Re: A question about algebraic numbers using Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg62794] Re: [mg62762] A question about algebraic numbers using Mathematica
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 5 Dec 2005 13:40:57 -0500 (EST)
- References: <200512050837.DAA08323@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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
- References:
- A question about algebraic numbers using Mathematica
- From: Kent Holing <KHO@statoil.com>
- A question about algebraic numbers using Mathematica