Calculate the square root of a number in $\mathbb{Q}(\sqrt[3]{7})$

Question

How can I calculate the root of a number in extension field $\mathbb{Q}(\sqrt[3]{7})=\{\alpha + \beta\sqrt[3]{7} + \gamma\sqrt[3]{7^2} | \alpha,\beta,\gamma \in \mathbb{Q}\}$?

Any number in $\mathbb{Q}(\sqrt[3]{7})$ is a vector with the elements $\alpha \beta \gamma$.

So multiplication of a number with itself results in $(\alpha^2 + 14\beta\gamma) + (7\gamma^2 + 2\alpha\beta)\sqrt[3]{7} + (\beta^2 + 2\alpha\gamma)\sqrt[3]{7^2}$.

Reversing this would mean solving the system of equations above for the vector to be square rooted. But how?

Answer 1

HINT:

Produce a cubic equation for the square root. This can be done by factoring polynomial with rational coefficients. Once we have the equation, it is easy to find the element (thanks for @Lubin: for providing the explicit form of the cubic equation -- makes the last step straightforward)

An element $\alpha + \beta \sqrt[3]{7} + \gamma \sqrt[3]{7^2}$ in $\mathbb{Q}[\sqrt[3]{7}]$ is a root of a polynomial with rational coefficients. $$P(x)=\prod_{\omega^{3}=1}( x- (\alpha + \beta \sqrt[3]{7}\omega + \gamma \sqrt[3]{7^2}\omega^2))=\\=x^3−3\alpha x^2 +(3\alpha^2 −21\beta \gamma)X−(\alpha^3+7 \beta^3 +49 \gamma^3− 21\alpha \beta \gamma)$$ Its square root $\sqrt{\alpha + \beta \sqrt[3]{7} + \gamma \sqrt[3]{7^2}}$ will satisfy the equation $P(y^2)=0$. If this element is itself in the field $$\sqrt{\alpha + \beta \sqrt[3]{7} + \gamma \sqrt[3]{7^2}}= u + v \sqrt[3]{7}+w\sqrt[3]{7^2}$$ then the polynomial $P(y^2)$ of degree $6$ factors into two polynomials of degree $3$ with rational coefficients : $P(y^2)=Q(y)\cdot Q(-y)$.

So now we know the cubic polynomial

$$y^3- 3 u y^2+(3u^2-21 v w)y-(u^3 + 7v^3 + 49 w^3- 21 u v w)$$ and we get $u,v,w$.

ADDED: Let's take an example: find the square root of $469+ 259\sqrt[3]{7}+ 106\sqrt[3]{7^2}$. So $\alpha = 469$, $\beta=259$, $\gamma=106$.

1. Set up the polynomial $$P(x) = x^3−3\alpha x^2 +(3\alpha^2 −21\beta \gamma)X−(\alpha^3+7 \beta^3 +49 \gamma^3− 21\alpha \beta \gamma)=\\=x^3 - 1407 x^2 + 83349 x - 12744900$$

2. Produce the polynomial $$P(y^2)=y^6 - 1407 y^4 + 83349 y^2 - 12744900$$

3. Factor the polynomial $P(y^2)$ $$y^6 - 1407 y^4 + 83349 y^2 - 12744900=\\=(y^3 - 21 y^2 - 483 y - 3570) (y^3 + 21 y^2 - 483 y + 3570)$$

4. Solve the system $$ 3u = 21 \\ 3 u^2 - 21 v w = - 483\\ u^3 + 7 v^3 + 49 w^3 - 21 u v w =3570$$

Get $u=7$, $v=6$, $w=5$. ( if we solved for the other factor we would get $(-7,-6,-5)$ ).