Show that the splitting fields of $x^3 - 2$ and $x^3 - 3$ are not equal

Question

I'm trying to solve a problem which is concerned with the size of the intersection of $H_1 = \mathbb{Q}(\sqrt[3]{2}, \zeta_3)$ and $H_2 = \mathbb{Q}(\sqrt[3]{3}, \zeta_3)$. If I can show $H_1 \not= H_2$ then I have the rest figured out, but I'm not sure how to do that.

Answer 1

We can work a little bit by hand:

Assume $\sqrt[3]{3} \in \mathbb Q(\sqrt[3]{2},\zeta_3)$.

Galois correspondence tells us that the intermediate fields of $\mathbb Q(\sqrt[3]{2},\zeta_3)$ of degree $3$ over $\mathbb Q$ are precisely

$$\mathbb Q(\sqrt[3]{2}),\mathbb Q(\sqrt[3]{2}\zeta_3),\mathbb Q(\sqrt[3]{2}\zeta_3^2)$$

By our assumption $\mathbb Q(\sqrt[3]{3})$ is one of these fields. Since it is contained in the reals, it must be equal to $\mathbb Q(\sqrt[3]{2})$, so we obtain

$$\sqrt[3]{3} = a+b\sqrt[3]{2}+c\sqrt[3]{4}$$ with some $a,b,c \in \mathbb Q$. Comparing the traces shows $a=0$. Take the cube of the equation $\sqrt[3]{3} = b\sqrt[3]{2}+c\sqrt[3]{4}$ and this will immediately show $b=c=0$, using the linear independence of $1,\sqrt[3]{2},\sqrt[3]{4}$

Answer 2

There are many ways of doing this, and many, I’m sure, that are more elementary than what I have in mind.

Take the field $F=\Bbb Q(\omega)$, where I call $\zeta_3=\omega$ just for ease of typing. The integers of this field, $\Bbb Z[\omega]$, are known to have unique factorization. Now, since $F$ contains the cube roots of unity, the Galois cubic extensions are, by Kummer theory, in one-to-one correspondence with the cyclic subgroups of $F^*/(F^*)^3$, crudely speaking with the elements of $F^*$ modulo cubes. But $2/3$ is not a cube, so that the extensions $F(\sqrt[3]2)$ and $F(\sqrt[3]3)$ are different. Why is $2/3$ not a cube? Because $2$ is a prime in $F$, and $3$ is, modulo units, the square of a prime.