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R/I as a bimodule - Tensor Products - D&F page 366 - Example 2 - R/I as a bimodule

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Dummit and Foote Section 10.4: Tensor Products of Modules. I would appreciate some help in understanding Example 2 on page 366 concerning viewing the quotient ring \(\displaystyle R/I \) as an \(\displaystyle (R/I, R) \)-bimodule.

Example (2) D&F page 366 reads as follows:

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"(2) Let I be an ideal (two sided) in the ring \(\displaystyle R\). Then the quotient ring \(\displaystyle R/I \) is an \(\displaystyle (R/I, R) \)-bimodule. ... ... ...

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Now for \(\displaystyle R/I \) to be a \(\displaystyle (R/I, R) \)-bimodule we require that :

1. \(\displaystyle R/I \) is a left \(\displaystyle R/I \)-module

2. \(\displaystyle R/I \) is a right \(\displaystyle R \)-module

3. (a + I) ( (b+ I) r) = ( (a + I) (b+ I) ) r where a+I, b+I belong to R/I and r is in R.

I have problems with the meaning and rules governing operations on elements in 2 above, and a similar problem with the operations in 3.


Consider now, \(\displaystyle R/I \) as a right \(\displaystyle R \)-module

Following Dummit and Foote's definition of a module on page 337 (see attachment) and following the definition closely and carefully (and adjusting for a right module rather than a left module), for \(\displaystyle M = R/I\) to be a right \(\displaystyle R\)-module we require

(1) \(\displaystyle R/I \)to be an abelian group under the operation +, which is achieved under the normal definition of addition of cosets, visually:

\(\displaystyle (a + I) + (b + I) = (a+b) + I \)

(2) an action of \(\displaystyle R\) on \(\displaystyle R/I \) (that is a map \(\displaystyle R/I \times R \to R/I \)) denoted by \(\displaystyle ( a + I ) r \) for all \(\displaystyle (a + I) \in R/I \text{ and for all } r \in R \) which satisfies:

(a) \(\displaystyle ( a + I ) ( r + s) = (a + I) r + (a + I) s \text{ where } (a + I) \in R/I \text{ and } r, s \in R\)

... ... and so on for conditions (b), (c) and (d) - see D&F page 337 (see attachment)


My question is as follows:

How do we interpret the action \(\displaystyle ( a + I ) r \) , and also how do we interpret, indeed form/calculate expressions like \(\displaystyle (a + I) r \) in expressions (a) above ... also actually in (b), (c), (d) as well

I would appreciate some help.

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
It is natural to set:

$(a + I)r = ar + I$

(since $I$ is a 2-sided ideal, $Ir = I$).