20170217, 02:47  #12 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·5^{2}·43 Posts 
Sage, may or may not do what you need. You will probably be able to figure it better than me.
http://www.sagemath.org/tour.html 
20170217, 02:57  #13 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
probably why you can make your own scripts in PARI/GP. also you can use Pol to convert a vector of coefficients to a polynomial. you could also maybe use the new powers command.

20170217, 16:19  #14 
Feb 2017
Nowhere
2^{3}·7·89 Posts 
I figured out the idea to this: polsubcyclo(n, d) will give ALL polynomials with the same number field as the cyclotomic polynomial d if and only if d  phi(n).
The same number field, you say? I'll ask Pari for a cursory description: ? ?polsubcyclo polsubcyclo(n,d,{v=x}): finds an equation (in variable v) for the dth degree subfields of Q(zeta_n). Output is a polynomial or a vector of polynomials is there are several such fields, or none.[/I] The dth degree subfields are not the same as the whole cyclotomic field if d < phi(n). (17:57) gp > polsubcyclo(2201, 5) [...] I figured out the idea to this: polsubcyclo(n, d) will give ALL polynomials with the same number field as the cyclotomic polynomial d if and only if d  phi(n). I've organized the new polynomials. Dr. Sardonicus should take a look at these ones: x^5  x^4  880*x^3 + 176*x^2 + 179584*x + 26624 x^5 + x^4  28*x^3 + 37*x^2 + 25*x + 1 x^5  x^4  880*x^3 + 6779*x^2 + 14509*x  112039, x^5 + x^4  12*x^3  21*x^2 + x + 5 x^5  x^4  880*x^3 + 15583*x^2  95541*x + 196101 x^5  x^4  880*x^3  2025*x^2 + 49725*x  112039 All 5 appear to have the same number field properties as x^5+x^4+x^3+x^2+x+1 "The same number field as the cyclotomic polynomial d" you say? Eh? And it gives polynomials for all subfields of degree d, whether there are any or not (if there are none, it gives an empty vector). I also don't know what you mean, "the same number field properties." Do you? EDIT: The mistake is the polynomials are degree 5, not 4! That's not the only mistake. x^5 + x^4 + x^3 + x^2 + x + 1 is not irreducible. It is the product of the cyclotomic polynomials for the primitive 2nd, 3rd, and 6th roots of unity. The splitting field is the field Q(zeta_3), which is of degree 2. Let's see here, 2201 factors as 31*71. Q(zeta_2201) contains Q(zeta_31) and Q(zeta_71), so also the degree5 subfield of Q(zeta_31), and the degree5 subfield of Q(zeta_71). These will have discriminants of 31^4 and 71^4, respectively. There are also 4 other degree5 subfields, whose discriminants are divisible by both 31 and 71 (probably all 31^4 * 71^4, but I'm too lazy to check). If you want polynomials defining the same field as any given one (which I assume is monic and irreducible), it's very easy to find any number of them, but the methods I know do NOT involve specifying coefficients in advance. You might, after all, specify things beyond the realm of possibility. Besides, solving highdegree multivariate polynomial equations is, at best, time consuming even if there happen to be solutions. I would suggest that you learn the basics of the "theory of equations" before proceeding further. Last fiddled with by Dr Sardonicus on 20170217 at 16:21 
20170218, 19:46  #15  
"Sam"
Nov 2016
5^{2}×13 Posts 
Quote:
This time, using logic, and trial and error x^6+14x^4+21x^2+7 discriminant D = 2^6*7^5*13^4 defines the same (or very similar) field as x^6+x^5+x^4+x^3+x^2+x+1 discriminant D = 7^5. Other methods should find more degree 6 polynomials defining the same field as the cyclotomic polynomial for 7. 

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