Prove $\mathbb Q[\sqrt[3]{2}]$={$a+b\cdot\sqrt[3]{2}$} is not a field

Question

Suggestion: Prove that you can't write $\mathbb Q[\sqrt[3]{2}]$ like $a+b\cdot\sqrt[3]{2}$

Answer 1

$\mathbb{Q}[\sqrt[3]2]$ is in fact a field. However, it is true you can't write represent it as $\{ a + b \sqrt[3] 2 \mid a, b \in \mathbb{Q}\}$. Note that $\sqrt[3] 2$ is an element of $\mathbb{Q}[\sqrt[3]2]$, and since fields are closed under multiplication, $\sqrt[3] 2^2$ is also in the field. However, you cannot write this in form $a + b \sqrt[3] 2$ because $\sqrt[3] 2^2$ is linearly independent to both over the rationals.

Answer 2

$\mathbb Q[\sqrt[3]2]\cong \mathbb Q[X]/(X^3-2)$ and since $X^3-2$ is irreducible (by Eisenstein criterion), and $\mathbb Q[X]$ is principal (because $\mathbb Q$ is a field) it follow that $(X^2-2)$ is maximal. Therefore $\mathbb Q[\sqrt[3]2]$ is a field.

Answer 3

$[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. So $\mathbb{Q}(\sqrt[3]{2})=\{a + b \sqrt[3]{4} + c \sqrt[3]{2} | a, b, c \in \mathbb{Q}\}$.

Answer 4

Well, the ring extension ${\Bbb Q}[\sqrt[3] 2]$ equals the field extension ${\Bbb Q}(\sqrt[3] 2)$. Indeed, the first one equals the quotient ring ${\Bbb Q}[x]/\langle x^3-2\rangle$. Since $x^3-2$ is irreducible over $\Bbb Q$, this quotient ring is a field and therefore equals the field extension ${\Bbb Q}(\sqrt[3] 2)$.