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- Jun 22, 2012

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I am reading Dummit and Foote Section 10.4: Tensor Products of Modules.

I would appreciate some help in understanding Example (8) on page 366 concerning viewing the quotient ring \(\displaystyle R/I \) as an \(\displaystyle (R/I, R) \)-bimodule.

Example (8) D&F page 370 reads as follows: (see attachment)

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"Let \(\displaystyle R\) be a ring (not necessarily commutative). Let \(\displaystyle I\) be a two-sided ideal in \(\displaystyle R\), and let \(\displaystyle N\) be a left \(\displaystyle R\)-module. Then, as previously mentioned, \(\displaystyle R/I\) is an \(\displaystyle (R/I, R)\)-bimodule, so the tensor product is a left \(\displaystyle R/I \) module. ... ... ...

... ... The tensor product is generated as an abelian group by simple tensors

\(\displaystyle (r \ mod \ I ) \otimes n = r(1 \otimes n) \text{ for } r \in R \text{ and } n \in N \) ... ... ... ... (1)

viewing the \(\displaystyle R/I\) module tensor product as an \(\displaystyle R\)-module on which \(\displaystyle I\) acts trivially" (???)

... ... ...

----------------------------------------------------------------------------

I am seeking help in order to understand how and why

\(\displaystyle (r \ mod \ I ) \otimes n = r(1 \otimes n) \text{ for } r \in R \text{ and } n \in N \) ... ... ... ... (1)

is the case, and also what D&F mean by "viewing the \(\displaystyle R/I\) module tensor product as an \(\displaystyle R\)-module on which \(\displaystyle I\) acts trivially"

My understanding of how the LHS of (1) may be transformed into something like the RHS is as follows:

Following D&F page 367 (see attachment) after the definition of the standard \(\displaystyle R\)-module structure we read (approximately) the following:

For a situation where \(\displaystyle N\) is a left \(\displaystyle R\)-module and \(\displaystyle M\) is an \(\displaystyle (S, R)\)-bimodule, we have that the \(\displaystyle (S, R)\)-bimodule structure of \(\displaystyle M\) implies that

\(\displaystyle s( \sum_{Finite} m_i \otimes n_i ) = \sum_{Finite} s m_i \otimes n_i \) ... ... ... (10.8)

gives a well-defined action of S on \(\displaystyle M \otimes_R N \) as a left S-module.

Rewriting (10.8) for a simple or elementary tensor (and switching LHS and RHS) we have:

\(\displaystyle sm \otimes n = s(m \otimes n) \) ... ... ... ... (10.8)'

Now, in Example 8 we have:

\(\displaystyle N\) is a left \(\displaystyle R\)-module

and

\(\displaystyle M\) = \(\displaystyle R/I \) is an \(\displaystyle (R/I, R)\)-bimodule

and

\(\displaystyle R/I \otimes_R N \) is a left \(\displaystyle R/I\)-module

Thus in Example 8 we have:

\(\displaystyle (r \ mod \ I ) \otimes n = (r \ mod \ I ) \cdot 1_{R/I} \otimes n \)

and so applying (10.8)' we get

\(\displaystyle ( r \ mod \ I ) \otimes n = ((r \ mod \ I ) \cdot 1_{R/I}) \otimes n = (r \ mod \ I )( 1_{R/I} \otimes n) \) ... ... ... ... (2)

Would appreciate help with the above.

Peter

I would appreciate some help in understanding Example (8) on page 366 concerning viewing the quotient ring \(\displaystyle R/I \) as an \(\displaystyle (R/I, R) \)-bimodule.

Example (8) D&F page 370 reads as follows: (see attachment)

-------------------------------------------------------------------------------

"Let \(\displaystyle R\) be a ring (not necessarily commutative). Let \(\displaystyle I\) be a two-sided ideal in \(\displaystyle R\), and let \(\displaystyle N\) be a left \(\displaystyle R\)-module. Then, as previously mentioned, \(\displaystyle R/I\) is an \(\displaystyle (R/I, R)\)-bimodule, so the tensor product is a left \(\displaystyle R/I \) module. ... ... ...

... ... The tensor product is generated as an abelian group by simple tensors

\(\displaystyle (r \ mod \ I ) \otimes n = r(1 \otimes n) \text{ for } r \in R \text{ and } n \in N \) ... ... ... ... (1)

viewing the \(\displaystyle R/I\) module tensor product as an \(\displaystyle R\)-module on which \(\displaystyle I\) acts trivially" (???)

... ... ...

----------------------------------------------------------------------------

I am seeking help in order to understand how and why

\(\displaystyle (r \ mod \ I ) \otimes n = r(1 \otimes n) \text{ for } r \in R \text{ and } n \in N \) ... ... ... ... (1)

is the case, and also what D&F mean by "viewing the \(\displaystyle R/I\) module tensor product as an \(\displaystyle R\)-module on which \(\displaystyle I\) acts trivially"

My understanding of how the LHS of (1) may be transformed into something like the RHS is as follows:

Following D&F page 367 (see attachment) after the definition of the standard \(\displaystyle R\)-module structure we read (approximately) the following:

For a situation where \(\displaystyle N\) is a left \(\displaystyle R\)-module and \(\displaystyle M\) is an \(\displaystyle (S, R)\)-bimodule, we have that the \(\displaystyle (S, R)\)-bimodule structure of \(\displaystyle M\) implies that

\(\displaystyle s( \sum_{Finite} m_i \otimes n_i ) = \sum_{Finite} s m_i \otimes n_i \) ... ... ... (10.8)

gives a well-defined action of S on \(\displaystyle M \otimes_R N \) as a left S-module.

Rewriting (10.8) for a simple or elementary tensor (and switching LHS and RHS) we have:

\(\displaystyle sm \otimes n = s(m \otimes n) \) ... ... ... ... (10.8)'

Now, in Example 8 we have:

\(\displaystyle N\) is a left \(\displaystyle R\)-module

and

\(\displaystyle M\) = \(\displaystyle R/I \) is an \(\displaystyle (R/I, R)\)-bimodule

and

\(\displaystyle R/I \otimes_R N \) is a left \(\displaystyle R/I\)-module

Thus in Example 8 we have:

\(\displaystyle (r \ mod \ I ) \otimes n = (r \ mod \ I ) \cdot 1_{R/I} \otimes n \)

and so applying (10.8)' we get

\(\displaystyle ( r \ mod \ I ) \otimes n = ((r \ mod \ I ) \cdot 1_{R/I}) \otimes n = (r \ mod \ I )( 1_{R/I} \otimes n) \) ... ... ... ... (2)

**Can someone please confirm (or otherwise) that the above is correct?**

Assuming I am correct in what I have done so far, then can someone help me with how we get from (2) to relation (1)?

... ... and, of course, how do we justify the steps from (2) to (1)?

Assuming I am correct in what I have done so far, then can someone help me with how we get from (2) to relation (1)?

... ... and, of course, how do we justify the steps from (2) to (1)?

Would appreciate help with the above.

Peter

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