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Short Exact Sequences and at Tensor Product

  1. Jul 14, 2014 #1

    WWGD

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    Hi,let:

    0->A-> B -> 0

    ; A,B Z-modules, be a short exact sequence. It follows A is isomorphic with B.

    . We have that tensor product is

    right-exact , so that, for a ring R:

    0-> A(x)R-> B(x)R ->0

    is also exact. STILL: are A(x)R , B(x)R isomorphic?

    I suspect no, if R has torsion. Anyone have an example of

    A(x)R , B(x)R non-isomorphic, but A,B isomorphic? Thanks.
     
  2. jcsd
  3. Jul 14, 2014 #2
    Yes, they are. You should look for a proof.
     
  4. Jul 14, 2014 #3

    WWGD

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    Sorry, that is not what I meant to ask, instead , I am looking for an example where:

    A(x)R , B(x)R are isomorphic,

    but A,B are not isomorphic.

    Thanks.
     
  5. Jul 16, 2014 #4
    Oops, I never saw your reply. Weird. Anyway, consider

    [tex]\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Z}_2\cong \mathbb{Z}_2 \cong \mathbb{Z}_2 \otimes_\mathbb{Z}\mathbb{Z}_2[/tex]

    In general, we have the relation

    [tex]\mathbb{Z}_n\otimes_\mathbb{Z} \mathbb{Z}_m \cong \mathbb{Z}_\text{gcd(m,n)}[/tex]

    This result will also give you a wealth of counterexamples (note that the above still holds if we define ##\mathbb{Z}_0 = \mathbb{Z}##)
     
  6. Jul 19, 2014 #5

    WWGD

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    A, thanks, nice.
     
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