Re: A question about algebraic numbers using Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg62798] Re: [mg62762] A question about algebraic numbers using Mathematica*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 5 Dec 2005 13:41:02 -0500 (EST)*References*: <869C3FFA9CA97944A4EEF55353DE74245D4B91@ST-EXCL05.statoil.net>*Sender*: owner-wri-mathgroup at wolfram.com

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. 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 > > -----Original Message----- > From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl] To: mathgroup at smc.vnet.net > Sent: 5. desember 2005 15:25 > To: Kent Holing > Subject: [mg62798] Re: [mg62762] A question about algebraic numbers using > Mathematica > > > On 5 Dec 2005, at 17:37, Kent Holing wrote: > >> I want to find the inverse of 2 - r in Q[r] where r is a root of the >> equation >> x^4 - 2c x^3 + (c^2 - 2a^2) x^2 + 2a^2 c x - a^2 c^2 = 0 for a, b and >> c integers. >> >> Can this be done for general a, b and c? (I know how to do it for >> specific given numerical values of a, b and c.) >> >> Kent Holing >> > > > I think this can be done using Groebner basis. Suppose r satisfies the > above equation and let v be 2-r. Let us first find the algebraic > equation satisfied by v. > > In[1]:= > poly1 = First[GroebnerBasis[{r^4 - 2*c*r^3 + (c^2 - 2*a^2)*r^2 + > 2*a^2*c*r - a^2*c^2, v - (2 - r)}, {a, b, c, v}, > {r}]] > > Out[1]= > v^4 + 2*c*v^3 - 8*v^3 - 2*a^2*v^2 + c^2*v^2 - 12*c*v^2 + 24*v^2 + > 8*a^2*v - 4*c^2*v - 2*a^2*c*v + 24*c*v - 32*v - > 8*a^2 - a^2*c^2 + 4*c^2 + 4*a^2*c - 16*c + 16 > > From this we can easily see what the inverse of v is as a > polynomial in > v with rational coefficients, and by substituting 2-r for v we can > also > get a polynomial in r. We proceed as follows. First we find the free > term (not involving v) in poly1: > > In[2]:= > free = poly1 /. v -> 0 > > Out[2]= > (-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16 > > Next we subtract the free term from poly1 and divide by v: > > In[3]:= > poly2 = Cancel[(poly1 - free)/v] > > Out[3]= > v^3 + 2*c*v^2 - 8*v^2 - 2*a^2*v + c^2*v - 12*c*v + 24*v + 8*a^2 - > 4*c^2 - 2*a^2*c + 24*c - 32 > > Now substitute 2-r for v: > > In[4]:= > q = Expand[poly2 /. v -> 2 - r] > > Out[4]= > -r^3 + 2*c*r^2 - 2*r^2 + 2*a^2*r - c^2*r + 4*c*r - 4*r + 4*a^2 - > 2*c^2 - 2*a^2*c + 8*c - 8 > > The inverse ought to be the polynomial poly3 given by: > > In[5]:= > poly3 = Expand[(q/free)*(r - 2)] > > Out[5]= > -(r^4/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16)) + > (2*c*r^3)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + > (2*a^2*r^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (c^2*r^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (2*a^2*c*r)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (8*a^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + > (4*c^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + > (4*a^2*c)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (16*c)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + 16/((- > c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) > > Finally we check that this is indeed the inverse of r-2 modulo the > original polynomial p: > > In[6]:= > FullSimplify[PolynomialReduce[poly3, r^4 - 2*c*r^3 + (c^2 - 2*a^2) > *r^2 + 2*a^2*c*r - a^2*c^2, r]] > > Out[6]= > {{1/(a^2*((c - 4)*c + 8) - 4*(c - 2)^2)}, 1} > > We get the remainder 1, as we should. > > Andrzej Kozlowski > > > ------------------------------------------------------------------- > The information contained in this message may be CONFIDENTIAL and is > intended for the addressee only. 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**Follow-Ups**:**Re: Re: A question about algebraic numbers using Mathematica***From:*Daniel Lichtblau <danl@wolfram.com>