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Abstract Algebra: Finite Field

  1. Nov 26, 2012 #1
    Show that every finite field with p+1 elements, where p is a prime number, is commutative.

    I know this has something to do with composite numbers, but I'm not quite sure how to show this.
     
  2. jcsd
  3. Nov 26, 2012 #2

    jgens

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    Just a friendly FYI that the term field usually refers to a commutative ring where every non-zero element is a unit. Some older texts use the term field to mean what nowadays would be called a division ring or skew field. For reference see here: http://en.wikipedia.org/wiki/Field_(mathematics [Broken])

    Now assuming you have a skew field with p+1 elements and that you want to show that the multiplication is commutative, all you need to do is look at the order of the group of units. In this case, the order uniquely determines the group.
     
    Last edited by a moderator: May 6, 2017
  4. Nov 26, 2012 #3

    micromass

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    An interesting result is Wedderburn's little theorem. It states that every finite skew field (and more general: finite domain) is in fact commutative. What you need to prove in the OP is a special (and much easier) case for p+1.

    http://en.wikipedia.org/wiki/Wedderburn's_little_theorem

    Sorry, I just wanted to mention this.
     
  5. Nov 27, 2012 #4
    Maybe you can make use of thie following: Every finite field must consist of a prime-power number of elements, i.e. some number qn with a prime q and a positive integer n. (If you haven't seen this before, you might need to prove it before you use it.) What possibilities do have get if qn=p+1 with primes p and q?
     
  6. Nov 27, 2012 #5

    micromass

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    I don't think you really need this in the proof. The reason is that you are working with a skew field and you need to show that it is a field. You don't know if your things are fields yet.
     
  7. Nov 27, 2012 #6
    I meant "field" to mean "commutative or non-commutative division ring", since I often work with "non-commutative fields". Sorry for the misunderstanding.
     
  8. Nov 27, 2012 #7

    micromass

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    OK, but then showing that every finite field (commutative or noncommutative) has order [itex]p^n[/itex] is quite difficult to show. I think that result is actually quite a lot harder than the problem in the OP.
     
  9. Nov 27, 2012 #8
    Try proof by contradiction. If the skewfield F (I'll use that word if it's more common) has the order q, and q has more than one prime factor, it would have to be something like q = pp'r where p,p' are the smallest prime factors of q, and r is a positive integer. If for any integer n we write F(n)=(1+...+1) with n ones, we get

    F(p) F(p') F(r) = F(pp'r) = 0

    And so F(p)=0 in F, just like F(p')=0 (since F is a skewfield). But this means

    0 = F(p) - F(p') = F(p-p'),

    i.e. the order q of F must be divisible by p-p'. But that's impossible, since p and p' are the smallest divisors of q, and |p-p'| is less than both p and p'.
     
  10. Nov 28, 2012 #9

    Hurkyl

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    Maybe I'm being silly, but doesn't its prime ring have to be a field, and doesn't it have to be a vector space over it's prime ring?
     
  11. Nov 28, 2012 #10

    micromass

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    Hmmm, so I guess it wasn't so difficult to show after all. I should have thought a bit about it before answering...
     
  12. Nov 28, 2012 #11
    None of what I have read has made sense...
     
  13. Nov 28, 2012 #12

    jgens

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    Just follow up on this hint:
     
  14. Nov 28, 2012 #13
    What have you tried on your own so far? You wrote: "I know this has something to do with composite numbers". I suggested that a field's order must be a rather special kind of composite number, i.e. a prime power qn. Since your (skew)field is supposed to have p+1 elements with a prime p, you'll need qn=p+1. Do you know anything about prime numbers that you could use in this equation? (Hint: Is p odd or even? What about q?)
     
  15. Nov 28, 2012 #14

    jgens

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    I have not given this approach much thought, but this seems a whole lot more complicated than just showing that the group of units is abelian since it has prime order. Maybe I am missing something.
     
  16. Nov 28, 2012 #15
    Maybe we're both missing something. I don't think one needs the concept of a "unit" at all, since all non-zero elements of a skewfield are invertible anyway, so the group of all units is just the group of the p non-zero elements of the skewfield.

    What really seems necessary, though, is the proof that a prime-order group is commutative.
     
  17. Nov 28, 2012 #16

    jgens

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    This is just a matter of taste. Group of units is just a concise way to describe the object I want to talk about.

    The proof is trivial. I would expect that most people in a course on elementary field theory / ring theory (as the OP seems to be) have seen enough group theory to know that prime order groups are abelian. But maybe not.
     
  18. Nov 28, 2012 #17
    I agree. But that would mean the original problem was rather meaningless. It could even be answered quite simply by

    If the skewfield has p+1 elements with a prime number p, then its multiplicative group has p elements and so must be commutative. D'uh.​

    If there's any point at all to the original question, could it be something apart from what we both agree is trivial?
     
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