I'm curious how the expressions for the coefficients of the
reciprocal change if the question had been about the field
Q(tau) if tau = 2^(1/3) * exp(2πi/3), rather than about the
field Q(2^(1/3)).
—Dan
Rich Schroeppel wrote:
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The product of (all) the conjugates is the norm.
The reciprocal of one field element is the product of the other
conjugates, divided by the norm.
For a cube root w, the other two conjugates of a0 + a1w + a2ww
are, (using u^3=1 and recalling that u+uu = -1):
a0 + a1wu + a2wwuu and a0 + a1wuu + a2wwu.
The norm is a0^3 + 2a1^3 + 4a2^3 - 6a0a1a2.
[side puzzle: prove this is never zero for rational a0,a1,a2 unless
a0=a1=a2=0.]
The "rational" part (the w^0 term) of the product of the other two
conjugates is (selecting three terms from the nine term product)
a0^2 + a1wu * a2wwu + a1wuu * a2wwuu,
which, after using w^3=2, uuuu=u, and u+uu = -1,
becomes a0^2 + a1 a2 w^3 (uu + uuuu), and then a0^2 - 2a1a2.
b0 is just this expression divided by the norm.
b1w and b2ww come from a circular substitution a0 -> a2ww -> a1w -> a0,
which, of course, is equivalent to Henry's answer.
Higher roots follow the same principles, but are more complicated.
Full polynomials, with intermediate degree terms, are another step up in
complexity, but the same method will work.
I was working with the 2^(1/5) field, and needed to compute reciprocals.
I worked out the complete formula, which came out fairly large.
I found a better computational method was to work with only the first
half of the radical terms, using F + G 5rt2 + H 5rt4, which has a much
simpler reciprocal formula and norm formula. For a general number
with all five terms, A + B 5rt2 + C 5rt4 + D 5rt8 + E 5rt16,
it was easy to come up with a multiplier X + Y 5r2 + Z 5r4 which zeroed
out the 5r8 and 5r16 terms in the product. This gives the equation
(A + B 5rt2 + C 5rt4 + D 5rt8 + E 5rt16) * (X + Y 5r2 + Z 5r4) =
P + Q 5r2 + R 5r4.
Then I used the shorter reciprocal formula to compute 1 / (P + Q 5r2 + R 5r4),
and multiply that by X + Y 5r2 + Z 5r4, getting the reciprocal of the
full width field element A + B 5rt2 + C 5rt4 + D 5rt8 + E 5rt16.
This idea also works for the cbrt2 field, but doesn't give any computing
advantage. It is helpful for tribonacci fields.
Rich
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Quoting Henry Baker <hbaker1(a)pipeline.com>:
> BTW, "maxima" is free...
>
> 2
> 2 a1 a2 - a0
> (%o2) [[b0 = - --------------------------------,
> 3 3 3
> 4 a2 - 6 a0 a1 a2 + 2 a1 + a0
> 2
> 2 a2 - a0 a1
> b1 = --------------------------------,
> 3 3 3
> 4 a2 - 6 a0 a1 a2 + 2 a1 + a0
> 2
> a0 a2 - a1
> b2 = - --------------------------------]]
> 3 3 3
> 4 a2 - 6 a0 a1 a2 + 2 a1 + a0
>
>
> At 08:32 AM 4/5/2018, Allan Wechsler wrote:
>> Suppose w is the real cube root of 2, w^3 = 2.
>>
>> Then for a0, a1, a2 rational, numbers of the form a0 + a1*w + a1*w^2 form a
>> field. What I want is the formula for the multiplicative inverse in this
>> field. It's elementary algebra, but it is voluminous and I keep making
>> errors when I try to do it by hand. Any computer algebra system could
>> probably instantly solve for b0, b1, b2 subject to (a0 + a1w + a2ww)(b0 +
>> b1w + b2ww) = 1. Can somebody do me a favor and give me formulas for the
>> b's in terms of the a's?
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