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ℤ and ℤ[x] isomorphism

  1. Oct 4, 2011 #1
    There exists none. What's the easiest way to prove this?
    Can we state that all elements of ℤ are in ℤ[x] but not the other way around?
     
  2. jcsd
  3. Oct 4, 2011 #2

    Office_Shredder

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    Just because you came up with a bad attempt at an isomorphism (the obvious inclusion of the integers) doesn't mean a better one doesn't exist.
     
  4. Oct 4, 2011 #3

    lavinia

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    I think a ring isomorphism must map 1 to 1 in Z[x] so Z must be mapped to Z in Z[x].

    h(m) = h(1.m) = h(1)h(m). So h(1) = 1. I think this is right.
     
  5. Oct 6, 2011 #4

    Deveno

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    if there WAS such an isomorphism, we would also have one from Z[x] to Z, say φ.

    now 1 ( = 1 + 0x + 0x^2 +....) in Z[x] is a multiplicative identity,

    so say φ(f(x)) = k.

    so k = φ(f(x)) = φ(1f(x)) = φ(1)φ(f(x)) = φ(1)k.

    so φ(1) = 1.

    by induction, for n ≥ 1, φ(n) = φ(1+1+...+1) (n times)

    = φ(1) + φ(1) + ...+ φ(1) (n times)

    = 1 + 1 +...+ 1 (n times)

    = n.

    similarly, φ(-n) = -φ(n) = -n.

    given this, and the fact that we must assign an integer value to φ(x),

    how can φ be injective?
     
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