Hey guys.
I'm struggling with the notion of a tensor product of modules. My textbook starts off with some easy exercises, but since I'm having trouble with the definition, even the easy exercises are tough for me.
In particular, we have
I can prove the first part, albeit not easily. But the second part is giving me more trouble. I can prove that if is zero then the whole tensor product is zero, i.e. . But how do I derive a contradiction from this? Or is there an altogether easier way of showing it?Originally Posted by Dummit & Foote, Abstract Algebra, 3rd Ed, p375
Any help would be much appreciated. Thanks!
Unfortunately I can't check that belongs to the denominator, so to speak, of the underlying quotient group, because I don't know how to perform operations on that quotient group. In particular, I don't know how to perform operations on the numerator of that quotient group, which is perhaps the SOURCE of my problem in understanding the definition of a tensor product.
According to my definition, we construct by considering the free -module over . But is supposed to be an abelian group in its own right, and I don't understand how. For example, one idea would be to say that since and are both abelian groups individually, then we can just carry over the addition operations from each of them so that . Yet clearly this is not the case. But if not that, then how do we perform operations inside the group ?
Here's what my textbook has to say on the subject: link
if is a commutative ring with unity, is an ideal of and is an -module, then this is very easy to see:
define the map by and the map by obviously is -bilinear and thus it induces an -module homomorphism
defined by It is easy to see that is well-defined (prove it!). finally, both and are identity maps.
so is an -module isomorphism. so if and only if . now, apply this to your question.
As you can see, your book (a very good one, by the way) says relevant stuff many pages before 10.4.
In particular, in this case, , with
,
, and thus
in our tensor product iff , so if we can find a characteristic in
the elements of that isn't present in w we're done.
Hint: i) you can't "take away" the 2 and write [tex] (why?)
ii) In the generating elements of N, m is always 0 or 1 (mod 2)
Tonio
This might give you some intuitive idea.
Since is a right -module and is a left -module, it satisfies for , , and .
Thus, in . However, this rule does not applies to , since 1 is not an element of . Further verfications show that is a non-zero element in .
Let S be an R-module and N be a left R-module. You may already know that is a left R-module such that for all and .Unfortunately I can't check that belongs to the denominator, so to speak, of the underlying quotient group, because I don't know how to perform operations on that quotient group. In particular, I don't know how to perform operations on the numerator of that quotient group, which is perhaps the SOURCE of my problem in understanding the definition of a tensor product.
According to my definition, we construct by considering the free -module over . But is supposed to be an abelian group in its own right, and I don't understand how. For example, one idea would be to say that since and are both abelian groups individually, then we can just carry over the addition operations from each of them so that . Yet clearly this is not the case. But if not that, then how do we perform operations inside the group ?
Here's what my textbook has to say on the subject: link
It is abelian (by the definition of R-module) in the sense that , where and are elements in R-module .
The previous post gives a good explanation.
For an actual PROOF that , try this:
Define the map by .
You can prove (unless I'm wrong about this, which is possible..) that is a linear map; thus, it corresponds to a homomorphism . Also notice that , so , which implies .
EDIT: Could also do this: use the result of example 8 to see that and, as you noted earlier, .