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I know this has something to do with composite numbers, but I'm not quite sure how to show this.

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- Thread starter S_Manifesto
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- #1

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I know this has something to do with composite numbers, but I'm not quite sure how to show this.

- #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.

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.

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

Sorry, I just wanted to mention this.

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^{n}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 q^{n}=p+1 with primes p and q?

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.

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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.

I meant "field" to mean "commutative or non-commutative division ring", since I often work with "non-commutative fields". Sorry for the misunderstanding.

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I meant "field" to mean "commutative or non-commutative division ring", since I often work with "non-commutative fields". Sorry for the misunderstanding.

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.

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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.

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'.

- #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?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.

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None of what I have read has made sense...

- #12

jgens

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None of what I have read has made sense...

Just follow up on this hint:

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.

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None of what I have read has made sense...

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 q

- #14

jgens

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I suggested that a field's order must be a rather special kind of composite number, i.e. a prime power q^{n}. Since your (skew)field is supposed to have p+1 elements with a prime p, you'll need q^{n}=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?)

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.

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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.

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.

- #16

jgens

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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.

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

What really seems necessary, though, is the proof that a prime-order group is commutative.

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.

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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.

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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