Elements of $\mathbb Q(2^{1/3})$

$\mathbb R$ is an extension of $\mathbb Q$.

Why the elements of $\mathbb Q(2^{1/3})$ have the form $a+b\sqrt[3]2$?

More specifically why $\mathbb Q(2^{1/3})=\{a+b\sqrt[3]2+c\sqrt[3]4:a,b,c \in\mathbb Q\}$?

Solution 1:

They don't have the form $a + b2^{1/3}$.

But you are right that $\mathbb{Q}(2^{1/3}) = \{a + b2^{1/3} + c4^{1/3}: a, b, c \in \mathbb{Q}\}$.

To see this, note that the smallest field extension of $\mathbb{Q}$ containing $2^{1/3}$ must also contain all polynomials in $2^{1/3}$ with rational coefficients. This set forms a ring $\mathbb{Q}[2^{1/3}]$ which turns out to be a field (can you see why?).

Note also that the minimal polynomial of $2^{1/3}$ over $\mathbb{Q}$ is $X^3 - 2$, with degree 3 (how would you prove this?). So higher powers of $2^{1/3}$ are linear combinations of $1, 2^{1/3}, 4^{1/3}$ over $\mathbb{Q}$, and these form a basis for $\mathbb{Q}(2^{1/3})/\mathbb{Q}$.