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Math Help - Short exact sequences

  1. #1
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    Short exact sequences

    I am claiming the following is not a short exact sequence
    0\rightarrow Z\stackrel{i}{\rightarrow} Z \oplus Z \oplus Z \stackrel{j}{\rightarrow} Z \rightarrow 0

    I am thinking of it this way: If it was a short exact sequence, then i would be injective and j would be onto. Also, (Z \oplus Z \oplus Z )/ i(Z)  would be isomorphic to Z .

    Now i(1)= (a,b,c) \neq 0 for some a,b,c \in Z. Now, is it true that
     i(Z)=im(i)=Z<(a,b,c)> ? And that (Z \oplus Z \oplus Z )/ i(Z)  = Z/aZ \times Z/bz \times Z/cZ ?If yes, how do you show this is not isomorphic to Z ? If not, what am I doing wrong and how do you prove my claim?
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  2. #2
    Senior Member Tinyboss's Avatar
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    That sequence can't be exact. If you have 0->A->B->C->0, then C is isomorphic to B/A', where A' is the image of A under the map A->B. Because of exactness, that map is injective, so that A' is isomorphic to A. You can't take Z^3, mod out by something isomorphic to Z, and get Z.
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  3. #3
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    Ok, that makes sense, but what would be a good argument to show (Z \oplus Z \oplus Z) / A' is not isomorphic to Z?
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  4. #4
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    Can we say, }  Z \oplus Z \oplus Z has 3 generators, now ( Z \oplus Z \oplus Z)/i(Z)  is isomorphic to ( Z \oplus Z \oplus Z) / Z which has at least 2 generators, but  Z only has one generator?
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  5. #5
    Senior Member Tinyboss's Avatar
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    You have to be careful when talking about the number of generators, because it's not fixed--for instance, the integers are generated by (1) or by (2,3). I think the free rank of an abelian group might be helpful here.
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  6. #6
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    Or just say it's not exact by definition. To be exact, we would need \mathrm{im}(i)=\mathrm{ker}(j).

    Of course, the other issue is still a good question in and of itself, so don't let this stop that discussion.
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  7. #7
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    I am not sure I understand how   im(i) = Ker(j) helps show Z^3/i(Z) is not isomorphic to Z. Can you elaborate on that?

    Also @ Tinyboss: How, do I use the free rank of an abelian group to prove that? My algebra is very rusty, and I tried to look it up, but I don't think I understand how to apply the definition here...
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  8. #8
    Senior Member Tinyboss's Avatar
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    Sorry, what I thought would work, doesn't. I asked a buddy and he said to tensor the sequence with Q, yielding the same sequence except with Q's in place of Z's. Since Q is a field, there is a well-defined notion of dimension, and so that sequence can't be exact. And because tensoring preserves exact sequences, this implies the original sequence wasn't.

    I don't understand it 100%, but I'm putting it here since my first guess was wrong.
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  9. #9
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    Quote Originally Posted by math8 View Post
    I am not sure I understand how   im(i) = Ker(j) helps show Z^3/i(Z) is not isomorphic to Z. Can you elaborate on that?

    Also @ Tinyboss: How, do I use the free rank of an abelian group to prove that? My algebra is very rusty, and I tried to look it up, but I don't think I understand how to apply the definition here...
    I don't know that it does. I was just giving another way of answering the original question, about the sequence not being exact.
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