On Sun, Feb 8, 2009 at 12:24 PM, shreevatsa <[email protected]> wrote:
>
> Hello,
>
> A friend of mine wanted to know all the conjugates of a particular
> algebraic number, and I was trying to figure out the right way of
> finding them with Sage. Sorry if this is too trivial a question.
>
> This is what I tried:
>
> sage: t = sqrt(2-sqrt(2)) + i*sqrt(sqrt(2)-1)
Since you're lucky and all conjugates are expressible in terms of
radicals, the following
easily gives all 8:
sage: t = sqrt(2-sqrt(2)) + i*sqrt(sqrt(2)-1)
sage: f = t.minpoly(); f
x^8 - 12*x^6 + 6*x^4 - 12*x^2 + 1
sage: solve(SR(f)==0,x)
[x == -sqrt(2*2^(1/4)*sqrt(2*sqrt(2) + 3) + 2*sqrt(2) + 3), x ==
sqrt(2*2^(1/4)*sqrt(2*sqrt(2) + 3) + 2*sqrt(2) + 3), x ==
-sqrt(-2*2^(1/4)*sqrt(2*sqrt(2) + 3) + 2*sqrt(2) + 3), x ==
sqrt(-2*2^(1/4)*sqrt(2*sqrt(2) + 3) + 2*sqrt(2) + 3), x ==
-sqrt(2*2^(1/4)*sqrt(3 - 2*sqrt(2))*I - 2*sqrt(2) + 3), x ==
sqrt(2*2^(1/4)*sqrt(3 - 2*sqrt(2))*I - 2*sqrt(2) + 3), x ==
-sqrt(-2*2^(1/4)*sqrt(3 - 2*sqrt(2))*I - 2*sqrt(2) + 3), x ==
sqrt(-2*2^(1/4)*sqrt(3 - 2*sqrt(2))*I - 2*sqrt(2) + 3)]
> sage: p = t.minpoly(); p
> x^8 - 12*x^6 + 6*x^4 - 12*x^2 + 1
> sage: K.<a> = NumberField(p)
> sage: a.galois_conjugates(K)
> [a, -a, a^7 - 12*a^5 + 6*a^3 - 12*a, -a^7 + 12*a^5 - 6*a^3 + 12*a]
>
> I copied this result into a text editor to replace 'a' with 'x', then
> did
>
> sage: ts = [q(t).exp_simplify() for q in [x, -x, x^7 - 12*x^5 +
> 6*x^3 - 12*x, -x^7 + 12*x^5 - 6*x^3 + 12*x]]; ts
> [sqrt(sqrt(2) - 1)*I + sqrt(2 - sqrt(2)),
> -sqrt(sqrt(2) - 1)*I - sqrt(2 - sqrt(2)),
> sqrt(sqrt(2) - 1)*I - sqrt(2 - sqrt(2)),
> sqrt(2 - sqrt(2)) - sqrt(sqrt(2) - 1)*I]
>
> which are indeed conjugates, so it's good. But what about the other 4
> roots? I guess the problem is that NumberField(p) is not big enough to
> contain all the conjugates. Trying
>
> sage: a.galois_conjugates(CC)
>
> does give all 8 conjugates, but in decimal numbers, not in terms of
> radicals, while one can get all 8 by using:
>
> sage: a.galois_conjugates(QQ[t,i])
> [a^7 + (-12)*a^5 + 6*a^3 + (-12)*a,
> (-1/8*I - 3/8)*a^7 + (11/8*I + 35/8)*a^5 + (5/8*I - 5/8)*a^3 +
> (17/8*I + 21/8)*a,
> (1/8*I - 3/8)*a^7 + (-11/8*I + 35/8)*a^5 + (-5/8*I - 5/8)*a^3 +
> (-17/8*I + 21/8)*a,
> (-1)*a,
> a,
> (-1/8*I + 3/8)*a^7 + (11/8*I - 35/8)*a^5 + (5/8*I + 5/8)*a^3 +
> (17/8*I - 21/8)*a,
> (1/8*I + 3/8)*a^7 + (-11/8*I - 35/8)*a^5 + (-5/8*I + 5/8)*a^3 +
> (-17/8*I - 21/8)*a,
> (-1)*a^7 + 12*a^5 + (-6)*a^3 + 12*a]
>
> after which replacing the 'a's with 'x' and pasting it back as above
> gives all the conjugates:
>
> sage: ts = [q(t).exp_simplify() for q in (--replaced expression,
> snipped--) ]; ts
>
> [sqrt(sqrt(2) - 1)*I - sqrt(2 - sqrt(2)),
> (-sqrt(2) - 1)*sqrt(sqrt(2) - 1) + (-sqrt(2) - 1)*sqrt(2 - sqrt
> (2)),
> sqrt(sqrt(2) - 1)*(sqrt(2) + 1) + (-sqrt(2) - 1)*sqrt(2 - sqrt
> (2)),
> -sqrt(sqrt(2) - 1)*I - sqrt(2 - sqrt(2)),
> sqrt(sqrt(2) - 1)*I + sqrt(2 - sqrt(2)),
> sqrt(2 - sqrt(2))*(sqrt(2) + 1) + (-sqrt(2) - 1)*sqrt(sqrt(2) -
> 1),
> sqrt(sqrt(2) - 1)*(sqrt(2) + 1) + sqrt(2 - sqrt(2))*(sqrt(2) +
> 1),
> sqrt(2 - sqrt(2)) - sqrt(sqrt(2) - 1)*I]
>
> So Sage completely helped with the question, which is great. :-)
>
> My question is whether there is a better way to do this. In
> particular:
>
> 1. Is there a way to avoid manually replacing 'a' with 'x' as above?
> 2. Here, K=NumberField(p) was not a big enough field, but K[i] was. Is
> it possible to automatically choose an appropriately big field? [I
> tried a.galois_conjugates(K.galois_closure('a1')) as in the
> documentation, but wasn't sure how to interpret the result.]
> 3. It seems that while p.roots() returns an empty list and
> p.complex_roots() returns the roots in decimals,
>
> sage: (x^8 - 12*x^6 + 6*x^4 - 12*x^2 + 1).roots()
>
> does find all roots, in terms of radicals. So might it be possible to
> write a conjugates() function in cases where the polynomial *can* be
> solved with radicals?
> >
>
--
William Stein
Associate Professor of Mathematics
University of Washington
http://wstein.org
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