Correct AlgebraicElement.denominator

[skirpichev]

For example,

In [22]: a = RootOf(4*x**2 + 2*x + 1, 0)

In [23]: A = QQ.algebraic_field(a)

In [24]: A.from_expr(a)
Out[24]: 
        ___  
  1   ╲╱ 3 ⋅ⅈ
- ─ - ───────
  4      4   

In [25]: _.denominator
Out[25]: mpz(4)

In [26]: a
Out[26]: 
        ___  
  1   ╲╱ 3 ⋅ⅈ
- ─ - ───────
  4      4   

Correct answer is 2. E.g. see the Mathematica:

In[1]:= AlgebraicNumberDenominator[-1/4 - 1/4*3^(1/2)*I]

Out[1]= 2

Refs Arno1996alg has correct implementation.

It worth also investigating if making the primitive element to be an algebraic integer - a good idea.

[skirpichev]

Or we might try to drop the need for this (numerator/denominator are for the rootisolation, which invokes clear_denoms()).