Is Eisenstein all powerful?
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From: Bill Dubuque on 5 Aug 2010 21:17 quasi <quasi(a)null.set> wrote: > > <various questions about the Eisenstein irreducibility test> To better understand the Eisenstein and related irreducibility tests you should learn about Newton polygons. It's the "master theorem" behind all these related results. A good place to start is Filaseta's notes - see the links below from one of my old posts here. They may be stale but you can probably get a copy by emailing him if need be. --Bill Dubuque [1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/ [2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf [3] Newton Polygon Applet http://www.math.sc.edu/~filaseta/newton/newton.html [4] Abhyankar, Shreeram S. Historical ramblings in algebraic geometry and related algebra. Amer. Math. Monthly 83 (1976), no. 6, 409-448. http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAGA%3E
From: quasi on 5 Aug 2010 23:00 On 05 Aug 2010 21:17:55 -0400, Bill Dubuque <wgd(a)nestle.csail.mit.edu> wrote: >quasi <quasi(a)null.set> wrote: >> >> <various questions about the Eisenstein irreducibility test> > >To better understand the Eisenstein and related irreducibility tests >you should learn about Newton polygons. It's the "master theorem" behind >all these related results. A good place to start is Filaseta's notes - >see the links below from one of my old posts here. They may be stale >but you can probably get a copy by emailing him if need be. > >--Bill Dubuque > >[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/ > >[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf > >[3] Newton Polygon Applet >http://www.math.sc.edu/~filaseta/newton/newton.html > >[4] Abhyankar, Shreeram S. >Historical ramblings in algebraic geometry and related algebra. >Amer. Math. Monthly 83 (1976), no. 6, 409-448. >http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAGA%3E Thanks, Bill -- I'll investigate those leads when I get a chance. quasi
From: Roman B. Binder on 7 Aug 2010 03:55 > On Wed, 4 Aug 2010 11:09:18 -0700 (PDT), Arturo > Magidin > <magidin(a)member.ams.org> wrote: > > >On Aug 4, 1:07 pm, quasi <qu...(a)null.set> wrote: > >> On Wed, 04 Aug 2010 01:07:43 -0500, quasi > <qu...(a)null.set> wrote: > >> >On Tue, 03 Aug 2010 16:27:52 -0500, Robert Israel > >> ><isr...(a)math.MyUniversitysInitials.ca> wrote: > >> > >> >>quasi <qu...(a)null.set> writes: > >> > >> >>> On Tue, 03 Aug 2010 13:04:31 -0500, quasi > <qu...(a)null.set> wrote: > >> > >> >>> >Prove or disprove: > >> > >> >>> >If f in Z[x] with deg(f) > 1 is irreducible > then there exists g in > >> >>> >Z[x] with deg(g) >= 1 such that the > polynomial f(g(x)) is irreducible > >> >>> >by the Eisenstein criterion. > >> > >> >>> To clarify ... > >> > >> >>> It's immediate that if f(g(x)) is irreducible > and deg(g)>=1, then f(x) > >> >>> is also irreducible. > >> > >> >>> But the problem asks, for irreducible f with > deg(f) > 1, is there > >> >>> always such a g for which the Eisenstein > criterion could be _directly_ > >> >>> applied to prove the irreducibility of f(g(x)) > (and thus indirectly > >> >>> establish the irreducibility of f). > >> > >> >>> quasi > >> > >> >>Try f(x) = x^2 + 4. Suppose Eisenstein worked > for > >> >>g(x) = sum_{j=0}^n a_j x^j, > >> >>f(g(x)) = (a_0^2 + 4) > >> >> + sum_{k=1}^{2n} > sum_{i=max(0,k-n)}^{min(k,n)} a_i a_{k-i} x^k > >> > >> >>Since a0^2 + 4 == 0 or 1 mod 4, p can't be 2. > >> >>Now p does not divide a_n, but (considering the > case k=2n-1) does divide > >> >>a_{n-1}, and (similarly by looking at k=2n-2, > ..., k=n) must divide > >> >>a_{n-2}, ..., a_0. But then p can't divide > a_0^2 + 4, contradiction. > >> > >> >A nice, simple counterexample. > >> > >> >Thanks. > >> > >> At this point, it appears that the Eisenstein > criterion can only be > >> regarded as a special case method. It works > beautifully when it works, > >> but its lack of general power is exposed by its > apparent inability to > >> establish the irreducibility of x^2 + 4. > >> > >> However, let's boost the power of Eisenstein and > try again ... > >> > >> An amplified Eisenstein criterion: > >> > >> Let f in Z[x] with deg(f) = n > 1 be given by > >> > >> f(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_0 > >> > >> Suppose there is a prime p, and a positive integer > k such that > >> > >> (1) p does not divide a_n > >> (2) p^k divides a_i for i = 0 ... (n-1) > >> (3) p^(k+1) does not divide a_0 > >> > >> Then f is irreducible. > >> > >> Proof: Omitted for now, but almost line by line (I > think) the same as > >> the proof of the ordinary Eisenstein criterion. I > hope it's correct. > > > >Seems hard to claim that the proof will work line > for line, given that > >one of the first things one does in the usual proof > is note that since > >p^2 