Tensor product of modules

1. May 8, 2012

Arian.D

I'm reading about tensor product of modules, there's a theorem in the book that leaves parts of the proof to the reader. I've attached the file, I didn't put this in HW section because first of all I thought this question was more advanced to be posted in there and also because I want to discuss something else with people on this forum.

I'm thinking what happens when a simple tensor becomes zero? I mean suppose that we have $x\otimes y = 0$. Does it mean that x or y must be zero? or it's possible that one of them nonzero but still their tensor product turns out to be zero?
If possible, please check my proof in the file as well.

I know I'm asking too much, but please help me as quickly as possible because tomorrow I'll have a conference about tensors, the professor is also the head of the math. department of our university and I'm the only under-graduate student in his class. so I'm very determined to have a successful conference tomorrow in the class and I need your help very much guys.

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2. May 8, 2012

micromass

Staff Emeritus

The answer to your other question is no. Consider the $\mathbb{Z}$-module $\mathbb{Z}\otimes \mathbb{Z}_2$, then

$$2\otimes 1= 2(1\otimes 1)=1\otimes 2=1\otimes 0=0,$$

but neither 2 or 1 are zero.

In general, in the R-module $M\otimes N$ it holds that $m\otimes n=0$ if and only if there exists elements $m^\prime_j\in M$ and $a_j\in R$ such that

$$\sum_j a_jm_j^\prime = m$$

and

$$a_jn=0$$

for all j.

3. May 8, 2012

Arian.D

I see. Thanks.
Then why my proof stands correct? Because at one point I've said that y or (r+I) must be zero. Isn't that false? I don't get it.

4. May 8, 2012

micromass

Staff Emeritus
Oh right, that part is incorrect of course.

Can you prove the following equalities:

$$Im(\phi(j\otimes 1))=\phi(Im(j\otimes 1))=\phi(Ker(\pi\otimes 1))=Ker((\pi\otimes 1)\phi^{-1})$$

5. May 8, 2012

Arian.D

Actually I've proved the theorem in another way, this time it's tedious and quite unusual.
I hope you enjoy it.

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6. May 8, 2012

micromass

Staff Emeritus
Your proof that f is well-defined is very weird. It's just a chain of equalities, I can't seem to make anything about it.

I don't really get why you're making it so hard. Your original proof was very easy, except for that one gap that you should prove.

7. May 8, 2012

mathwonk

Maybe I am making it too easy by assuming R is commutative, but then it is pretty easy to see that the obvious maps taking (a)tens(m)-->[am], and [m]-->(m)tens([1]),

are well defined and inverse to each other. done.

8. May 8, 2012

Arian.D

Okay then. Assume f is well-defined and check the rest please.
My original proof is very easy, but it's not very easy to fill that gap. Unless something comes to your mind right now, because to me it doesn't look very obvious.

But I have to prove it in the general non-commutative case tomorrow. :(

9. May 9, 2012

mathwonk

Then I suggest trying to provide all the reasons for my steps, and then see how they go in the non commutative case. probably they are exactly the same. My answer is like a cab from the airport that drops you off on the opposite side of the street from your hotel, but expects you to cross the street yourself.

10. May 10, 2012

Arian.D

Yup, but I think I already got a proof myself. The professor liked my proof and said it was correct. so I'm now kinda sure that my proof is correct now. But it wouldn't hurt if others check it too.
I personally like my proof so much, because it uses a lot of ideas in Algebra which are new to me and it was the first time that I've employed them in such an advanced algebraic topic. (advanced in my level I mean, not in the level of people like you and micromass). So if you validate my proof I'll be very appreciative.

11. May 10, 2012

Arian.D

By the way, I got a general question.
Suppose that A,B and C are 3 R-modules. and each one of them are R-isomorphic respectively to A', B', and C'. Here's the question.
if $A\to B \to C$ is an exact sequence, does it mean that the sequence $A' \to B' \to C'$ is exact as well? I'm saying this because when two algebraic structures are isomorphic their algebraic properties will be preserved because we're just relabeling things in the two sets and keep every thing else the same. Is that right?

12. May 10, 2012

micromass

Staff Emeritus
How do you define the maps between A', B' and C' ??

