Re: Re: A question about algebraic numbers using Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg62814] Re: [mg62798] Re: [mg62762] A question about algebraic numbers using Mathematica*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Tue, 6 Dec 2005 00:03:10 -0500 (EST)*References*: <869C3FFA9CA97944A4EEF55353DE74245D4B91@ST-EXCL05.statoil.net> <200512051841.NAA21123@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote: > On 5 Dec 2005, at 23:38, Kent Holing wrote: > > >> thanks >>What about the following (I have thought more about the question)? >>Since in my case of interest the quartic q(x) and 2-x are irreducible >>over Q we must have that GCD[q(x),2-x] = 1 so there exist >>polynomials u >>and v such that u q + v (2-x) = 1. >>Since u is constant (the degree of u < the degree 2-x) and q(r) = 0 we >>have that v(2-r) = 1, so >>v is the inverse of 2-r in Q[r]. Is this an easier way of determing >>the >>inverse of 2-r? >>By the way, could you provide me with the inverse of 2c-r? > > > Yes, you are right of course, but I think you would want Mathematica > do this for you? Mathematica has a function PolynomialExtendedGCD in > the package Algebra`PolynomialExtendedGCD` and this function will do > exactly what you need when a,b and c are integers but will not work > with symbols. So although mathematically your argument is correct, I > do not think Mathematica can be of help in this. This will work after the next release. <<Algebra`PolynomialExtendedGCD` poly = r^4 - 2*c*r^3 + (c^2-2*a^2)*r^2 + 2*a^2*c*r - a^2*c^2; In[5]:= {g, {u,v}} = PolynomialExtendedGCD[poly, 2*c-r, r]; The inverse should be v: In[7]:= InputForm[v] Out[7]//InputForm= (-2*a^2*c)/(-5*a^2*c^2 + 4*c^4) + (2*c^3)/(-5*a^2*c^2 + 4*c^4) - (2*a^2*r)/(-5*a^2*c^2 + 4*c^4) + (c^2*r)/(-5*a^2*c^2 + 4*c^4) + r^3/(-5*a^2*c^2 + 4*c^4) We check that we have a reciprocal relation. In[8]:= PolynomialMod[v*(2*c-r), poly] Out[8]= 1 > As for your last question, I think it can be done in the same way. > Here are all the steps again: > > > First define the polynomial: > In[1]:= > poly = r^4 - 2*c*r^3 + (c^2 - 2*a^2)*r^2 + 2*a^2*c*r - > a^2*c^2; > > Next find the polynomial poly1 satisfied by (2c -r): > > In[2]:= > poly1 = First[GroebnerBasis[{poly, v - (2*c - r)}, > {a, b, c, v}, {r}]] > > Out[2]= > 4*c^4 - 12*v*c^3 - 5*a^2*c^2 + 13*v^2*c^2 - 6*v^3*c + > 6*a^2*v*c + v^4 - 2*a^2*v^2 > > Next find the part free of v: > > In[3]:= > free = poly1 /. v -> 0 > > Out[3]= > 4*c^4 - 5*a^2*c^2 > > The polynomial q below is our inverse: > > In[4]:= > poly2 = Cancel[(poly1 - free)/v] > > Out[4]= > -12*c^3 + 13*v*c^2 + 6*a^2*c - 6*v^2*c + v^3 - 2*a^2*v > > In[5]:= > q = Expand[poly2 /. v -> 2*c - r]/free > > Out[5]= > (-2*c^3 - r*c^2 + 2*a^2*c - r^3 + 2*a^2*r)/ > (4*c^4 - 5*a^2*c^2) > > Finally, we check that it really is the inverse, that is, that q*(2c - > r) gives remainder 1 on division by poly: > > In[6]:= > FullSimplify[PolynomialReduce[q*(2*c - r), poly, r]] > > Out[6]= > {{1/(c^2*(4*c^2 - 5*a^2))}, -1} > > Somehow I got -1 so the inverse is actually -q. (However as it is > already after midnight here I shall leave it at that). > > Andrzej Kozlowski > [...] The GroebnerBasis approach is a bit easier if one works with v*(2*c-r)-1 as a generator. In[12]:= InputForm[v2 = vv /. First[Solve[GroebnerBasis[{poly, vv*(2*c - r)-1}, {vv,r}, CoefficientDomain->RationalFunctions][[2]]==0, vv]]] Out[12]//InputForm= -((2*a^2*c - 2*c^3 + 2*a^2*r - c^2*r - r^3)/(c^2*(-5*a^2 + 4*c^2))) We check that this agrees with our previous result. In[13]:= Together[v2-v] Out[13]= 0 This method is basically how the Algebra` add-on function PolynomialPowerMod does negative powers. Daniel Lichtblau Wolfram Research

**References**:**Re: A question about algebraic numbers using Mathematica***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>