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.

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) ?

share|improve this question
Would this help? $$\eqalign{ & {\left( {x - 1} \right)^3} = 2{\left( {\root 3 \of 2 + 1} \right)^3} = 2\left( {2 + 3\root 3 \of 2 + 3\root 3 \of 4 + 1} \right) = 2\left( {3 + 3\left( {x - 1} \right)} \right) = 6x \cr & {\left( {x - 1} \right)^3} - 6x = 0 \cr} $$ –  Pedro Tamaroff Sep 9 '12 at 17:47
@PeterTamaroff - Is $x=\alpha$ ? –  Belgi Sep 9 '12 at 17:56
Yes, of course. –  Pedro Tamaroff Sep 9 '12 at 17:59

5 Answers 5

up vote 11 down vote accepted

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 $3.$

share|improve this answer
How did you get this equality ? (I can't verify it) –  Belgi Sep 9 '12 at 17:49
@Belgi Have you tried putting $x=\sqrt[3]{2}$ into $x^2+x+1-r$ and seeing what you get? –  Ragib Zaman Sep 9 '12 at 17:52
Thanks, I understand what you did –  Belgi Sep 9 '12 at 17:54

Let $\beta=\sqrt[3]2$. Then $\alpha=1+\beta+\beta^2$, hence $\mathbb Q(\alpha)\subseteq \mathbb Q(\beta)$. Clearly, the minimal polynomial of $\beta$ is $X^3-2$, hence $[\mathbb Q(\beta):\mathbb Q]=3$. As a vector space, $\mathbb Q(\beta)$ has $1, \beta, \beta^2$ as a basis, hence $\alpha$ is irrational (the only way to write $\alpha=a+b\beta+c\beta^2$ with rational coefficients is $\alpha=1+\beta+\beta^2$). Thus $1<[\mathbb Q(\alpha):\mathbb Q]|3$, i.e. and $[\mathbb Q(\alpha):\mathbb Q]=3$.

You can find the minimal polynomial of $\alpha$ itself: $$\alpha = 1+\beta+\beta^2$$ $$\alpha^2 = (1+\beta+\beta^2)^2=1+2\beta+3\beta^2+2\beta^3+\beta^4 = 5+4\beta+3\beta^2$$ $$\alpha^3 = (1+\beta+\beta^2)(5+4\beta+3\beta^2)=5+9\beta+12\beta^2+7\beta^3+3\beta^4=19+15\beta+12\beta^2$$ Find a combination that eliminates all $\beta$ and $\beta^2$: $$\alpha^2-3\alpha=2+\beta$$ $$\alpha^3-4\alpha^2=-1-\beta$$ $$\Rightarrow\quad \alpha^3-3\alpha^2-3\alpha-1=0$$

share|improve this answer
How did you deduce that $\alpha$ is irrational ? –  Belgi Sep 9 '12 at 17:46
So its degree is 2*3=6 right ? –  mick Sep 9 '12 at 17:51
@Belgi: I clarified this ($1,\beta,\beta^2$ are linearly independant). –  Hagen von Eitzen Sep 9 '12 at 17:59
You can type faster than I can Hagen! –  Mark Bennet Sep 9 '12 at 17:59
@mick: No the degree of $\mathbb Q(\alpha)=\mathbb Q(\beta)$ is 3 because we have only added the real roots, not the two complex roots. –  Hagen von Eitzen Sep 9 '12 at 17:59

Your second method is the right one. To see that $\alpha$ is not rational, note that $$\frac{a}{b}=\alpha\implies \frac{a-b}{b\sqrt[3]{2}}=(1+\sqrt[3]{2})\implies \frac{(a-b)^3}{2b^3}=3\alpha\implies(a-b)^3=6ab^2$$ and note that this last equation is homogeneous, so if it has a solution it in $\mathbb Z$ it has a solution with $a$ and $b$ coprime. But $6|(a-b)^3$ so $6|(a-b)$, and so $6^2|ab^2$ so either $6|a$ or $6|b$, but either way since $6|(a-b)$ we get $6|a$ and $6|b$, hence $a$ and $b$ are not coprime. Thus no solution exists, so $\alpha$ is irrational.

share|improve this answer

Denote $\alpha = 1 + \sqrt[3]{2} + \sqrt[3]{4}$. Since $\alpha \in \mathbb{Q}(\sqrt[3]{2})$, we get $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt[3]{2})$. Now

$\sqrt[3]{2} + \sqrt[3]{4} \in \mathbb{Q}(\alpha)$

$(\sqrt[3]{4} + \sqrt[3]{2})^2 = 2 \sqrt[3]{2} + \sqrt[3]{4} + 4 \in \mathbb{Q}(\alpha)$

Subtracting the first element from the second implies $\sqrt[3]{2} \in \mathbb{Q}(\alpha)$, and therefore $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt[3]{2})$.

share|improve this answer
Can you elaborate further on this? –  KangHoon You Jan 28 at 1:48
Yes, if you can elaborate on what is not clear. –  Mikko Korhonen Jan 28 at 7:56

If $\alpha$ is rational, then so is $(2^{1/3} + 4^{1/3})$, and so the cube of that expression is rational; expanding, we see that the cube of that expression looks like a rational number plus $3*(32)^{1/3} = 6*4^{1/3}$; standard arguments tell you that this is irrational.

share|improve this answer
How did you get this ? my calculations gives Wolfram result with both $2^{1/3}$ and $4^{1/3}$ ?. link:wolframalpha.com/input/?i=%282^%281%2F3%29%2B4^%281%2F3%29%29^3 –  Belgi Sep 9 '12 at 17:52
Actually, if $r = 2^{1/3}$, $(\alpha - 1)^3 = 6 + 6 r + 6 r^2 = 6 \alpha$, so $\alpha$ is a root of the polynomial $(z - 1)^3 - 6 z = z^3 - 3 z^2 - 3 z - 1$, which has no rational roots. –  Robert Israel Sep 9 '12 at 17:56

Your Answer


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.