MHB Understanding D&F Example 2: R/I Bimodule on Page 366

[Math Amateur]

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 $$R/I $$ as an $$ (R/I, R) $$-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 $$R$$. Then the quotient ring $$R/I $$ is an $$ (R/I, R) $$-bimodule. ... ... ...

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

1. $$ R/I $$ is a left $$ R/I $$-module

2. $$ R/I $$ is a right $$ R $$-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, $$ R/I $$ as a right $$ R $$-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 $$M = R/I$$ to be a right $$R$$-module we require

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

$$ (a + I) + (b + I) = (a+b) + I $$

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

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

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

I would appreciate some help.

Peter

 

[Deveno]

It is natural to set:

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

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