This is an exercise for the book Abstract Algebra by Dummit and Foote
(pg. 530): Find the degree of $\alpha:=1+\sqrt[3]{2}+\sqrt[3]{4}$
over $\mathbb{Q}$
My efforts:
I first try to find the minimal polynomial by writing
$\alpha=1+\sqrt[3]{2}+\sqrt[3]{4}\implies\alpha-1=\sqrt[3]{2}(1+\sqrt[3]{2})\implies(\alpha-1)^{3}=2(1+\sqrt[3]{2})^{3}$
but I didn't manage to get the minimal polynomial from this (which
is, according to Wolfram, of degree $3$).
I also tried another method that failed: I noted that $\mathbb{Q}(\alpha)\subset\mathbb{Q}(\sqrt[3]{2})$
hence is of degree $\leq3$, moreover, since it is a subfield and
$3$ is prime it only remains to show that $\alpha$ is not rational
(which I can't prove).
Can someone pleases help me show that the degree is $3$ (preferably
is one of the two methods I tried) ?
Best Answer
If $1+ \sqrt[3]{2} + \sqrt[3]{4}=r$ with $r$ rational, then $\sqrt[3]{2}$ would satisfy $x^2+x+1-r=0$, contradicting the fact that is has degree $3.$