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- Jun 22, 2012

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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 \(\displaystyle {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b \) 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 \(\displaystyle M \otimes_R \text{ with } r,s \in R, m \in M \text{ and } n \in N \), we can always rearrange

\(\displaystyle (rm) \otimes n = r(m \otimes n) = m \otimes (rn) \) ... ... ... (1)

Also for \(\displaystyle r, s \in R \),

\(\displaystyle (r + s)(m \otimes n) = rm \otimes n + sm \otimes n \) ... ... ... (2)

... ... "

Now on page 396 (Example 27.3.2 - see attachment) Garret gives the solution to the following exercise:

Show \(\displaystyle {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b \) 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 \(\displaystyle m \in {\mathbb{Z}}_a \text{ and } n \in {\mathbb{Z}}_b \) we have the following:

\(\displaystyle m \otimes n \)

\(\displaystyle = 1 \cdot (m \otimes n) \)

\(\displaystyle (ra + sb)(m \otimes n) \)

\(\displaystyle ra(m \otimes n) + sb(m \otimes n) \) [here I think Garrett has a typo!]

Now at this point (correcting for the typo) Garrett seems to me to reason as follows:

\(\displaystyle ra(m \otimes n) = ar(m \otimes n) = a(rm \otimes n) \) by the commutativity of \(\displaystyle \mathbb{Z} \) and by (1) above.

Garrett then seems to reason as follows:

\(\displaystyle a(rm \otimes n) = a.0 = 0 \)

and also similarly

\(\displaystyle b(m \otimes sn) = b.0 = 0 \)

Now, since I did not understand Garrett's solution, I looked for another way to show what was required and proceeded as follows:

\(\displaystyle ra(m \otimes n) = r(am \otimes n) = r(0 \otimes n) \) since \(\displaystyle am = 0 \)

and then continue to reason that \(\displaystyle r(0 \otimes n) = r.0 = 0 \)

BUT ... I am really uncertain as to whether \(\displaystyle (0 \otimes n) = 0 \)

However, if I am correct then a similar chain of reasoning then proceeds and follows:

\(\displaystyle sb(m \otimes n) = s(m \otimes bn) = r(m \otimes 0) \) since \(\displaystyle bn = 0 \)

and so then \(\displaystyle ra(m \otimes n) + sb(m \otimes n) \) = 0 + 0 = 0

Can someone please clarify the above for me?

Peter

In a chat room discussion ThePerfectHacker suggested I show that \(\displaystyle {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b \) 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 \(\displaystyle M \otimes_R \text{ with } r,s \in R, m \in M \text{ and } n \in N \), we can always rearrange

\(\displaystyle (rm) \otimes n = r(m \otimes n) = m \otimes (rn) \) ... ... ... (1)

Also for \(\displaystyle r, s \in R \),

\(\displaystyle (r + s)(m \otimes n) = rm \otimes n + sm \otimes n \) ... ... ... (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 \(\displaystyle {\mathbb{Z}}_a \otimes_\mathbb{Z} {\mathbb{Z}}_b \) 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 \(\displaystyle m \in {\mathbb{Z}}_a \text{ and } n \in {\mathbb{Z}}_b \) we have the following:

\(\displaystyle m \otimes n \)

\(\displaystyle = 1 \cdot (m \otimes n) \)

\(\displaystyle (ra + sb)(m \otimes n) \)

\(\displaystyle ra(m \otimes n) + sb(m \otimes n) \) [here I think Garrett has a typo!]

Now at this point (correcting for the typo) Garrett seems to me to reason as follows:

\(\displaystyle ra(m \otimes n) = ar(m \otimes n) = a(rm \otimes n) \) by the commutativity of \(\displaystyle \mathbb{Z} \) and by (1) above.

Garrett then seems to reason as follows:

\(\displaystyle a(rm \otimes n) = a.0 = 0 \)

and also similarly

\(\displaystyle b(m \otimes sn) = b.0 = 0 \)

**BUT why are \(\displaystyle (rm \otimes n) \) and \(\displaystyle (m \otimes sn) \) 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:

\(\displaystyle ra(m \otimes n) = r(am \otimes n) = r(0 \otimes n) \) since \(\displaystyle am = 0 \)

and then continue to reason that \(\displaystyle r(0 \otimes n) = r.0 = 0 \)

BUT ... I am really uncertain as to whether \(\displaystyle (0 \otimes n) = 0 \)

However, if I am correct then a similar chain of reasoning then proceeds and follows:

\(\displaystyle sb(m \otimes n) = s(m \otimes bn) = r(m \otimes 0) \) since \(\displaystyle bn = 0 \)

and so then \(\displaystyle ra(m \otimes n) + sb(m \otimes n) \) = 0 + 0 = 0

Can someone please clarify the above for me?

Peter

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