exists $N_x$ such that for $n>N_x,\ p|x_n$ . So for $n>N_x$ , $\hat{x}_n=\hat{0}$ . So S/pS contains only finite combinations of $\hat{e}_k$ 's ( $\hat{e}_k$ is the element of M/pM that has $\hat{1}$ on the $k^{th}$ place and $\hat{0}$ in the rest). Obviously $e_k \in S\ \forall k\in\mathbb{N}$ , so $\hat{e}_k\in S/pS\ \forall k\in\mathbb{N}$ .
This proves that S/pS is isomorphic to the required direct sum which happens to be countable generated.

The problem is over if we prove the implication: S not countable generated as a $\mathbb{Z}$ module implies S/pS not countable generated as a $\mathbb{Z}/p\mathbb{Z}$ module. In Kaplanski's book, this is given as "manifestly" type implication.

**Posted:** Fri May 28, 2004 1:43 pm

Soarer

**Posted:** Thu Mar 20, 2008 6:36 am

_________________
Hallo! Mein name ist Soarer und ich komme aus Hong Kong. Ich spreche Chinesisch, Englisch und ein bisschen Deutsch. Ich mag Mathematik!
Please correct my English or German.

jmerry

See also here and the various links.
It's only the second topic from the top- why didn't you see it?

**Posted:** Thu Mar 20, 2008 1:02 pm

Soarer

I actually reached this thread from that link, and I started to go over the proof and got stuck at that place... I'm sorry if I'm missing something obvious.

**Posted:** Thu Mar 20, 2008 1:08 pm

_________________
Hallo! Mein name ist Soarer und ich komme aus Hong Kong. Ich spreche Chinesisch, Englisch und ein bisschen Deutsch. Ich mag Mathematik!
Please correct my English or German.

jmerry

Oops- I didn't check the date on the original post.

Basically, if you have an uncountably generated free module, the generators are distinct mod $p$ , and an uncountable abelian group isn't countably generated.

**Posted:** Thu Mar 20, 2008 11:35 pm

Soarer

Oops.. thanks.

**Posted:** Thu Mar 20, 2008 11:44 pm

_________________
Hallo! Mein name ist Soarer und ich komme aus Hong Kong. Ich spreche Chinesisch, Englisch und ein bisschen Deutsch. Ich mag Mathematik!
Please correct my English or German.