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When is a tensor 0?

  1. Aug 3, 2013 #1
    Let [itex]M[/itex] be a module over the commutative ring [itex]K[/itex] with unit 1. I want to prove that [itex]M \cong M \otimes K.[/itex] Define [itex]\phi:M \rightarrow M \otimes K[/itex] by [itex]\phi(m)=m \otimes 1.[/itex] This is a morphism because the tensor product is K-linear in the first slot. It is also easy to show that the map is surjective. This is where I get stuck.

    Suppose [itex]\phi(m)=\phi(n),[/itex] so that [itex]0 = m \otimes 1 - n \otimes 1 = (m-n) \otimes 1.[/itex] How do I prove that this implies that [itex]m=n[/itex] and thus the map is injective? More generally, how can you tell when a tensor is 0?
     
  2. jcsd
  3. Aug 3, 2013 #2

    micromass

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    Use the universal property to find an explicit inverse.
     
  4. Aug 3, 2013 #3
    Thanks, micromass! I think I've got it. If you wouldn't mind, please let me know if this looks alright.

    Define [itex]h: M \times K \rightarrow M[/itex] by [itex]h(m,k)=mk.[/itex] This map is bilinear, so it factors through the tensor map [itex]M \times K \rightarrow M \otimes K.[/itex] Thus there is a unique (irrelevant?) morphism [itex]f[/itex] such that [itex]f(m \otimes k)=mk.[/itex] This is the right and left inverse of the morphism [itex]\phi[/itex], implying that [itex]\phi[/itex] is an isomorphism.
     
  5. Aug 3, 2013 #4

    micromass

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    That looks alright! The uniqueness of the map is indeed irrelevant here, but it's nice to know.
     
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