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Could someone explain this to me please?

  1. Mar 27, 2012 #1
    Well, first of all I'm not quite sure If I'm posting my question in the right section so forgive me if it's in the wrong section. I'm studying module-theory on my own and in the chapter I'm in the book is discussing tensor product of modules (actually I decided to read this book only because of this chapter xD), there's a theorem in the book that is about taking the tensor product of a given exact sequence (the book doesn't say it this way, but this is how I memorize the whole theorem because It's kinda like that). I've attached the theorem and its proof, the book that I'm reading is not in English so I was forced to translate everything into English, if there are points that are ambiguous tell me to clarify them.

    I understand the whole proof, except two parts that I've highlighted them in red in the pdf.

    Thanks in advance

    Attached Files:

  2. jcsd
  3. Mar 27, 2012 #2
    OK. For the first part in read, you need to prove that [itex]\varphi_2[/itex] is well-defined.

    So take [itex]\pi(\sum x_n\otimes y_n)=\pi(\sum x_n^\prime\otimes y_n^\prime)[/itex]. You must show that [itex]\phi_2[/itex] of this is equal. Notice that

    [tex]\pi(\sum x_n\otimes y_n-\sum x_n^\prime\otimes y_n^\prime)=0[/tex]

    Now, use the definition of [itex]\pi[/itex] to show find an element of [itex]Im(1_T\otimes \varphi)[/itex].

    For the other one. You must show [itex]Ker(1_T\otimes \psi)\subseteq Im(1_T\otimes \varphi)[/itex]. So take [itex]\sum x_i\otimes y_i\in Ker(1_T \otimes \psi)[/itex]. What does this mean by definition?? Try to relate this to [itex]\varphi_2[/itex].
  4. Mar 27, 2012 #3
    This is the best that I could come up with. Please read it and see if it's OK.

    Attached Files:

  5. Mar 27, 2012 #4
    That looks ok to me!! Good work!
  6. Mar 30, 2012 #5
    Hey, today I read about the concept of an exact functor. Can we say that, according to this theorem, the tensor product is an exact functor? Is this what this theorem wants to tell us?
  7. Mar 30, 2012 #6
    No, the theorem wants to tell us that the tensor product is a right-exact functor.

    An exact functor sends an exact sequence

    [tex]0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0[/tex]

    to an exact sequence

    [tex]0\rightarrow FA\rightarrow FB\rightarrow FC\rightarrow 0[/tex]

    But the tensor product doesn't satisfy this. That is, the left arrow [itex]0\rightarrow FA[/itex] isn't there. Or: tensoring doesn't send an injective map to an injective map.

    We do have right-exactness. That is the sequence

    [tex]A\rightarrow B\rightarrow C\rightarrow 0[/tex]

    is send to

    [tex]0\rightarrow FA\rightarrow FB\rightarrow FC\rightarrow 0[/tex]

    That tensoring is not exact is a serious problem. To solve the problem, the notion of homology is invented.
  8. Mar 30, 2012 #7


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    Homework Helper

    If you like fancy arguments that make simple things hard, think about this: since

    Hom(Atens(*),**) ≈ Hom(A,Hom(*,**)), and since Hom(A,*) preserves injections, while Hom(*,Z) changes surjections into injections, it follows that Hom(A,Hom(*,Z)) changes surjections into injections, hence Hom(Atens(*),Z) changes surjections into injections, it can be deduced that Atens(*) preserves surjections.

    I.e. if you know that Hom is "left exact", it follows from the universal defining property of tensor product, that the latter functor is right exact, (but actually some of the properties stated above have to be used backwards as well).

    To build on what micromass said, it is therefore interesting to study those objects for which tensoring is exact, called "flat".

    Last edited: Mar 30, 2012
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