20171103, 06:36  #1 
"Sam"
Nov 2016
326_{10} Posts 
Modular Exponentiation results in PFGW
Long story short, I am working with PARI/GP to explore and discover more about cyclotmic field extensions as described here.
Here, polcyclo(n) is the nth cyclotomic polynomial. Code:
J = idealhnf(bnfinit(polcyclo(n)),p,xw); idealnorm(bnfinit(polcyclo(n)),J) Jp = bnfisprincipal(bnfinit(polcyclo(n)),J) u = nfbasistoalg(bnfinit(polcyclo(n)),Jp[2]) norm(u) Code:
(22:40) gp > J = idealhnf(bnfinit(polcyclo(5)),11,x3); (22:40) gp > idealnorm(bnfinit(polcyclo(5)),J) %402 = 11 (22:40) gp > Jp = bnfisprincipal(bnfinit(polcyclo(5)),J) %403 = [[]~, [1, 1, 1, 0]~] (22:40) gp > u = nfbasistoalg(bnfinit(polcyclo(5)),Jp[2]) %404 = Mod(x^2  x + 1, x^4 + x^3 + x^2 + x + 1) (22:40) gp > norm(u) %405 = 11 (22:40) gp > Code:
w = Mod(x,f=polcyclo(n));bnf = bnfinit(f); uu = bnfisintnorm(bnf,p); #uu Code:
(22:48) gp > w = Mod(x,f=polcyclo(5));bnf = bnfinit(f); (22:48) gp > uu = bnfisintnorm(bnf,11); #uu %407 = 4 In the field Kn, if there are no principal ideals of norm p, then its principal generators are of the form a/b. Here b is any prime q = 1 (mod n), or a product of prime factors congruent to 1 (mod n). Let b be a perfect kth power (This will be important later, look at ***). The resulting polynomial P(x) in Mod(P(x),polcyclo(n) will have terms of the form a/b*x^n where a, b, n are all integers. If we were to get rid of the "/b" part in P(x), we would have a polynomial Q(x) with integer coefficients, and terms of the form a*x^n, where a, n are integers. (***) If u = Mod(Q(x),polcyclo(n)), then the norm of u is p*b^(n*k2*k). To come up with all elements of norm p*b^(n*k2*k), we can refer to this code, Code:
w = Mod(x,f=polcyclo(n));bnf = bnfinit(f); uu = bnfisintnorm(bnf,p); #uu Here we can use: Code:
J = idealhnf(bnfinit(polcyclo(n)),p*b^(n*k2*k),xw); idealnorm(bnfinit(polcyclo(n)),J) Jp = bnfisprincipal(bnfinit(polcyclo(n)),J) u = nfbasistoalg(bnfinit(polcyclo(n)),Jp[2]) norm(u) c^(phi(D)/n) (mod D) = w This may seem like an easy task, but it is not. Let us work with polcyclo(23), n = 23, K23. Here we choose p = 139, which is nonprincipal, b = 461, k = 2, and base c = 3. In order to find w and solve this problem, we need to compute, 3^(phi(139*461^(23*22*2))/23) (mod 139*461^(23*22*2)) = w 3^(2760*461^41) (mod 139*461^42) = w I tried running it into PFGW using % to represent modulo, and attempted to compute 3^(2760*461^41) (mod 139*461^42) Code:
C:\Users\Documents\PFGW\PFGW>pfgw64.exe code.txt PFGW Version 3.8.3.64BIT.20161203.Win_Dev [GWNUM 28.6] ***WARNING! file code.txt may have already been fully processed. 3^(2760*461^41)%(139*461^42)  Evaluator failed C:\Users\Documents\PFGW\PFGW> 
20171103, 06:56  #2  
Sep 2002
Database er0rr
3^{2}·11·37 Posts 
Quote:
Code:
? Mod(3,139*461^42)^(2760*461^41) Mod(1, 1043406809315900616956999799105701752849906128374948710828030894314678039911009065037910659790911409501482999082019) 

20171103, 07:24  #3  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
24C9_{16} Posts 
Quote:
You can't do that for any sufficiently large (which is really quite small) number. You will run out of atoms in the universe to hold your intermediate result. 

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