I'm trying to prove that $\mathbb{Q}(\sqrt[3]{2})$ is isomorphic to $\mathbb{Q}(e^{2\pi i/3}\sqrt[3]{2})$. I want to use the first isomorphism theorem that say that if $\phi: A\to B$ is a homomorphism, then $\phi(A)$ is isomorphic to $A/\!\ker{\phi}$. I'm trying to build a homomorphism such that $\phi(\mathbb{Q}(\sqrt[3]{2}))=\mathbb{Q}(e^{2\pi i/3}\sqrt[3]{2})$ and $\ker{\phi}=\{0\}$.

Does such an homorphism exist or is there another way to prove this?

Thanks for your help.

  • 1
    $\begingroup$ Note that every non-trivial homomorphism from $Q(\sqrt[3]2)$ to another field is first of all injective (what are the ideals of a field?) and second, completely determined by where $\mathbb Q$ and $\sqrt[3]2$ go. So, all you need to specify is what a homomorphism must do to a rational and what is must do to $\sqrt[3]2$. Of course, a cursory look at the second field tells you what the image of $\mathbb Q$ and $\sqrt[3]2$ should be. Use this to formulate a map, and show that it is an isomorphism either by noting there is an inverse map, or showing surjectivity. $\endgroup$ – Teresa Lisbon Sep 12 '18 at 14:35

Another way to approach it: $$\Bbb Q(\sqrt[3]2)\cong\Bbb Q[x]/(x^3-2)$$ This follows from the evaluation homomorphism: $$\phi:\Bbb Q[x]\rightarrow\Bbb Q(\sqrt[3]2)\\f(x)\mapsto f(\sqrt[3]2)$$ And then the 1st isomorphism theorem.

What about $\Bbb Q(e^{2\pi i/3}\sqrt[3]2)$?


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