Is $\mathbb{Q}(i,\sqrt[3]{2})=\mathbb{Q}(i,\sqrt[3]{4})$?

Question

Is $\mathbb{Q}(i,\sqrt[3]{2})=\mathbb{Q}(i,\sqrt[3]{4})$?

$\mathbb{Q}(i,\sqrt[3]{2})\supseteq\mathbb{Q}(i,\sqrt[3]{4})$ holds because $\sqrt[3]{4}=(\sqrt[3]{2})^2$, but I don't know how to show the other inclusion. Or is there a better way of showing this?

Answer 1

Since $$ \sqrt[3]{4}\cdot\sqrt[3]{4} = \sqrt[3]{16}=\sqrt[3]{2^3\cdot 2} = 2\sqrt[3]{2}, $$ $\sqrt[3]{2}\in \mathbf Q(\sqrt[3]{4})$. Hence the two fields are equal.

Another way to see this is to consider the minimal polynomial of $\sqrt[3]{4}$ over $\mathbf Q$. By the rational roots test, if $p/q$ is a rational root of $f(X)=X^3-4$ in lowest terms, then $q$ divides $1$, and $p$ divides $4$. Hence the only options for roots over $\mathbf Q$ of $f(X)$ are $\pm 1,\pm 2,\pm 4$, and none of these is a root. Since a polynomial of degree $3$ is irreducible over $\mathbf Q$ if and only if it has no rational roots, $X^3-4$ is irreducible over $\mathbf Q$, hence $f(X)$ is the minimal polynomial of $\sqrt[3]{4}$. Thus $[\mathbf Q(\sqrt[4]3):\mathbf Q] = 3$, so since $\mathbf Q(\sqrt[3]{2})$ is also a degree-$3$ extension over $\mathbf Q$ and $\mathbf Q(\sqrt[3]4)\subset\mathbf Q(\sqrt[3]2)$ as you noted, we actually have equality $\mathbf Q(\sqrt[3]4)=\mathbf Q(\sqrt[3]2)$.

Answer 2

Hint: $\frac{\sqrt[3]{2}}{2}=(\sqrt[3]{4})^{-1}$, $\frac{\sqrt[3]{4}}{2}=(\sqrt[3]{2})^{-1}$