13. May 10, 2012

Arian.D

Ahh, right. Well, we'll define them in the most natural way, for example if in the first sequence we have the two mappings $\varphi$ and $\psi$ in the new sequence we will have the two new mappings $\varphi(f_1)$ and $\psi(f_2)$ where $f_1$ and $f_2$ are isomorphisms between A and B with A' and B' respectively. I guess now the ambiguity is clear.

By the way, how could I write a function above an arrow in Latex? I want to write down $\varphi$ above an arrow from A to B but I don't know how to do that :(

14. May 10, 2012

micromass

Staff Emeritus
I don't really get what you mean with $\varphi(f_1)$ and $\psi(f_2)$. The functions $\varphi$ and $\psi$ act on modules, but you make it seem like they act on maps

Did you perhaps mean: $f_2\circ \varphi\circ f_1^{-1}:A^\prime\rightarrow B^\prime$ as the first map and $f_3\circ \psi\circ f_2^{-1}$ for the second map??

15. May 10, 2012

micromass

Staff Emeritus
Code (Text):

A\xrightarrow{f} B
$$A\xrightarrow{f} B$$

16. May 10, 2012

Arian.D

Hmmm, let me tell you what is in my head then.
If A and A' are isomorphic, then there exists an isomorphism between them. this isomorphism takes an element of A and then sends it to an element in A' which is just the same element with a new name or sign or whatever. so in this way I could find a similar map for A' to B' as the one I've already been given between A and B. That's my idea. I don't know if I'm able to say what's in my head or not but I'll try again if I'm still unclear.

Thanks.

17. May 10, 2012

micromass

Staff Emeritus
I know what you mean, but if you want to prove things, then you should spell out the maps specifically. Look again at my post 14, are those the maps you mean??

18. May 10, 2012

Arian.D

Yup. They are the same maps I have in my mind.

19. May 10, 2012

micromass

Staff Emeritus
OK, so then we wish to prove that

$$Ker(f_3\circ \psi \circ f_2^{-1})=Im(f_2\circ \varphi \circ f_1^{-1})$$

I'll do one side of the inclusion:
Take x in the kernel, then $f_3(\psi(f_2^{-1}(x)))=0$. Since $f_3$ is an isomorphism, its kernel is 0, thus $\psi(f_2^{-1}(x))=0$. Thus $f_2^{-1}(x)\in Ker(\psi)$. By exactness, we have that $Ker(\psi)=Im(\varphi)$, thus there exists a $y\in A$ such that $\varphi(y)=f_2^{-1}(x)$. In other words, $f_2(\varphi(y))=x$. Write $z=f_1(y)$, then $f^{-1}_1(z)=y$. Thus
$$f_2(\varphi(f_1^{-1}(z)))=f_2(\varphi(y))=x$$
The left-hand side is certainly in the image, thus $x\in Im(f_2\circ \varphi\circ f_1^{-1})$.

Can you do the other inclusion??

20. May 10, 2012

Arian.D

Sure. It's easy.

Take x in $Im(f_2\circ \varphi \circ f_1^{-1})$. Because it's in image, then there exists y in the domain of $f_2\circ \varphi \circ f_1^{-1}$, i.e. in A', such that $f_2\circ \varphi \circ f_1^{-1}(y)=x$
Since $f_2$ is an isomorphism it's invertible, and we'll have:
$f_2^{-1}(x)=\varphi \circ f_1^{-1}(y)$
Now let's apply $\psi$ to both sides, we'll obtain:
$\psi \circ f_2^{-1}(x) = \psi \circ \varphi \circ f_1^{-1}(y) = 0$
Since $\psi \circ \varphi = o$, because anything that we put into $\varphi$ goes to the kernel of $\psi$ and therefore it becomes zero!
We're done! Since $f_3$ is a homomorphism, it sends 0 to 0, hence:
$f_3 \circ \psi \circ f_2^{-1} (x) = f_3(0) = 0$
and that says $x \in Ker(f_3\circ \psi \circ f_2^{-1})$

It's a very neat theorem, I think it'll make a lot of things easier in future for me, and the good thing is that it works on almost all algebraic structures that I know now, because nowhere in the proof we assumed anything else except the definition of an exact sequence, kernel and image of homomorphisms. All of them stay intact in groups, rings and modules. Thanks for your help micromass.