Math Amateur
Gold Member
MHB
- 3,920
- 48
I am reading Dummit and Foote Section 10.4: Tensor Products of Modules.
I am currently studying Example 2, page 363 (see attachment) and I am trying to closely relate the example to Theorem 8 and the D&F text on extension of the scalars preceding Theorem 8 on pages 359-362)
In Example 2 (see attachment) we have
$$ R = \mathbb{Z}, \ S = \mathbb{Q}, \text{ and } A
$$ is a finite abelian group.
Since A is an abelian group, it is the $$ \mathbb{Z} $$-module N in the notation of Theorem 8.
Given the above and following the construction of D&F pages 359-362 (and Theorem 8) - we can form the tensor product of S and N over R, so doing this we have:
$$ S \otimes_R N = \mathbb{Q} \otimes_\mathbb{Z} A $$
That is fine ... BUT ..
D&F say the following:
"In this case the $$ \mathbb{Q} $$-module $$ \mathbb{Q} \otimes_\mathbb{Z} A $$ is 0."
But D&F in saying this are calling the tensor product of S and N (in this case $$ \mathbb{Q} $$ and A) the extension of the scalars from N!
However, after some careful study and considerable generous help, I had formed the view that the extension of the scalars in Theorem 8 was, in fact, the S_module L and not the S-module $$ S \otimes_R N $$
Can someone please clarify this for me?
Further to the above, D&F go on to write:
"By Theorem 8, we see again that any homomorphism from a finite abelian group into a rational vector space is the zero map"
Is this statement referring to the map $$ \phi \ : \ N \to L $$? Why?
If not, then what map in Theorem 8 is it referring to? Why?
Hoping someone can help.
Peter
I am currently studying Example 2, page 363 (see attachment) and I am trying to closely relate the example to Theorem 8 and the D&F text on extension of the scalars preceding Theorem 8 on pages 359-362)
In Example 2 (see attachment) we have
$$ R = \mathbb{Z}, \ S = \mathbb{Q}, \text{ and } A
$$ is a finite abelian group.
Since A is an abelian group, it is the $$ \mathbb{Z} $$-module N in the notation of Theorem 8.
Given the above and following the construction of D&F pages 359-362 (and Theorem 8) - we can form the tensor product of S and N over R, so doing this we have:
$$ S \otimes_R N = \mathbb{Q} \otimes_\mathbb{Z} A $$
That is fine ... BUT ..
D&F say the following:
"In this case the $$ \mathbb{Q} $$-module $$ \mathbb{Q} \otimes_\mathbb{Z} A $$ is 0."
But D&F in saying this are calling the tensor product of S and N (in this case $$ \mathbb{Q} $$ and A) the extension of the scalars from N!
However, after some careful study and considerable generous help, I had formed the view that the extension of the scalars in Theorem 8 was, in fact, the S_module L and not the S-module $$ S \otimes_R N $$
Can someone please clarify this for me?
Further to the above, D&F go on to write:
"By Theorem 8, we see again that any homomorphism from a finite abelian group into a rational vector space is the zero map"
Is this statement referring to the map $$ \phi \ : \ N \to L $$? Why?
If not, then what map in Theorem 8 is it referring to? Why?
Hoping someone can help.
Peter