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First, thanks to both Deveno and ThePerfectHacker for helping me to gain a basic understanding of tensor products of modules.
In a chat room discussion ThePerfectHacker suggested I show that $$ {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b $$ where a and b are relatively prime integers - that is where a and b are such that (a,b) = 1 = ra + sb for some integers r,s.
I was aware of a solution to this exercise in Paul Garret's book Abstract Algebra - see Chapter 27 : Tensor Products, page 396 (see pages 395-396 attached) . However, I need help with Garrett's solution since I do not completely understand it ... so I am presenting his solution ... and my alternative solution ... I am hoping someone can indicate a correct solution ...
Note first that on page 395 in Section 27.3 First Examples (see attachment), Paul Garret writes:
" ... ... we emphasize that in $$ M \otimes_R \text{ with } r,s \in R, m \in M \text{ and } n \in N $$, we can always rearrange
$$ (rm) \otimes n = r(m \otimes n) = m \otimes (rn) $$ ... ... ... (1)
Also for $$ r, s \in R $$,
$$ (r + s)(m \otimes n) = rm \otimes n + sm \otimes n $$ ... ... ... (2)
... ... "
but how or why these relations are true, I have very little idea ... can someone please help me to see why these would hold given the basic structure of the tensor product?Now on page 396 (Example 27.3.2 - see attachment) Garret gives the solution to the following exercise:
Show $$ {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b $$ where a and b are relatively prime integers - that is where a and b are such that (a,b) = 1 = ra + sb for some integers r,s.
Garrett proceeds as follows:
For $$ m \in {\mathbb{Z}}_a \text{ and } n \in {\mathbb{Z}}_b $$ we have the following:
$$ m \otimes n $$
$$ = 1 \cdot (m \otimes n) $$
$$ (ra + sb)(m \otimes n) $$
$$ ra(m \otimes n) + sb(m \otimes n) $$ [here I think Garrett has a typo!]
Now at this point (correcting for the typo) Garrett seems to me to reason as follows:
$$ ra(m \otimes n) = ar(m \otimes n) = a(rm \otimes n) $$ by the commutativity of $$ \mathbb{Z} $$ and by (1) above.
Garrett then seems to reason as follows:
$$ a(rm \otimes n) = a.0 = 0 $$
and also similarly
$$ b(m \otimes sn) = b.0 = 0 $$
BUT why are $$ (rm \otimes n) $$ and $$ (m \otimes sn) $$ equal to zero? Can someone please help?
Now, since I did not understand Garrett's solution, I looked for another way to show what was required and proceeded as follows:
$$ ra(m \otimes n) = r(am \otimes n) = r(0 \otimes n) $$ since $$ am = 0 $$
and then continue to reason that $$ r(0 \otimes n) = r.0 = 0 $$
BUT ... I am really uncertain as to whether $$ (0 \otimes n) = 0 $$
However, if I am correct then a similar chain of reasoning then proceeds and follows:
$$ sb(m \otimes n) = s(m \otimes bn) = r(m \otimes 0) $$ since $$ bn = 0 $$
and so then $$ ra(m \otimes n) + sb(m \otimes n) $$ = 0 + 0 = 0
Can someone please clarify the above for me?
Peter
In a chat room discussion ThePerfectHacker suggested I show that $$ {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b $$ where a and b are relatively prime integers - that is where a and b are such that (a,b) = 1 = ra + sb for some integers r,s.
I was aware of a solution to this exercise in Paul Garret's book Abstract Algebra - see Chapter 27 : Tensor Products, page 396 (see pages 395-396 attached) . However, I need help with Garrett's solution since I do not completely understand it ... so I am presenting his solution ... and my alternative solution ... I am hoping someone can indicate a correct solution ...
Note first that on page 395 in Section 27.3 First Examples (see attachment), Paul Garret writes:
" ... ... we emphasize that in $$ M \otimes_R \text{ with } r,s \in R, m \in M \text{ and } n \in N $$, we can always rearrange
$$ (rm) \otimes n = r(m \otimes n) = m \otimes (rn) $$ ... ... ... (1)
Also for $$ r, s \in R $$,
$$ (r + s)(m \otimes n) = rm \otimes n + sm \otimes n $$ ... ... ... (2)
... ... "
but how or why these relations are true, I have very little idea ... can someone please help me to see why these would hold given the basic structure of the tensor product?Now on page 396 (Example 27.3.2 - see attachment) Garret gives the solution to the following exercise:
Show $$ {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b $$ where a and b are relatively prime integers - that is where a and b are such that (a,b) = 1 = ra + sb for some integers r,s.
Garrett proceeds as follows:
For $$ m \in {\mathbb{Z}}_a \text{ and } n \in {\mathbb{Z}}_b $$ we have the following:
$$ m \otimes n $$
$$ = 1 \cdot (m \otimes n) $$
$$ (ra + sb)(m \otimes n) $$
$$ ra(m \otimes n) + sb(m \otimes n) $$ [here I think Garrett has a typo!]
Now at this point (correcting for the typo) Garrett seems to me to reason as follows:
$$ ra(m \otimes n) = ar(m \otimes n) = a(rm \otimes n) $$ by the commutativity of $$ \mathbb{Z} $$ and by (1) above.
Garrett then seems to reason as follows:
$$ a(rm \otimes n) = a.0 = 0 $$
and also similarly
$$ b(m \otimes sn) = b.0 = 0 $$
BUT why are $$ (rm \otimes n) $$ and $$ (m \otimes sn) $$ equal to zero? Can someone please help?
Now, since I did not understand Garrett's solution, I looked for another way to show what was required and proceeded as follows:
$$ ra(m \otimes n) = r(am \otimes n) = r(0 \otimes n) $$ since $$ am = 0 $$
and then continue to reason that $$ r(0 \otimes n) = r.0 = 0 $$
BUT ... I am really uncertain as to whether $$ (0 \otimes n) = 0 $$
However, if I am correct then a similar chain of reasoning then proceeds and follows:
$$ sb(m \otimes n) = s(m \otimes bn) = r(m \otimes 0) $$ since $$ bn = 0 $$
and so then $$ ra(m \otimes n) + sb(m \otimes n) $$ = 0 + 0 = 0
Can someone please clarify the above for me?
Peter
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