Show $\Bbb Q[\sqrt[3]2]\cong \Bbb Q[x]/(x^3-2)$ via First Isomorphism Theorem

Question

I am asked to prove that $\mathbb{Q}[x]/(x^3-2)\cong R$, where $R$ is the set of numbers $a+b\sqrt[3]2+c\sqrt[3]4$ with $a,b,c \in\mathbb{Q}$.

By, first isomorphism theorem of rings:

I have defined my map as $\mathbb{Q}[x] \rightarrow R$ by $(f(x)) \mapsto f(\sqrt[3]2)$.

I am not sure how to show surjective or how to go about the kernel. Would proving containment be best for the kernel?

Answer 1

Your map $\theta$ is a good choice. For surjectivity, note that $f(x) = a + bx + cx^2$ is mapped to $a+b\sqrt[3]{2}+c\sqrt[3]{4}$.

Your intuition for how to prove that $\ker \theta = \langle x^3-2\rangle$ is also good.

By definition of $\theta$, $\theta( x^3-2) = \sqrt[3]{2}^3 - 2 = 0$, so $\langle x^3-2\rangle \subset \ker \theta$.

For the other direction, if $f(x) \in \ker \theta$, then $f(\sqrt[3]{2})=0$. As such, $\sqrt[3]{2}$ is a root of $f$, and so the minimal polynomial of $\sqrt[3]{2}$ must divide $f$. But that minimal polynomial is $x^3-2$, and $(x^3-2) | f$ is exactly what it means to say that $f \in \langle x^3-2\rangle$.