Re: A question about algebraic numbers using Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg62804] Re: A question about algebraic numbers using Mathematica*From*: dh <dh at metrohm.ch>*Date*: Mon, 5 Dec 2005 13:41:12 -0500 (EST)*References*: <dn0ug5$8b7$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Kent, I assume by Q[r] you mean the field extension of the rationals by r. Because of the extension field minimal polynomial p[x]: x^4 - 2c x^3 + (c^2 - 2a^2) x^2 + 2a^2 c x - a^2 c^2 p[r]==0 r^4 can be written as r^4== -(- 2c x^3 + (c^2 - 2a^2) x^2 + 2a^2 c x - a^2 c^2) This shows that all numbers can be written as polynomials in r with dgeree <=3. Therefore, to get the inverse of 2-r we need to determine a number y: y= y0 + y1 r + y2 r^2 + y3 r^3 so that y (2-r) ==1 or y(2-r)-1 ==0 Towards this aim, we multiply and expand y(2-r)-1, then replace r^4 by lower powers and finally collect coefficients of factors of r: y= y0 + y1 r + y2 r^2 + y3 r^3; rep= r^4-> -(- 2c r^3 + (c^2 - 2a^2) r^2 + 2a^2 c r - a^2 c^2); res1= CoefficientList[Expand[y (2-r)-1] /. rep ,r] Now, because of y(2-r)-1 ==0, all these coefficients must er zero: Solve[ Thread[res1=={0,0,0,0}], {y0,y1,y2,y3} ] this gives the inverse of 2-r Daniel 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 >