does not divide a_0, then p does not divide the > constant term of > >one of the putative factors of f... How will that be > translated here? > > > >Or, one works by going to Z/pZ[x] via reduction, in > which case you > >have that if f(x)=g(x)h(x), then reducing modulo p > you have a_nx^n = > >G(x)H(x), and from the fact that Z/pZ is an integral > domain you > >conclude that G(x) and H(x) both have zero constant > term, giving a > >contradiction; this does not work in Z/p^kZ. > > Yes, I see the issue clearly now. My error was a > trivial arithmetic > one (a failure to divide by 2), resulting in inflated > divisors. > > >Perhaps you ought to try to write it out carefully > first. > > No point now, since the result is false. > > But yes, I posted too quickly based on a visualized > proof, done all in > my head. > > Sorry. > > quasi Hi quasi, I think, the most simple, once we'd like to check proper results of Eisenstein criterion is to come back to general proof of it: Let (a2)x^2 + (a1)x + (a0) = f(x) some p#(a2) but p|(a1) and (a0) but p^2#(ao)......(B.C,) suppose we have some factorization: [(b1)x +(b0)][(c1)x + (c0)= f(x) Now: (b1)(c1)x^2 +[(b1)(c0) + (b0)(c1)]x + (b0)(c0) = f(x) Now once (b0)(c0) = (a0) and b;c of gcd=1..,(*) so only (b0) or (c0) could be divided by p (B.C) Let p|(b0) but p#(c0) ; Now in the therm (a1) = (b1)(c0) + (b0)(c1) the division by p could be completed only if p|(b1) and once (a2) = (b1)(c1) in (B.C.) used to be defined as not divisible by p so it is contradiction: Therefore Eisenstein conditions are true here: only once p^2|(a0) we'll able to find some factorization... But once (a1)=0 You can see as some inductive chain is missed here and therefore: x^2 +4 is not reducible once x^2 -4 is reducible. Generally for such polynomials, where inductive chain is broken and can not be obtain such simple inductive proof contradicting p#(an), Eisenstein criterion can not work properly. (*)it is to noticed that (an);(an-1);..(ak);..(a2);(a1) can not have some common factor to obtain some trivial factorization: f(x) = hf'(x) so simple (b)(c) of gcd=1 Thank You for Your attention ! P.S. How are You ! Nowadays I am trying to find sufficient condition not allowing F(x)/H(x) completely. It looks once coefficients of the oldest therm are of gcd=1 and once also absolute terms are of gcd=1 so there is the deal: when we suppose some polynomial G(x) to complete the division; let F(x) is n degree; H(x) of n-k degree so G(x) of k degree and should be: (fn)=[h(n-k)](gk) and (f0)=(h0)(g0); then once: {(fn);[h(n-k)]} of gcd=1 and [(f0);(h0)] of gcd=1....(**) so such division for primitive polynomials F(x) and H(x) is not possible. ( it looks even once the one of these two simple condition fails such division could not be completed: (suppose we can vary (fn);(f0);[h(n-k);(h0); to be the most simple numbers eg. 1 or 2 but one of (**) conditions still holds ? Thanks Again. With The Best Regards Ro-Bin
From: Bill Dubuque on 11 Aug 2010 12:58
quasi <quasi(a)null.set> wrote: > Bill Dubuque <wgd(a)nestle.csail.mit.edu> wrote: >>quasi <quasi(a)null.set> wrote: >>> >>> <various questions about the Eisenstein irreducibility test> >> >>To better understand the Eisenstein and related irreducibility tests >>you should learn about Newton polygons. It's the "master theorem" behind >>all these related results. A good place to start is Filaseta's notes - >>see the links below from one of my old posts here. They may be stale >>but you can probably get a copy by emailing him if need be. >> >>--Bill Dubuque >> >>[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/ >> >>[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf >> >>[3] Newton Polygon Applet >>http://www.math.sc.edu/~filaseta/newton/newton.html >> >>[4] Abhyankar, Shreeram S. >>Historical ramblings in algebraic geometry and related algebra. >>Amer. Math. Monthly 83 (1976), no. 6, 409-448. >>http://links.jstor.org/sici?sici=0002-9890(197606/07)83:6%3C409:HRIAGA%3E > > Thanks, Bill -- I'll investigate those leads when I get a chance. Btw, you may also find this of interest 2001i:11103 11K65 (11K99 13B25) Dobbs, David E.(1-TN); Johnson, Laura E.(1-TN) On the probability that Eisenstein's criterion applies to an arbitrary irreducible polynomial. * Advances in commutative ring theory (Fez, 1997), 241--256, Lecture Notes in Pure and Appl. Math., 205, Dekker, New York, 1999. The authors consider monic random trinomials f = X^m + a X + b with integer coefficients. They consider the following question: What is the probability that Eisenstein's criterion can be applied to f (for some prime number p)? For such polynomials with height at most n, they prove that this probability p_n satisfies 0.2294 < p_n < 0.2784 when n is large enough. The proof involves a careful study of some series related to classical arithmetical functions. |