Find $a, b, c$ such that element $x=a\alpha^2+b\alpha+c \in \mathbb{Q}[x]/\langle x^3+x+11 \rangle$

Question

I'm trying to generate a fundamental unit of the number field $K=\mathbb{Q}(\alpha)$, where $\alpha^3+\alpha+11$.
I found a non-trivial unit and I need to find $a,b,c\in\Bbb{Q}$ such that $$\frac{(-5\alpha^2-4\alpha+8)(6\alpha^2+6)(10\alpha^2+10\alpha)}{(8\alpha^2+8\alpha)^6(9\alpha^2+9\alpha)^{11}(10\alpha^2)}= a\alpha^2+b\alpha+c.$$ Does anyone know how to compute $a, b, c$ on Wolfram alpha or Sagemath?

Answer 1

A quick google search leads to the "Number field element" page of the Sage documentation, which shows that the following code does the trick in SageMath:

K.<a> = NumberField(x^3 + x + 11)
f = a.coordinates_in_terms_of_powers()
f((-5*a^2-4*a+8)*(6*a^2+6)*(8*a^2+8)^(-6)*(9*a^2+9*a)^(-11)*(10*a^2)^(-1)*(10*a^2+10*a))

Answer 2

If we define the number field K defined by the degree three polynomial in the question, with a as a generator, then Sage automatically expresses elements in K as polynomials in a.

sage: K.<a> = NumberField(x^3+x+11)
sage: n = (-5*a^2 - 4*a + 8) * (6*a^2 + 6) * (10*a^2 + 10*a)
sage: d = (8*a^2 + 8)^6 * (9*a^2 + 9*a)^11 * (10*a^2)
sage: u = n/d
sage: u
-37655851421063/7501767430252463720585645819465760768*a^2
+ 3549027946847/340989428647839260026620264521170944*a
- 399196880586881/15003534860504927441171291638931521536

The answer by @Servaes is useful to get a list of coefficients instead of a polynomial in a.

Note that in that list, the coefficients are ordered by increasing powers of a: first the coefficient of a^0 (i.e. the constant coefficient), then of a^1, then of a^2.

Another way to get those, without using the coordinates_in_terms_of_powers method, is as follows:

sage: u.polynomial().coefficients()
[-399196880586881/15003534860504927441171291638931521536,
 3549027946847/340989428647839260026620264521170944,
 -37655851421063/7501767430252463720585645819465760768]