From: Franz Lemmermeyer Subject: Re: infinity of "primes" Date: Mon, 24 Apr 2000 16:37:19 +0200 Newsgroups: sci.math Summary: [missing] Rajarshi Ray wrote: > I'm trying to show that there are an infinity of "primes", i.e. > irreducible elements in F[x], the linear algebra of polynomials. If F is infinite, this is of course trivial: take x-a for a in F. > There > seems to be a way to do this similar to how Euler's zeta function is > used to show there an infinite number of the usual primes. Can somebody > help me? If F is finite, Landau's student Kornblum generalized Dirichlet's theorem to F_p[X]: see Math. Zeitschr. 5, 100-111 (1919). Artin gave a similar proof in his thesis: Math. Ztschr. 19, 207-246 (1924). Of course everything simplifies considerably if you're only interested in the infinity of primes; you would need to show that the zeta function has a pole in s=1 and use the Unique Factorization Property to derive a contradiction via the Euler factorization. For the definition of the zeta function of function fields, see e.g. the books of Stichtenoth or Lorenzini. franz ============================================================================== From: Richard Carr Subject: Re: infinity of "primes" Date: Sun, 23 Apr 2000 23:16:03 -0400 Newsgroups: sci.math On Sun, 23 Apr 2000, Rajarshi Ray wrote: :Date: Sun, 23 Apr 2000 22:59:37 GMT :From: Rajarshi Ray :Newsgroups: sci.math :Subject: infinity of "primes" : :Hi, : :I'm trying to show that there are an infinity of "primes", i.e. :irreducible elements in F[x], the linear algebra of polynomials. There You could try f+x (or 1+fx, if you prefer). If F is infinite this would work. If F is only finite I guess you'd need a few more. Then you'd need to find that there are irreducible elements of arbitrarily high degree. Suppose not- then all irreducible elements have degree at most n, say. The polynomials of degree at most n over F are finite in number, since F is finite and thus a splitting field, K, for the whole bunch can be found that is finite over F but then every polynomial in F[x] splits over K. This in fact is enough to make K the algebraic closure of F (the usual process would iterate the finding roots of polynomials to get fields F_0=F, F_1, F_2, etc. but it turns out that F_1 is the algebraic closure anyway). Algebraic closures are infinite. Thus K is infinite and finite at the same time, a contradiction. Thus there are an infinite number of primes as required. (The algebraic closure is infinite as in general, x^n-1, say, will have lots of roots. If n is coprime to p=char(F), then x^n-1 and nx^{n-1} are coprime so x^n-1 doesn't have repeated roots, so it has at least n roots.) :seems to be a way to do this similar to how Euler's zeta function is :used to show there an infinite number of the usual primes. Can somebody :help me? : :Thanks : : : :-- :Reality is that which refuses to go away when I stop believing in it. : : Phillip K. Dick ============================================================================== From: Richard Carr Subject: Re: infinity of "primes" Date: Mon, 24 Apr 2000 02:17:39 -0400 Newsgroups: sci.math On Sun, 23 Apr 2000, trichy wrote: :Date: Sun, 23 Apr 2000 22:23:22 -0700 :From: trichy :Newsgroups: sci.math :Subject: Re: infinity of "primes" : :Richard is correct, and his proof gives some nice insight into :good ways to think about :field theory. If you want something a little less high powered, Thanks. I didn't think it was that good myself, but it was the best I could think of at the time. :consider the following. Let's first show that there are :infinitely many prime numbers (we are working in the integers). :Suppose not. Then let the primes :be p1, p2, ... , pn. Look at 1+p1*p2*...*pn. None of the primes :divide it, so it's a prime, which is a contradiction. With a :little thought, it should be clear that this argument will :generalize to F[x] regardless of F. : :Tom : :* Sent from RemarQ http://www.remarq.com The Internet's Discussion Network * :The fastest and easiest way to search and participate in Usenet - Free! : :