Comment #59 on issue 1598 by mattpap: New polynomials manipulation module
http://code.google.com/p/sympy/issues/detail?id=1598
There is actually quite a lot more you can do:
1. You can factor a polynomial over a non trivial domain:
In [1]: factor(x**2-2, extension=sqrt(2)+sqrt(3))
Out[1]:
⎛ ⎽⎽⎽⎞ ⎛ ⎽⎽⎽⎞
⎝x + ╲╱ 2 ⎠⋅⎝x - ╲╱ 2 ⎠
and you arrived with a result you could get in QQ(sqrt(2)). But you can
perform
formal factorization and see the difference:
In [2]: a = Symbol('alpha')
In [3]: factor(x**2-2, extension=AlgebraicNumber(sqrt(2)+sqrt(3), alias=a))
Out[3]:
⎛ 3⎞ ⎛ 3⎞
⎜ 9⋅α α ⎟ ⎜ 9⋅α α ⎟
⎜x + ─── - ──⎟⋅⎜x - ─── + ──⎟
⎝ 2 2 ⎠ ⎝ 2 2 ⎠
We obtained two linear factors in terms of sqrt(2)+sqrt(3) (alpha). This
can be
verified by substitution:
In [4]: (-a**3/2 + 9*a/2).subs(a, sqrt(2)+sqrt(3)).expand()
Out[4]:
⎽⎽⎽
-╲╱ 2
In [5]: (a**3/2 - 9*a/2).subs(a, sqrt(2)+sqrt(3)).expand()
Out[5]:
⎽⎽⎽
╲╱ 2
2. We can compute primitive elements of two or more extensions, e.g.:
In [6]: primitive_element(sqrt(2), sqrt(3))
Out[6]:
⎛ ⎽⎽⎽ ⎽⎽⎽⎞
⎝Poly(_x**4 - 10*_x**2 + 1, _x, domain='QQ'), ╲╱ 2 + ╲╱ 3 ⎠
3. We can compute a function with translates an algebraic field into
another field if
possible, e.g.:
In [7]: field_isomorphism(sqrt(2), sqrt(2)+sqrt(3))
Out[7]: [1/2, 0, -9/2, 0]
Which means QQ(sqrt(2)) and QQ(sqrt(2)+sqrt(3)) are isomorphic, the first
is a
subfield of the other, and we got explicit isomorphism as a side effect.
However,
QQ(sqrt(2)) is not isomorphic to QQ(sqrt(3)) (there is no way to express
sqrt(2) in
terms of sqrt(3)):
In [8]: field_isomorphism(sqrt(2), sqrt(3))
(the result is None).
In [9]: from sympy.polys.algebratools import QQ
Actually we can use __contains__ (in) to get a boolean result:
In [10]: sqrt(2) in QQ.algebraic_field(sqrt(2)+sqrt(3))
Out[10]: True
In [11]: 2+3*sqrt(3) in QQ.algebraic_field(sqrt(2)+sqrt(3))
Out[11]: True
In [12]: sqrt(2) in QQ.algebraic_field(sqrt(3))
Out[12]: False
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