# Tensor product of modules

by Arian.D
Tags: modules, product, tensor
 Mentor P: 18,345 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??
P: 101
 Quote by micromass 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??
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.
P: 659
 Quote by 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.
In general, the issue is whether your modules being tensored have torsion or not. If they have torsion elements, then a non-zero product can be zero, and the answer is no otherwise.

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