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A mapping from an integral domain to non-negative integers, Abstract Algebra

  1. Dec 14, 2012 #1
    So just had this question as extra credit on a final:

    Let D be an integral domain, and suppose f is a non-constant map from D to the non-negative integers, with f(xy) = f(x)f(y). Show that if a has an inverse in D, f(a) = 1.


    Couldn't figure it out in time. I was thinking the way to go about it was assume f(a) > 1 and show then the function would be constant, somehow using the fact that any y in D could be factored as a(a^- * y).

    Maybe: Suppose f(a) = n /= 1. Then f(a*a^-*y) = n*m*f(y), n*m /= 1 since we are in the integers and f(n or m) /=1 by assumption. But in D, a*a^-1* y = y, but n*m*f(y) /= f(y), a contradiction.

    Is this right? It doesn't make use of the hypothesis that f is non-constant I don't think.
     
  2. jcsd
  3. Dec 14, 2012 #2
    I think there is some more to it than that. What happens if f(a) = 0? Remember f is mapping to the non-negative integers, so you need to consider this case. This will help you complete the proof, I think.
     
  4. Dec 14, 2012 #3
    Okay, so you know that a has an inverse. Let us suggestively call it [itex] a^{-1} [/itex]. Then [itex] f(1) = f(a a^{-1}) = f(a) f(a^{-1}). [/itex] so that f(a) divides f(1). Maybe you should try to determine f(1). In particular, does the fact that [itex] 1 = 1\cdot 1 [/itex] help you? You should be able to show that either [itex] f(1) = 1 [/itex] or [itex] f(1) = 0[/itex]. However, one of these is not allowed by hypothesis (which one? why?). Given that the other one must be true, you are done (again, why?).
     
  5. Dec 14, 2012 #4
    I would approach it as follows:

    [itex]
    f(1_D) = f(1_D \cdot 1_D)=f(1_D) \cdot f(1_D)
    [/itex]

    Since the images we're considering must be non-negative integers, we have that

    [itex] f(1_D) = 0 [/itex] or [itex] f(1_D) = 1 [/itex]

    Cases:

    i) Suppose [itex]f(1_D)=0[/itex]

    Let [itex] a \in D [/itex] . It follows immediately that

    [itex] f(a) = f(1_D \cdot a) = f(1_D) \cdot f(a) = 0 \cdot f(a) = 0 [/itex], for any [itex] a \in D [/itex].

    This of course implies that f is constant (namely the 0 map), we have a contradiction, since we assumed f was non-constant.

    ii) Suppose [itex]f(1_D)=1[/itex]
    Let [itex] a \in D [/itex], further suppose [itex] \exists \ a^{-1} \in D [/itex] such that [itex] a \cdot a^{-1} = 1_D [/itex].
    Then:
    [itex]f(1_D) = f( a \cdot a^{-1} ) = f(a) \cdot f(a^{-1}) = 1 [/itex]
    Again, since these images under f are non-negative integers, it follows that
    [itex] f(a) = f(a^{-1}) = 1 [/itex]

    We have our desired result. Any element in D having an inverse in D will necessarily map to 1.
     
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