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Tensor Products - Example 8 - Dummit and Foote - Section 10.4, page 370

Peter

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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 (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)?


Would appreciate help with the above.

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
First let's examine the statement:

"....$R/I$ tensor product as an $R$-module on which $I$ acts trivially."

We have a simple tensor, say:

$(r + I) \otimes n$.

If $x \in I$, then $x + I = 0 + I = I$, whence:

$(x + I)((r + I) \otimes n = [(x + I)(r + I)] \otimes n$

$= [(0 + I)(r + I)] \otimes n = (0r + I) \otimes n$

$=0_{R/I} \otimes n = 0$.

This is what is meant by "$I$ acts trivially", it induces a 0-map.

Recall that for any ring homomorphism $f: R \to S$ we can define an $S$-module $M$ as an $R$-module, by setting:

$rm = f(r)m$, and the fact that $f$ is a ring homomorphism automatically assures us the module axioms will be satisfied.

There is nothing "out of the ordinary" using the ring homomorphism $R \to R/I$.

So our $R$-action on $R/I \otimes_R N$ is given by:

$r(r' + I)\otimes n = [(r + I)(r' + I)]\otimes n = (rr' + I)\otimes n$.

If $r = x \in I$ this becomes:

$(xr' + I) \otimes n = (0 + I)\otimes n = 0$, since $xr' \in I$ (since $I$ is an ideal).

Given $r \in R$ it is also clear that:

$r + I = r1_R + I = r(1_R + I) = r(1_{R/I})$

Hence we have:

$r \text{ mod } I\otimes n = r(1_{R/I} \otimes n)$.

(Dummit and Foote really ought to be clear about "which 1" they are using, since we have 2 different rings floating around).

Note that they replace "bilinear" with "balanced map" which is the "corresponding idea" to a bilinear map when the $R$-action is on "two different sides", and $R$ is not necessarily commutative (such maps are also called "middle linear").

The universal mapping property of the tensor product actually applies most generally to $R$-balanced maps, and the $R$-bilinear maps most often given as examples are "special cases".
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
First let's examine the statement:

"....$R/I$ tensor product as an $R$-module on which $I$ acts trivially."

We have a simple tensor, say:

$(r + I) \otimes n$.

If $x \in I$, then $x + I = 0 + I = I$, whence:

$(x + I)((r + I) \otimes n = [(x + I)(r + I)] \otimes n$

$= [(0 + I)(r + I)] \otimes n = (0r + I) \otimes n$

$=0_{R/I} \otimes n = 0$.

This is what is meant by "$I$ acts trivially", it induces a 0-map.

Recall that for any ring homomorphism $f: R \to S$ we can define an $S$-module $M$ as an $R$-module, by setting:

$rm = f(r)m$, and the fact that $f$ is a ring homomorphism automatically assures us the module axioms will be satisfied.

There is nothing "out of the ordinary" using the ring homomorphism $R \to R/I$.

So our $R$-action on $R/I \otimes_R N$ is given by:

$r(r' + I)\otimes n = [(r + I)(r' + I)]\otimes n = (rr' + I)\otimes n$.

If $r = x \in I$ this becomes:

$(xr' + I) \otimes n = (0 + I)\otimes n = 0$, since $xr' \in I$ (since $I$ is an ideal).

Given $r \in R$ it is also clear that:

$r + I = r1_R + I = r(1_R + I) = r(1_{R/I})$

Hence we have:

$r \text{ mod } I\otimes n = r(1_{R/I} \otimes n)$.

(Dummit and Foote really ought to be clear about "which 1" they are using, since we have 2 different rings floating around).

Note that they replace "bilinear" with "balanced map" which is the "corresponding idea" to a bilinear map when the $R$-action is on "two different sides", and $R$ is not necessarily commutative (such maps are also called "middle linear").

The universal mapping property of the tensor product actually applies most generally to $R$-balanced maps, and the $R$-bilinear maps most often given as examples are "special cases".
Thanks for the help, Deveno.

Just a minor clarification:

You write:

"Given $r \in R$ it is also clear that:

\(\displaystyle r + I = r1_R + I = r(1_R + I) = r(1_{R/I}) \)

I am uncertain regarding how you justify the step:

\(\displaystyle r1_R + I = r(1_R + I) = r(1_{R/I}) \)

I can see that \(\displaystyle r 1_R + I = (r +I)(1_R + I) \) but I am unsure from there ... unless we are replacing r with r + I ... ???

Can you please explain/clarify?

Another, but I suspect related issue ... I find it disconcerting when you write:

\(\displaystyle r(r' + I)\otimes n = [(r + I)(r' + I)]\otimes n \)

since in (2), if we interpret it literally we have r = r +I, which seems a little weird ...

Can you clarify ...

