Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
In your "quote," the phrase "this extension" seems to refer to two different fields. Can you provide the direct quote? Are the authors claiming that because $[\Bbb Q(\sqrt{-3}):\Bbb Q]=2$ that $[\Bbb Q(\sqrt[3]{2},\sqrt{-3}):\Bbb Q(\sqrt[3]{2})]$ must be $\le2$ (and subsequently $=2$ by ruling out $=1$)? –  blue Apr 24 at 1:12
add comment

2 Answers 2

If we have any diamond of finite field extensions

$\hskip 3in$ diamond

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.