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        riddles >> putnam exam (pure math) >> A Characterization of finite fields?
        
(Message started by: ecoist on Jun 2^nd, 2006, 9:25pm)

Title: A Characterization of finite fields?
Post by ecoist on Jun 2^nd, 2006, 9:25pm

Let R be a ring with the property that every function f from R into R is a polynomial of the form f(x)=a₀+a₁x¹+...+a_nxⁿ, where n is some nonnegative integer and the a_i, for i=0,...,n, are elements of R. When n=0, we mean the constant function f(x)=a₀. Must R be a finite field?

Title: Re: A Characterization of finite fields?
Post by Eigenray on Jun 3^rd, 2006, 3:56pm

Yep yep.
[hideb]First off, there are |R|^|R| > |R| functions from R->R, and only |R|+|R|²+... = |R| + aleph₀ polynomials, so R is finite.

Second, let x be any non-zero element of R, and let f be a function with f(0)=1, f(x)=0. Then if f is a polynomial, we have
1 = -(a₁+a₂x+...+a_nx^n-1)x,
so every non-zero element has a left inverse. It follows every element has a right inverse: if yx=1, and zy=1, then x = zyx = z, so y is also the right inverse of x. So R is a division ring.

By [link=http://planetmath.org/?op=getobj&from=objects&id=3473]Wedderburn's theorem[/link], R is a field.[/hideb]

Title: Re: A Characterization of finite fields?
Post by ecoist on Jun 3^rd, 2006, 4:13pm

Nice. But why does R have a multiplicative identity?

Title: Re: A Characterization of finite fields?
Post by ecoist on Jun 11^th, 2006, 7:49pm

Hey, Eigenray! Are you aware that your proof is not complete? After 8 days I found a tortured proof that R has a multiplicative identity (which you assumed in your proof), but I'm sure you have a better solution.

Title: Re: A Characterization of finite fields?
Post by Eigenray on Jun 12^th, 2006, 9:25am

on 06/11/06 at 19:49:25, ecoist wrote:

I'm sure you have a better solution.

R has a multiplicative identity because that's part of my definition of ring. :) (At least I didn't assume it was commutative!) Sorry to disappoint you.

Title: Re: A Characterization of finite fields?
Post by ecoist on Jun 12^th, 2006, 2:40pm

That is disappointing. I'm still recovering from your stunning proof for the Classroom Dilemma problem. I'd hate to have to complete your lovely proof of this problem with my pitiful argument.

Title: Re: A Characterization of finite fields?
Post by Michael_Dagg on Jun 12^th, 2006, 3:11pm

on 06/12/06 at 14:40:09, ecoist wrote:

...your stunning proof for the Classroom Dilemma problem.

I second that!

Title: Re: A Characterization of finite fields?
Post by ecoist on Jun 20^th, 2006, 6:55pm

Ok, here is my proof R has a multiplicative identity.

[hide]Let R* be the set of nonzero elements of R. Using Eigenray's trick, if r and s are any elements of R* then there exists an element x in R* such that xr=s. Hence right multiplication by r is a mapping from R* onto R*, whence the mapping is a permutation of R*, and also of R since 0r=0. Since R* is finite, some finite power of this mapping is the identity map, whence e=r^k is a right identity for R, for some positive integer k.

(At this point I had hoped that this was enough to show that R* is a group, but the following semigroup is a counterexample.

. | a b c
---------
a | a a a
b | b b b
c | c c c

So I have more work to do.)

To show that e is also a left identity, and thus an identity for R, let r be any element of R* and form er-r. Then

e(er-r)=eer-er=er-er=0.

If er-r is not 0, then, since 0(er-r)=e(er-r)=0 and e is not 0, it follows that right multiplication by er-r is not 1-1, a contradiction. Hence er=r and e is an identity for R. The gap in Eigenray's proof is now filled.[/hide]