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
>