#14239: symbolic radical expression for algebraic number
-------------------------------------+-------------------------------------
       Reporter:  gagern             |        Owner:  davidloeffler
           Type:  enhancement        |       Status:  needs_work
       Priority:  major              |    Milestone:  sage-6.4
      Component:  number fields      |   Resolution:
       Keywords:                     |    Merged in:
        Authors:  Martin von Gagern  |    Reviewers:  Marc Mezzarobba,
Report Upstream:  N/A                |  Jeroen Demeyer
         Branch:                     |  Work issues:
  u/gagern/ticket/14239              |       Commit:
   Dependencies:  #17495, #16964     |  09683145989c437b59e02c02409d1dac05ac379d
                                     |     Stopgaps:
-------------------------------------+-------------------------------------

Comment (by gagern):

 Replying to [comment:82 jdemeyer]:
 > Can you please say where this example comes from?

 I was looking for numbers with a minpoly of degree more than four which
 allow for radical expressions but where computing said expression might
 take some time. I intended to use this in a performance comparison. This
 particular one I found using

 {{{
 sage: R.<x,y>=QQ[]
 sage: (x^2 - 5*x + 2 - y).resultant(y^3 + y - 4, y)
 -x^6 + 15*x^5 - 81*x^4 + 185*x^3 - 163*x^2 + 65*x - 6
 sage: _.univariate_polynomial().roots(QQbar, False)
 [0.1274914752251046?,
  4.872508524774896?,
  0.5703693585365878? - 0.4035749775931049?*I,
  0.5703693585365878? + 0.4035749775931049?*I,
  4.429630641463412? - 0.403574977593105?*I,
  4.429630641463412? + 0.403574977593105?*I]
 sage: _[1].minpoly()
 x^6 - 15*x^5 + 81*x^4 - 185*x^3 + 163*x^2 - 65*x + 6
 sage: K.<a> = NumberField(_, embedding=4.9)
 }}}

 I've essentially been toying with the original two polynomials until I got
 something real and with reasonable conversion to symbolic. There is no
 deeper meaning behind the actual values of the coefficients. The
 conversion to symbolic is possible because the polynomial in `y` is cubic,
 so one can express `y` using radicals, and once you know `y` the
 polynomial in `x` is quadratic so that, too, can be expressed using
 radicals. I was somewhat surprised that the symbolic expression could
 avoid complex numbers along the way, since usually solving cubic equations
 requires those.

 I then did my performance test using `SR(a + k)` for some integer `k`. And
 at some point decided that I should try more complicated elements of that
 field as well, tried the `a + a^3` and was really surprised by the result.

--
Ticket URL: <http://trac.sagemath.org/ticket/14239#comment:83>
Sage <http://www.sagemath.org>
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