Tensor Products and the Free Z-module - Bland Propostion 2.2.3 .... ....

[Math Amateur]

Tensor Products and the Free Z-module - Bland Proposition 2.2.3 ... ...

I am reading Paul E. Bland's book "Rings and Their Modules ...

Currently I am focused on Section 2.3 Tensor Products of Modules ... ...

I need some help in order to fully understand the nature of the free Z-module mentioned in Proposition 2.3.3 ...

Proposition 2.3.3 reads as follows:http://mathhelpboards.com/attachments/linear-abstract-algebra-14/5596-tensor-products-free-z-module-bland-propostion-2-2-3-a-bland-proposition-2-3-3-pngMy question is as follows:

Why do the elements of \(\displaystyle F\) have the form \(\displaystyle \sum n_{ (x,y) } (x, y)\) ?

... ... why don't they simply have the form \(\displaystyle \sum (x, y)\) ?

... where does the \(\displaystyle n_{ (x,y) }\) come from ?Hope someone can help ...

Peter

*** EDIT ***

After some reflection and paying more attention to Bland's notation ... I have had a slight re-framing of my question ... same basic problem though ...

Basic problem ... given the free $$\mathbb{Z}$$-module $$F = \mathbb{Z}^{ (M \times N) }$$ on $$M \times N$$

... ... ... how do we determine the elements of F ... ... or more particularly ... why are the elements of the form given by Bland ... namely ... $$\sum n_{ (x, y) } (x, y)$$ ... ...Now ... considering the notation, we are given that ... ... :

\(\displaystyle M^{ ( \Delta ) } \ = \ \bigoplus_{ \Delta } M_{ \alpha } \text{ where } \alpha \in \Delta \)... so this means that ... ...$$F = \mathbb{Z}^{ (M \times N) } \ \ = \ \ \bigoplus_{ M \times N } \mathbb{Z}_{ (x, y) }$$ where $$(x, y) \in M \times N$$Now $$\bigoplus_{ M \times N } \mathbb{Z}_{ (x, y) }$$ seems to me now to mean a copy of $$\mathbb{Z}$$ for each element of $$M \times N$$

... ... BUT ... ... it is more than copies of $$\mathbb{Z}$$ ... it has terms of the nature $$(x, y)$$ in it ... so my problem now is ... how did the $$(x, y)$$ get in there ... ?

... BUT ...

... ... basically the problem remains as follows:

... given the free $$\mathbb{Z}$$-module $$F = \mathbb{Z}^{ (M \times N) }$$ on $$M \times N$$ ...

... ... ... how do we determine the elements of F ... ... or more particularly ... why are the elements of the form given by Bland ... namely ... $$\sum n_{ (x, y) } (x, y)$$ ... ...

Hope someone can clarify the situation ...

Peter

*** NOTES ***

(1) An important note from Bland regarding notation for the above is as follows:http://mathhelpboards.com/attachments/linear-abstract-algebra-14/5597-tensor-products-free-z-module-bland-propostion-2-2-3-a-bland-important-note-notation-png

==================================================

(2) Bland's text above refers to the Remark following Proposition 2.2.6 ... this Remark reads as follows:http://mathhelpboards.com/attachments/linear-abstract-algebra-14/5598-tensor-products-free-z-module-bland-propostion-2-2-3-a-bland-remarks-following-proposition-2-2-6-png

 

[Deveno]

There is a (natural) isomorphism:

$\Bbb Z^{(M \times N)}$ and $\bigoplus\limits_{|M\times N|} \Bbb Z$.

The elements of the former are functions:

$f: M \times N \to \Bbb Z$ with finite support, which we add "point-wise":

$(f+g)(x,y) = f(x,y) + g(x,y)$ (same $x,y$ for each function), and the isomorphism is defined by:

$f \mapsto \sum\limits_{x,y} f(x,y)(x,y)$

(that is the summand *coefficients* $n_{(x,y)}$ are the *images* of the function $f$).

The reason for the "subscript" in the coefficient is so we only add "within the designated corresponding copy" of $\Bbb Z$ and don't "change lanes".

 

[Math Amateur]

Re: Tensor Products and the Free Z-module - Bland Proposition 2.3.3 ... ...

There is a (natural) isomorphism:

$\Bbb Z^{(M \times N)}$ and $\bigoplus\limits_{|M\times N|} \Bbb Z$.

