Finding a polynomial in $\mathbb{Z}[x]$.

Question

Let $r$ be a root of the polynomial $p(x)=(\sqrt{3}-\sqrt{2})x^3+\sqrt{2}x-\sqrt{3}+1.$ Find another polynomial $q(x)\in \mathbb{Z}[x] $ s.t $q(r)=0.$

I am unable to start. Please give me a hint so that I can proceed.

Thanks.

Answer 1

This is a polynomial with coefficients in $\mathbb{Q}[\sqrt{2},\sqrt{3}]$, so we can apply all automorphisms of $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ over $\mathbb{Q}$ to the coefficients. In particular, since the minimal polynomial of $\sqrt{2}$ is $x^2-2$ and the minimal polynomial of $\sqrt{3}$ is $x^2-3$, we have $\sigma_2$ which takes $\sqrt{2}$ to $-\sqrt{2}$ and $\sigma_3$ which takes $\sqrt{3}$ to $-\sqrt{3}$

If $f$ is your original polynomial, then \begin{align*} f&=(\sqrt{3}-\sqrt{2})x^3+\sqrt{2}x-\sqrt{3}+1\\ \sigma_2(f)&=(\sqrt{3}+\sqrt{2})x^3-\sqrt{2}x-\sqrt{3}+1\\ \sigma_3(f)&=(-\sqrt{3}-\sqrt{2})x^3+\sqrt{2}x+\sqrt{3}+1\\ \sigma_2\sigma_3(f)&=(-\sqrt{3}+\sqrt{2})x^3-\sqrt{2}x+\sqrt{3}+1\\ \end{align*}

The product of these polynomials has integral coefficients. The idea is that any automorphism of $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ will only commute the factors of the product and the product remains unchanged. Therefore, the coefficients must be in $\mathbb{Q}$.

Answer 2

Alternative solution, by eliminating the radicals with elementary manipulations:

$$\sqrt{3}(x^3-1) - \sqrt{2}(x^3 - x) + 1 = 0$$

$$\sqrt{3}(x^3-1) = \sqrt{2} x (x^2 - 1) - 1$$

Squaring:

$$3(x^3-1)^2 = 2 x^2 (x^2 - 1)^2 - 2 \sqrt{2} x(x^2-1) + 1$$

Rearranging:

$$3(x^3-1)^2 - 2 x^2 (x^2 - 1)^2 - 1 = - 2 \sqrt{2} x(x^2-1)$$

Squaring again:

$$\left(3(x^3-1)^2 - 2 x^2 (x^2 - 1)^2 - 1\right)^2 = 8 x^2(x^2-1)^2$$

$$\left(3(x^3-1)^2 - 2 x^2 (x^2 - 1)^2 - 1\right)^2 - 8 x^2(x^2-1)^2 = 0$$

After expanding and collecting, the LHS of the latter is a $12^{th}$ degree polynomial in $x\,$, with integer coefficients, having among its roots all roots of the original polynomial.