Arguments for Galois closure of $\mathbb{Q}(\sqrt[3]{2})$

Question

The standard argument for why $K = \mathbb{Q}(\sqrt[3]{2})$ is not the splitting field of $f = x^3 - 2$ relies on us implicitly choosing a complex embedding of $K$, or in other words choosing the real third root of $2$ as a solution for $f$. Then we argue that since $K$ is a real field it cannot contain the other two roots which are complex. This sounds like a bit of cheating, though, since a priori, until we choose an embedding, the roots of $f$ are indistinguishable.

So my question is whether there is a similar argument for $\mathbb{Q}(\alpha)$ not being the splitting field for $\alpha$ a non-real root of $f$, e.g. $\alpha = \omega \sqrt[3]{2}$ where $\omega$ is a primitive root of unity. Or in general, whether the not-a-splitting-field argument must necessarily rely on a particular choice of a complex embedding.

Answer 1

I will show that an extension $K$ of $\mathbb{Q}$ of degree $3$ that has one root of $x^3-2$ cannot contain any other roots, and therefore cannot be a Galois extension of $\mathbb{Q}$. (Since if $K$ is a Galois extension of $\mathbb{Q}$, and $f$ is an irreducible polynomial in $\mathbb{Q}[x]$ that has at least one root in $K$, then it must split over $K$).

Let $K$ be an extension of degree $3$. Let $\alpha\in K$ be a root of $x^3-2$. If $K$ contains another root $\beta\neq\alpha$ of $x^3-2$, then $K$ also contains $\frac{\alpha}{\beta}\neq 1$, but $$\left(\frac{\alpha}{\beta}\right)^3 = \frac{\alpha^3}{\beta^3} = \frac{2}{2}=1,$$ so $\frac{\alpha}{\beta}$ is a root of $x^3-1 = (x-1)(x^2+x+1)$. Since $\frac{\alpha}{\beta}\neq 1$, then $\frac{\alpha}{\beta}$ is a root of $x^2+x+1$, which is irreducible over $\mathbb{Q}$. Therefore, $[\mathbb{Q}(\frac{\alpha}{\beta}):\mathbb{Q}]=2$, but then $$ 3= \left[K:\mathbb{Q}\vphantom{\frac{\alpha}{\beta}}\right] = \left[K:\mathbb{Q}\left(\frac{\alpha}{\beta}\right)\right]\left[\mathbb{Q}\left(\frac{\alpha}{\beta}\right):\mathbb{Q}\right] = 2\left[K:\mathbb{Q}\left(\frac{\alpha}{\beta}\right)\right].$$ This is impossible, so no such second root $\beta$ can lie in $K$.

Answer 2

Arturo's answer is nicer in my opinion, but here is another solution I found:

First identify the roots of $x^3-2$ over $\Bbb C$. These are \begin{align} a&=\sqrt[3]2\\ b&=\sqrt[3]2\omega=-\frac{1}{2^{2/3}}+\frac{\sqrt3}{2^{2/3}}i\\ c&=\sqrt[3]2\omega^2=-\frac{1}{2^{2/3}}-\frac{\sqrt3}{2^{2/3}}i,\\ \end{align}

where $\omega=-\frac12+\frac{\sqrt3}{2}i$.

Let $L$ be one of the extensions $\Bbb Q[a]$, $\Bbb Q[b]$, or $\Bbb Q[c]$. Suppose $L$ contains two or more of $a$, $b$, or $c$. There are $3$ cases:

• Case 1: $L$ contains both $a$ and $b$
• Case 2: $L$ contains both $a$ and $c$.
• Case 3: $L$ contains both $b$ and $c$.

If $L$ contains both $a$ and $b$, then $L$ contains $a^2b=-1+\sqrt 3i$, so it also contains $\sqrt 3i$. But $\sqrt 3i$ is a root of $x^2+3$, which is irreducible over $\Bbb Q$. Hence $$3=[L:\Bbb Q]=[L:\Bbb Q[\sqrt 3i]][\Bbb Q[\sqrt 3i]:\Bbb Q],$$ which implies $2|3$. This shows cases $1$ and $2$ are impossible. For case $3$, you can similarly show that $b^2c=-1+\sqrt 3i$ (this isn't so bad using $b=\sqrt[3]2\omega$ and $c=\sqrt[3]2\omega^2$).