       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:

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
>

```

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