The elements of the former are functions:

$f: M \times N \to \Bbb Z$ with finite support, which we add "point-wise":

$(f+g)(x,y) = f(x,y) + g(x,y)$ (same $x,y$ for each function), and the isomorphism is defined by:

$f \mapsto \sum\limits_{x,y} f(x,y)(x,y)$

(that is the summand *coefficients* $n_{(x,y)}$ are the *images* of the function $f$).

The reason for the "subscript" in the coefficient is so we only add "within the designated corresponding copy" of $\Bbb Z$ and don't "change lanes".

I am reviewing the above post ... and have to confess that I am still puzzling over it a bit ...

Can you please give more detail on how the isomorphism you mentioned is formed and operates ...Specifically ... the isomorphism:$f \mapsto \sum\limits_{x,y} f(x,y)(x,y)$asssumes that elements of $\bigoplus\limits_{|M\times N|} \Bbb Z$ can be put into the form\(\displaystyle \sum\limits_{x,y} f(x,y)(x,y)\)How/why is this the case ... why/how can we do this ...

Peter

 

[Deveno]

Let's look at a finite case, so we can be more explicit.

We take $M = \{1,2,3\}$ (any 3-element set would do, but this particular 3-element set allows for easy "naming" by indices)

and $N = \{1,2\}$ (the same considerations apply).

So we have 6 elements in $M \times N$, namely:

$(1,1),(1,2),(2,1),(2,2),(3,1),(3,2)$.

Here is an element of $\Bbb Z^{M\times N}$:

$f(1,1) = 4$
$f(1,2) = -3$
$f(2,1) = 0$
$f(2,2) = 17$
$f(3,1) = 1$
$f(3,2) = -2$.

Now let's look at $\bigoplus\limits_6 \Bbb Z$, by which we mean the direct sum of 6 copies of the integers.

We have a basis of 6 elements, namely:

$e_1 = (1,0,0,0,0,0)$
$e_2 = (0,1,0,0,0,0)$
$e_3 = (0,0,1,0,0,0)$
$e_4 = (0,0,0,1,0,0)$
$e_5 = (0,0,0,0,1,0)$
$e_6 = (0,0,0,0,0,1)$.

Here, the subscripts are just "placeholders" and since $|M \times N| = 6$, we could re-name them:

$e_1 = e_{(1,1)}$
$e_2 = e_{(1,2)}$
$e_3 = e_{(2,1)}$
$e_4 = e_{(2,2)}$
$e_5 = e_{(3,1)}$
$e_6 = e_{(3,2)}$

Again, the letter "e" does nothing for us, here, we could just regard these as "abstract things" we label by their index:

$e_1 \to (1,1)$
$e_2 \to (1,2)$
$e_3 \to (2,1)$
$e_4 \to (2,2)$
$e_5 \to (3,1)$
$e_6 \to (3,2)$.

So we are going to just take "formal ($\Bbb Z$-) linear combinations" of the elements of $M \times N$, so an element of $\bigoplus\limits_6 \Bbb Z$ might be:

$4(1,1) - 3(1,2) + 0(2,1) + 17(2,2) + 1(3,1) - 2(3,2)$, or, what is probably a more *familiar* form:

$(4,-3,0,17,1,-2)$

Now here, $M$ and $N$ are finite sets, so we can arrange then in a linear sequence. If $M = N = \Bbb R$, for example, no such "linear listing" would be possible, and the best we could do is "tag" each coordinate by its corresponding point in the real plane, which is what the $n_{(x,y)}$ are for. Since $M \times N$ is "just a set" (even a set we might be able to impose some kind of algebraic structure on by other means), the elements

$n_{(x,y)}(x,y)$

are just "formal scalar multiples of formal basis elements".

The free nature of the free abelian group means that the $(x,y)$ have "no particular meaning" (they are just set elements) and we could (if we desired) replace them with symbols like $e_{(x,y)}$ to obtain finite linear combinations of the form:

$\sum n_{(x,y)}e_{(x,y)}$

In my opinion, most writers make too much of this roundabout way of representing tuples with functions, the crucial thing to take away from this is that:

$M^{(\Delta)} \subseteq M^{\Delta}$ and that if $|\Delta| < \infty$ they are equal, and if $\Delta$ is infinite, they are not.