Verify that $\mathbb{Q}[\sqrt[3]2]=\{a+b\sqrt[3]2+c\sqrt[3]2^2\,\mid\, a,b,c \in \mathbb{Q}\}$ is a subfield of $\mathbb{R}$

Question

Verify that $\mathbb{Q}[\sqrt[3]2]=\{a+b\sqrt[3]2+c\sqrt[3]2^2\,\mid\, a,b,c \in \mathbb{Q}\}$ is a subfield of $\mathbb{R}$.

I checked that $0$ and $1$ $\in \mathbb{Q}[\sqrt[3]2]$, and also checked that $x-y \in \mathbb{Q}[\sqrt[3]2]$

Then I take $x=a+b\sqrt[3]2+c\sqrt[3]2^2$ and $y=d+e\sqrt[3]2+f\sqrt[3]2^2$, with $a,b,c,d,e,f \in \mathbb{Q}$, but I can't verify if $x.y^{-1} \in \mathbb{Q}[\sqrt[3]2]$, because I can't find the conjugate to multiply.

Answer 1

Alternatively, consider the ring homomorphism $\varphi: \mathbb Q[x] \to \mathbb R$ induced by $x \mapsto \sqrt[3] 2$. Then prove:

• the image of $\varphi$ is $\mathbb{Q}[\sqrt[3] 2]$ – by definition

• $\mathbb{Q}[\sqrt[3] 2]=\{a+b\sqrt[3] 2+ b\sqrt[3] 4 \mid a,b,c\in\mathbb{Q}\}$ – reduce mod $3$ the exponents in powers of $\sqrt2$.

• $\ker\varphi = \langle x^3- 2\rangle$ – follows from polynomial division.

• $\ker\varphi = \langle x^3- 2\rangle$ is a maximal ideal – because $ x^3- 2$ is irreducible over $\mathbb Q$.

Answer 2

One way to do that is for double containment argument: $\{a+b\sqrt[3] 2+ b\sqrt[3] 4 \mid a,b,c\in\mathbb{Q}\}\subset \mathbb{Q}[\sqrt[3] 2]$ because $\mathbb{Q}[\sqrt[3] 2]$ is a field which contains $\sqrt[3]2$. In the other hand, $\mathbb{Q}[\sqrt[3] 2]$ is the smallest field which contains $\mathbb{Q}$ and $\sqrt[3] 2$ then $\mathbb{Q}[\sqrt[3] 2]\subset \{a+b\sqrt[3] 2+ b\sqrt[3] 4 \mid a,b,c\in\mathbb{Q}\}.$