Show that A = $\mathbb{Q} \sqrt[3]{2} $ = {a + b$\sqrt [3]{2}$ + c$({\sqrt [3]{2}})^2$; a, b, c $\in \mathbb{Q}$} is a field.

Question

I'm just having a trouble proving the existence of the inverse, can someone help me? Thanks in advance.

Answer 1

Hint

Let $a+b\sqrt[3]{2}+c\sqrt[3]{2}^2\in \mathbb Q(\sqrt[3]2)\setminus \{0\}$. You must find $\alpha ,\beta ,\gamma \in\mathbb Q$ s.t. $$(a+b\sqrt[3]{2}+c\sqrt[3]{2}^2)(\alpha +\beta \sqrt[3]{2}+\gamma \sqrt[3]{2}^2)=1,$$

which is a system of 3 equations with 3 variables. At the end, proving existence (and unicity) of $\alpha ,\beta ,\gamma \in\mathbb Q$ is sufficient, and rather straightforward from linear algebra.

Answer 2

Hint:

1. The algebraic field extension $\Bbb Q(\sqrt[3]{2})$ of $\Bbb Q$ contains $\Bbb Q$, $\sqrt[3]{2}$ and $(\sqrt[3]{2})^2$ and so the set $\{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\mid a,b,c\in\Bbb Q\}$.

2. The set $\{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\mid a,b,c\in\Bbb Q\}$ forms a field by considering the field axioms.