Degree of $\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$ over $\mathbb{Q}(\sqrt[3]{2})$

Question

In Dummit and Foote, Abstract Algebra, Sec 13.4 Example 3, The splitting field of $x^{3}-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$. Then he says, quote : Since $\sqrt{-3}$ satisfies the equation $x^{2}+3 =0$, the degree of this extension over $\mathbb{Q}$ is at most 2, hence must be 2 since we observe that $\mathbb{Q}(\sqrt[3]{2})$ is not the splitting field.

I do not understand why he considers this. Could somebody be more specific and explain me why this is so?

Thanks

Answer 1

If we have any diamond of finite field extensions

$\hskip 3in$

where $M=LK$ and the primitive element theorem applies, then

$$\color{Blue}{[M:K]}\le\color{Red}{[L:F]}.$$

Proof. By PET let $L=F(\theta)$ where $\theta$ has minimal polynomial $m(x)\in F[x]$. Then $M=K(\theta)$. Say the minimal polynomial of $\theta$ over $K$ is $n(x)\in K[x]$. Since $m(x)\in F[x]\subseteq K[x]$ and $m(\theta)=0$ we must have $n(x)\mid m(x)$ and therefore $\deg n(x)\le \deg m(x)$ but $\deg n=[M:K]$, $\deg m=[L:F]$ and so the inequality is satisfied.

In particular you have $M=\Bbb Q(\sqrt[3]{2},\sqrt{-3})$, $K=\Bbb Q(\sqrt[3]{2})$, $L=\Bbb Q(\sqrt{-3})$, $F=\Bbb Q$, which implies the inequality $[\Bbb Q(\sqrt[3]{2},\sqrt{-3}):\Bbb Q(\sqrt[3]{2})]\le[\Bbb Q(\sqrt{-3}):\Bbb Q]=2$. Since $\Bbb Q(\sqrt[3]{2},\sqrt{-3})$ is the splitting field of $x^3-2$ whereas $\Bbb Q(\sqrt[3]{2})$ isn't, we have $\Bbb Q(\sqrt[3]{2},\sqrt{-3})\ne\Bbb Q(\sqrt[3]{2})$ hence the degree is $>1$. Since the degree is $>1$ and $\le2$ it must be $2$. This is the reasoning involved.

Answer 2

I suppose you have no problem seeing that the degree is either 1 or 2. Your base field with cube root of 2 is a field of degree 3 over the rationals. If the answer is 1, it means that field will also contain square root of $-3$, a quadratic extension over the rationals. But a cubic field does not contain quadratic field.