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Abelian groups from the definition of a field

  1. Dec 4, 2007 #1
    Just a pregrad-level curiosity:

    I see often repeated (in the Wikipedia page defining "Field", for one) that, from the field's axioms, it can be deduced that F,+ and F\{0},* are both commutative groups.

    Yet, the closure property of * is only guaranteed on F, not necessarily on F\{0}. If I'm not mistaken, the finite field Z/6Z is a counter-example: 2*3=0 (mod 6), so {1,2,3,4,5} cannot be a group under multiplication. Is that correct?
     
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  3. Dec 4, 2007 #2

    Hurkyl

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    Z/6Z isn't a field...
     
  4. Dec 4, 2007 #3
    Hm, I think the trick here is that Z/6Z (assuming I understand your terminology... I think I would call this [tex]Z_6[/tex]) is not a finite field. The only finite fields of this type are of order p, where p is a prime number. When p is prime the multiplicative group will always be closed under multiplication.

    If you think about it, it would not make sense anyway for [tex]Z_6[/tex] to be a field. For example, if [tex]Z_6[/tex] is a field, then 2 would have to have a multiplicative inverse [tex]2^{-1}[/tex]. But what do we get when we multiply [tex]2^{-1}[/tex], whatever it is, by 0? We get... 0, not 3 as we should. I think there are other violations of the field axioms as well. [tex]Z_6[/tex] is only an additive group.
     
  5. Dec 4, 2007 #4

    mathwonk

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    so as you observe, it must be a theorem that in a field, F-{0} is closed for *. i recall that as one of the first lemmas proved in abstract algebra. i believe it uses the fact that every non zero element has an inverse, and the definition of a unit for multiplication, plus the distributive property of multiplication. ohh, and the fact that 0 and 1 are different elements.
     
  6. Dec 4, 2007 #5
    Thanks for all your answers; that the order of a finite field has to have some restrictions (being a prime is one) makes now more sense.

    I suppose the only think needed to show the closure of F\{0} under * is that, given two elements a,b of F, both non-zero, a*b cannot be zero. If a*b=0, since b is non-zero and therefore has an inverse b', then multiplying by b', a*b*b' = 0, and a=0, which contradicts the initial statement.
     
  7. Dec 4, 2007 #6

    morphism

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    I want to point out that the order of a finite field can be a power of a prime. However, Z/nZ is a field iff n is a prime. If p is a prime divisor of n but n/p isn't 1, then p*(n/p)=0 (mod n), i.e. Z/nZ admits a zero divisor. So Z/nZ can be a field only if n is a prime. The converse follows from the fact that nZ is a maximal ideal in Z when n is a prime (it's trivially a prime ideal and hence a maximal ideal because Z is a PID).
     
  8. Dec 4, 2007 #7
    It is a curious asymmetry; then there cannot be any isomorphism from a finite field F of order 9 to Z/9Z; if a is an element of F such that f(a)=3, a*a would be the 0 element of F. I'll try to look up examples of such finite field.
     
    Last edited: Dec 4, 2007
  9. Dec 4, 2007 #8

    JasonRox

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    Well, 9 is not a prime therefore a finite field of order 9 does not exist.
     
  10. Dec 5, 2007 #9

    Hurkyl

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    Except, of course, y'know, that one that does exist. :tongue:

    [tex]
    GF(3^2) \cong \mathbf{Z}[x] / \langle 3, x^2 + 1 \rangle
    [/tex]
     
    Last edited: Dec 5, 2007
  11. Dec 5, 2007 #10
  12. Dec 5, 2007 #11

    Hurkyl

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    [itex]\langle 3, x^2 + 1 \rangle[/itex] is the ideal generated by 3 and x² + 1. So, the example you linked is a ring of reduced representatives for the one I posted.

    (In fact, it turns out that all finite fields of order 9 are isomorphic)
     
  13. Dec 5, 2007 #12
    Yeah, when I said "The only finite fields of this type are of order p, where p is a prime number.", by "of this type" I meant the [tex]Z_n[/tex] fields, the ones obtained by applying a modulus to the integers. Sorry if I was unclear!
     
  14. Dec 5, 2007 #13

    JasonRox

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    Thanks for pointing that out. o:)
     
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