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Tensor Products - Dummit and Foote Section 10.4, Example 2, page 363

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 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

\(\displaystyle R = \mathbb{Z}, \ S = \mathbb{Q}, \text{ and } A
\) is a finite abelian group.

Since A is an abelian group, it is the \(\displaystyle \mathbb{Z} \)-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:

\(\displaystyle S \otimes_R N = \mathbb{Q} \otimes_\mathbb{Z} A \)

That is fine ... BUT ..

D&F say the following:

"In this case the \(\displaystyle \mathbb{Q} \)-module \(\displaystyle \mathbb{Q} \otimes_\mathbb{Z} A \) is 0."

But D&F in saying this are calling the tensor product of S and N (in this case \(\displaystyle \mathbb{Q} \) 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 \(\displaystyle S \otimes_R N \)

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 \(\displaystyle \phi \ : \ N \to L \)? Why?

If not, then what map in Theorem 8 is it referring to? Why?

Hoping someone can help.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
If a module is 0, any homomorphism to that module is the 0-map, and any composition which factors through a 0-map is likewise 0.

So since the tensor product $\Bbb Q \otimes_{\Bbb Z} A = 0$, the "extension of scalars" of the "integer scalars" of a finite abelian group (which are "actually" the scalars of some finite cyclic group, since a finite abelian group is a direct product of finite cyclic groups) results in sending these scalars to the only scalar in $\Bbb Q$ with finite additive order: 0.

This means $L = 0$, as well, because any bilinear map $B:\Bbb Q \times A \to L$ has to be the 0-map:

$a \in A$ has finite order, say $k$. Therefore, just from bilinearity (over $\Bbb Z$):

$B(q,a) = B(\frac{qk}{k},a) = kB(\frac{q}{k},a) = B(\frac{q}{k},ka) = B(\frac{q}{k},0)$

$= 0(B(\frac{q}{k},0)) = 0$

Since $B = \phi \otimes = 0$ it is clear that ALL these maps are trivial.

You can think of it this way: the tensor product lets us "multiply" finite abelian groups by rational numbers. But finite abelian group elements are annihilated by finite integers, and in $\Bbb Q$, all the non-zero elements are UNITS, so $k$ (the annihilator of $a \in A$ is "muliplcatively the same" as $1/k$ in $\Bbb Q$, that is: bilinearity lets us split 1 into $k$ and $1/k$, one factor goes into $\Bbb Q$ (where it "does nothing") and the other factor goes into $A$ where it wipes it out.

It is not hard to see that $F \otimes A = 0$ for ANY field $F$ of characteristic 0. You may wish to convince yourself, that indeed, if $R$ is a ring of finite characteristic, and $F$ is a field of characteristic 0 that the largest $R$-module $N$ we can embed in a vector space over $F$ is likewise the trivial quotient of $N$, using the same argument.

In fact, this has a lot to do with why some modules "don't have bases". If a module is generated by an element of finite (additive) order (along with perhaps other generators), then any set containing that finite order generator is provably $\Bbb Z$-linearly dependent, so also $R$-linearly dependent.

Therefore, one often seeks to choose "good rings" (like principal ideal domains) to make modules over, as they act "more like vector spaces". But this alone doesn't "fix" the problem above, we also have the problem of "torsion" to consider (that is, elements of finite order) which interferes with a module's "freeness" (its ability to have a basis).

Fields totally fix this problem: any $F$-module is free. Unfortunately, there's a "big gap" between "rings" and "fields", as the many kinds of localization of rings shows.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
If a module is 0, any homomorphism to that module is the 0-map, and any composition which factors through a 0-map is likewise 0.

So since the tensor product $\Bbb Q \otimes_{\Bbb Z} A = 0$, the "extension of scalars" of the "integer scalars" of a finite abelian group (which are "actually" the scalars of some finite cyclic group, since a finite abelian group is a direct product of finite cyclic groups) results in sending these scalars to the only scalar in $\Bbb Q$ with finite additive order: 0.

This means $L = 0$, as well, because any bilinear map $B:\Bbb Q \times A \to L$ has to be the 0-map:

$a \in A$ has finite order, say $k$. Therefore, just from bilinearity (over $\Bbb Z$):

$B(q,a) = B(\frac{qk}{k},a) = kB(\frac{q}{k},a) = B(\frac{q}{k},ka) = B(\frac{q}{k},0)$

$= 0(B(\frac{q}{k},0)) = 0$

Since $B = \phi \otimes = 0$ it is clear that ALL these maps are trivial.

You can think of it this way: the tensor product lets us "multiply" finite abelian groups by rational numbers. But finite abelian group elements are annihilated by finite integers, and in $\Bbb Q$, all the non-zero elements are UNITS, so $k$ (the annihilator of $a \in A$ is "multiplicatively the same" as $1/k$ in $\Bbb Q$, that is: bilinearity lets us split 1 into $k$ and $1/k$, one factor goes into $\Bbb Q$ (where it "does nothing") and the other factor goes into $A$ where it wipes it out.

It is not hard to see that $F \otimes A = 0$ for ANY field $F$ of characteristic 0. You may wish to convince yourself, that indeed, if $R$ is a ring of finite characteristic, and $F$ is a field of characteristic 0 that the largest $R$-module $N$ we can embed in a vector space over $F$ is likewise the trivial quotient of $N$, using the same argument.

In fact, this has a lot to do with why some modules "don't have bases". If a module is generated by an element of finite (additive) order (along with perhaps other generators), then any set containing that finite order generator is provably $\Bbb Z$-linearly dependent, so also $R$-linearly dependent.

