Let $F=\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3})$. Show that over $\mathbb{Q}$ this is an extension of degree 9.

I see that this is equivalent of showing that $\sqrt[3]{2}\notin \mathbb{Q}(\sqrt[3]{3})$. I don't know any other way of showing this except going through $\sqrt[3]{2}=a+b\sqrt[3]{3}+c\sqrt[3]{9}$ which doesn't lead me anywhere.


1 Answer 1


1/ If you would like to proceed that way then by contradiction, suppose that $K=\mathbb{Q}(\sqrt[3]{2}) = \mathbb{Q}(\sqrt[3]{3})$. Consider the splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega)$ of $x^3-2$ (where $\omega$ is the 3-th root of 1), then this is also the splitting field of $x^3-3$ (since we just adjoin $\omega$ to $K$).

The automorphism $\sigma$ such that $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}, \sigma(\omega \sqrt[3]{3}) = \omega^2 \sqrt[3]{3}, \sigma(\omega^2 \sqrt[3]{3}) = \sqrt[3]{3}$ also maps $\sqrt[3]{2}$ to a solution of $x^3-2$, i.e, $\sigma(\sqrt[3]{2})$ is either $\sqrt[3]{2},\omega \sqrt[3]{2}$ or $\omega^2 \sqrt[3]{2}$.

Now assume that $\sqrt[3]{2}=a+b\sqrt[3]{3}+c\sqrt[3]{9}$, by taking $\sigma$ both sides and considering case by case, you should get a contradiction.

2/ If you know about trace, however, there is a nicer proof in my opinion and this would help handle more general situation. Suppose that $K=\mathbb{Q}(\sqrt[3]{2}) = \mathbb{Q}(\sqrt[3]{3})$ as before. Then you can check that $$\mathrm{Trace}_{K/\mathbb{Q}}(\sqrt[3]{2}) = \mathrm{Trace}_{K/\mathbb{Q}}(\sqrt[3]{4}) = \mathrm{Trace}_{K/\mathbb{Q}}(\sqrt[3]{3})= \mathrm{Trace}_{K/\mathbb{Q}}(\sqrt[3]{9}) = 0$$ If we assume that $\sqrt[3]{2}=a+b\sqrt[3]{3}+c\sqrt[3]{9}$ then by taking trace both sides, we get $0=3a$ (since $a\in \mathbb{Q}$), hence $a=0$. Now multiplying both sides of $\sqrt[3]{2}=b\sqrt[3]{3}+c\sqrt[3]{9}$ by $\sqrt[3]{3}$ and $\sqrt[3]{9}$ and taking trace, we get $b=c=0$ as well. (Notice that $\mathrm{Trace}_{K/\mathbb{Q}}(\sqrt[3]{6}) = 0$ as well).


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