... ... Maybe it has something to do with notational complexities surrounding the definition of the action ... but anyway, as I said, I find it disconcerting since r may become r + 1 or vice versa at various points in the argument ... hope you can help ..

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
The expression:

$r(1 + I)$

comes from the NATURAL $R$-action of $R/I$ as an $R$-module, which is the NATURAL action of $R$ on an $S$-module $M$ when we have a homomorphism $f:R \to S$.

This doesn't make sense "in rings" but it DOES make sense in MODULES. So if we want to "mix and match" ring structures, modules are a natural setting. We do need some type of homomorphism connecting them, or else we cannot guarantee our "hybrid structure" will be well-behaved (in this case, we usually resort to the "default ring" $\Bbb Z$ where we always have some kind of compatibility).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
The expression:

$r(1 + I)$

comes from the NATURAL $R$-action of $R/I$ as an $R$-module, which is the NATURAL action of $R$ on an $S$-module $M$ when we have a homomorphism $f:R \to S$.

This doesn't make sense "in rings" but it DOES make sense in MODULES. So if we want to "mix and match" ring structures, modules are a natural setting. We do need some type of homomorphism connecting them, or else we cannot guarantee our "hybrid structure" will be well-behaved (in this case, we usually resort to the "default ring" $\Bbb Z$ where we always have some kind of compatibility).
Thanks Deveno ... but, I am not following you ... my apologies I am probably being somewhat slow ... sorry ...

To try to follow your remark:

""$r(1 + I)$

comes from the NATURAL $R$-action of $R/I$ as an $R$-module... ..."


I went to D&F's definition of a left R-module on page 337 ... following D&F we have ...

The (natural ?) R-action of R on M = R/I (that is a map

\(\displaystyle R \times R/I \to R/I \)) denoted by

\(\displaystyle rm = r(r' + I) \) where \(\displaystyle r, r' \in R \) and \(\displaystyle m = r' + I \in R/I \) ... ... "

so wouldn't the R-action of R/I as a left R-module be

\(\displaystyle rm = r(r' + I) \)?

Can you help?

I suspect I am missing something ... and also suspect I need to bring in the mapping \(\displaystyle f: R \to S \), but unsure ...

Peter


PS - EDIT

I suspect I need to carefully reflect on the remarks in D&F's Examples (1) and (2) on page 366, which I am now engaged upon in an attempt to clarify things ...
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Thanks Deveno ... but, I am not following you ... my apologies I am probably being somewhat slow ... sorry ...

To try to follow your remark:

""$r(1 + I)$

comes from the NATURAL $R$-action of $R/I$ as an $R$-module... ..."


I went to D&F's definition of a left R-module on page 337 ... following D&F we have ...

The (natural ?) R-action of R on M = R/I (that is a map

\(\displaystyle R \times R/I \to R/I \)) denoted by

\(\displaystyle rm = r(r' + I) \) where \(\displaystyle r, r' \in R \) and \(\displaystyle m = r' + I \in R/I \) ... ... "

so wouldn't the R-action of R/I as a left R-module be

\(\displaystyle rm = r(r' + I) \)?

Can you help?

I suspect I am missing something ... and also suspect I need to bring in the mapping \(\displaystyle f: R \to S \), but unsure ...

Peter


PS - EDIT

I suspect I need to carefully reflect on the remarks in D&F's Examples (1) and (2) on page 366, which I am now engaged upon in an attempt to clarify things ...
You're almost there. That is indeed the natural action, but clearly any element:

$r + I = (r + I)(1 + I)$, and $(r + I)(1 + I)$ is the "actual" action of:

$r$ on $(1 + I)$.

So in considering acting on elements $r \text{ mod } I\otimes n$, we just have to consider an $R$ action on $1 \text{ mod } I\otimes n$, and only $r \in R - I$ actually "do anything" (the elements of $I$ have trivial action, that is: annihilate).

*******

$R/I$ is not, by itself, an $R$-module, we need to DEFINE:

$r(r'+I)$.

But the definition:

$r(r'+I) = rr'+I$ is "natural" and satisfies all the usual axioms:

$r[(r'+I) + (r''+I)] = r[(r'+r'')+I] = r(r'+r'') + I = (rr'+rr'') + I$

$= [(rr') + I] + [(rr'') + I] = r(r' + I) + r(r'' + I)$ and

$(r + r')(r'' + I) = (r + r')r'' + I = (rr' + rr'') + I$

$= (rr' + I) + (rr'' + I) = r(r'' + I) + r'(r'' + I)$, as well as:

$r(r'(r''+I)) = r[(r'r'') + I] = r(r'r'') + I = (rr')r" + I = (rr')(r'' + I)$, and:

$1_R(r + I) = (1_Rr) + I = r + I$.