Therefore, one often seeks to choose "good rings" (like principal ideal domains) to make modules over, as they act "more like vector spaces". But this alone doesn't "fix" the problem above, we also have the problem of "torsion" to consider (that is, elements of finite order) which interferes with a module's "freeness" (its ability to have a basis).

Fields totally fix this problem: any $F$-module is free. Unfortunately, there's a "big gap" between "rings" and "fields", as the many kinds of localization of rings shows.


Thanks so much for the help and the information ... I am still studying this post ...

Just let me clarify a simple conceptual issue. In a previous thread in a post I asked you

"So, in Corollary 9, we are actually seeking to map a quotient of N into L? ... and further, the tensor product is only a step on the path to do this ...? Further, is the S-module where we have actually succeeded in "extending the scalars" actually L?"

and you answered "yes" from which I took, among other things, the view that conceptually speaking the S-module which was (or contained) the extension of the scalars was L (and not \(\displaystyle S \otimes_R N \)).

So when you write:

"So since the tensor product $\Bbb Q \otimes_{\Bbb Z} A = 0$, the "extension of scalars" of the "integer scalars" of a finite abelian group (which are "actually" the scalars of some finite cyclic group, since a finite abelian group is a direct product of finite cyclic groups) results in sending these scalars to the only scalar in $\Bbb Q$ with finite additive order: 0."

... you are not disagreeing with the view that L is the module we are creating exhibiting extension of the scalars. it is just in the case of the example

L = \(\displaystyle S \otimes_R N \) = 0

so we can (casually) refer to \(\displaystyle S \otimes_R N \) as the S-module exhibiting the extension of the scalars ... but in general the module exhibiting extension of the scalars will be L and not \(\displaystyle S \otimes_R N \)

Is that correct?

Would appreciate your further clarification.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Yes, $L$ is the beast we're after, in fact the tensor product may be "too big" for our intended purposes.

The tensor product acts as kind of a "filter" by limiting (via its universal bilinear property) the KIND of $S$-module we get. $L$ is, in fact, a quotient of this tensor product, but it is often the case that the homomorphism:

$\phi: S \otimes_R N \to L$ is injective so we can in this situation regard the tensor product as "the extension".

In the example in this thread, $\phi$ is the (injective) map which sends $0 \to 0$, so even if $L$ is "really big" the image of $\phi$ is "really small". In other words, even though:

$\psi: M \to M'$ may be our original mapping of two modules

what we are typically interested in is:

$\psi: M \to \psi(M)$

we are always free to consider a "target" module as lying inside a larger module, or to just consider the surjective map by only considering the image sub-module.

We have no such freedom with a domain of a mapping: the concept of function forces us to define every image of a domain element. With co-domains, we can "fudge" them somewhat. This is why we use arrows to indicate mappings, the domain and the co-domain do not play symmetric roles, the flow is "one-way".

Put another way: when we seek to "extend scalars" the tensor product limits "how faithfully" we can do so, but we might want (for some reason) to use a PARTICULAR module $L$ to map into, which could have the effect of "shrinking" the tensor product even MORE (homomorphisms either "isomorph" that is, preserve the structure element-by-element, or they "collapse" (by a factor of the kernel), those are the only options, they never "expand"...although we may be challenged to create a larger structure our original one "fits inside just perfect").

The 0-object of modules (the trivial module, unique up to isomorphism) is special because one you "kill" a module with a 0-map, there's no "resurrecting" it.

By the same token, if one has a bilinear map that is NOT a 0-map, then the tensor product has to be non-zero as well, because it's always "bigger".
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Yes, $L$ is the beast we're after, in fact the tensor product may be "too big" for our intended purposes.

The tensor product acts as kind of a "filter" by limiting (via its universal bilinear property) the KIND of $S$-module we get. $L$ is, in fact, a quotient of this tensor product, but it is often the case that the homomorphism:

$\phi: S \otimes_R N \to L$ is injective so we can in this situation regard the tensor product as "the extension".

In the example in this thread, $\phi$ is the (injective) map which sends $0 \to 0$, so even if $L$ is "really big" the image of $\phi$ is "really small". In other words, even though:

$\psi: M \to M'$ may be our original mapping of two modules

what we are typically interested in is:

$\psi: M \to \psi(M)$

we are always free to consider a "target" module as lying inside a larger module, or to just consider the surjective map by only considering the image sub-module.

We have no such freedom with a domain of a mapping: the concept of function forces us to define every image of a domain element. With co-domains, we can "fudge" them somewhat. This is why we use arrows to indicate mappings, the domain and the co-domain do not play symmetric roles, the flow is "one-way".

Put another way: when we seek to "extend scalars" the tensor product limits "how faithfully" we can do so, but we might want (for some reason) to use a PARTICULAR module $L$ to map into, which could have the effect of "shrinking" the tensor product even MORE (homomorphisms either "isomorph" that is, preserve the structure element-by-element, or they "collapse" (by a factor of the kernel), those are the only options, they never "expand"...although we may be challenged to create a larger structure our original one "fits inside just perfect").

The 0-object of modules (the trivial module, unique up to isomorphism) is special because one you "kill" a module with a 0-map, there's no "resurrecting" it.

By the same token, if one has a bilinear map that is NOT a 0-map, then the tensor product has to be non-zero as well, because it's always "bigger".

Thanks Deveno ... Still reading your post and thinking over the ideas ... But already I can see that it is right on the money in addressing my concerns ... Thanks!

Peter