Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

The following problem is part of my Algebra problem set:

Let $\alpha= \sqrt[3]{2}+\omega$ ,where $\omega=e ^{2\pi i /3} = -\frac{1}{2} + \frac{\sqrt{-3}}{2}$ is a primitive cube root of $1$.

$\alpha$ is a zero of $f(x) = 9 + 9x +3x^3 + 6x^4 + 3x^5 + x^6$. Show that $f(x)$ is irreducible.

** Eisenstein's Criterion fails here I realize that Gauss's Lemma is the best method of proving this although I am unclear as to whether proving the irreducibility of a polynomial in a subfield is sufficient to prove that it is irreducible. **

Please Also note that we have not had any Galois Theory.

Show that $\mathbb{Q}(\alpha)$ is the smallest field which contains all three zeros of $x^3 -2 $ (this is called the splitting field of $x^3 -2$).

Thank you in advance for your help!

share|cite|improve this question

marked as duplicate by T. Bongers, user7530, Hanul Jeon, Johannes Kloos, Dominic Michaelis Nov 4 '13 at 6:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

excuse me... i was a bit confused.... – Praphulla Koushik Oct 23 '13 at 9:31
Irreducible over $\Bbb Q$? – leo Oct 23 '13 at 15:37
This is a partial duplicate only (the first part). – Cameron Buie Oct 27 '13 at 16:14
Please, post only one question in one post. In particular, I suggest that you remove your first question from the post, as it has been previously answered, here. Posting several questions in the same post is discouraged and such questions may be closed, see meta. – Cameron Buie Oct 27 '13 at 17:46
@user: Agreed. I was going to post the above comment, regarding that, but then I got sidetracked. – Cameron Buie Oct 27 '13 at 17:48

We claim that the six degree polynomial you wrote down is irreducible by assuming that $\alpha = \sqrt[4]{2} + \omega$ is a root of it. We do this by contradiction. Suppose that it aint.

Claim 1: If $\alpha$ is not sextic over $\mathbb{Q}$ then it must be cubic over $\mathbb{Q}$.

As Paul said $\mathbb{Q}(\alpha)\subseteq \mathbb{Q}(\sqrt[3]{2},\omega)$ and since $[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6$ this means $d=[\mathbb{Q}(\alpha):\mathbb{Q}]$ must divide $6$. Now look at the divisors of six. Clearly $d\not = 1$ as $\alpha$ is not a rational number (why not?), also $d\not = 6$ as we are assuming that $\alpha$ is not sextic over $\mathbb{Q}$. Therefore, $d=2\text{ or }3$.

But $d$ cannot be $2$, because if it was then it would mean $\alpha$ would satisfy a quadratic polynomial $X^2+aX+b$ with $a,b\in \mathbb{Q}$. Now if $\alpha$ is a root of this polynomial then so is its complex conjugate, now $\overline{\alpha} = \overline{\omega} + \sqrt[3]{2} = \omega^2 + \sqrt[3]{2}$. Recall that the sum of the roots of $X^2 + aX+b$ add up to $-a$ and so $\alpha + \overline{\alpha} = - a \in \mathbb{Q}$ but $\alpha + \overline{\alpha} = \omega + \omega^2 + 2\sqrt[3]{2} = -1 + 2\sqrt[3]{2}$ and this number is not rational. Thus, it must be the case that $\alpha$ is cubic over $\mathbb{Q}$.

Claim 2: $\omega \not \in \mathbb{Q}(\alpha)$

Mr. Paul basically showed this above. But here is another proof. If $\omega \in \mathbb{Q}(\alpha)$ then the degree of $\omega$ over $\mathbb{Q}$ would divide the degree of $\mathbb{Q}(\alpha)/\mathbb{Q}$ (why?), but the degree of $\omega$ over $\mathbb{Q}$ is two which does not divide three.

Claim 3: $\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$

If it was then $\omega = \alpha - \sqrt[3]{2} \in \mathbb{Q}(\alpha)$ contradicting Claim 2.

Claim 4: $\omega^2\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$

This is a bit computational. Suppose that $\omega^2\sqrt[3]{2}\in \mathbb{Q}(\alpha)$ then:

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

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

Subtract (i) from (ii) to get $2\omega - \sqrt[3]{2}\in \mathbb{Q}(\alpha)$ but then $2\omega - \sqrt[3]{2} + (\omega + \sqrt[3]{2}) = 3\omega \in \mathbb{Q}(\alpha)$. This immediately would imply that $\omega \in \mathbb{Q}(\alpha)$ which contradicts Claim 2.

Claim 5: $\omega\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$.

This one is a similar computation but a bit trickier than Claim 4. Begin by assuming that $\omega\sqrt[3]{2} \in \mathbb{Q}(\alpha)$. First of all $\omega^2 = -1 - \omega$ (why?). Then, $(\sqrt[3]{2}+\omega)^2 = \sqrt[3]{4} + 2\sqrt[3]{2}\omega - 1 - \omega$. This implies that $\sqrt[3]{4} - \omega \in \mathbb{Q}(\alpha)$.


(iii) $(\sqrt[3]{4}-\omega)\omega\sqrt[3]{2} = 2\omega - \omega^2\sqrt[3]{2}$

(iv) $(\sqrt[3]{4}-\omega)(\omega\sqrt[3]{2})^2 = 2\omega^2\sqrt[3]{2} - \sqrt[3]{4}$

Add 2(iii) + (iv) to get that $4\omega - \sqrt[3]{4}\in \mathbb{Q}(\alpha)$ be we shown just above that $\sqrt[3]{4} - \omega \in \mathbb{Q}(\alpha)$ so when we add them we get $3\omega \in \mathbb{Q}(\alpha)$ which is a contradiction.

Claim 6: $\sqrt[3]{2}$ is cubic over $\mathbb{Q}(\alpha)$.

We know from Claim 3 that $\sqrt[3]{2}$ is not in the field and $\sqrt[3]{2}$ solves the polynomial $X^3-2$ so the degree of $\sqrt[3]{2}$ is either two or three. It cannot be two. Because if it degree two then $X^3-2$ must be reducible over $\mathbb{Q}(\alpha)$, in particular one of its roots: $\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}$ must be contained in $\mathbb{Q}(\alpha)$ but Claim 3,4,5 show that is impossible. Thus $\sqrt[3]{2}$ must be cubic.

Claim 7: CONTRADICTION ARISES from assuming $\alpha$ was cubic.

Let $K=\mathbb{Q}(\alpha)$. By the above claims $K(\omega)/K$ is quadratic field and $K(\sqrt[3]{2})/K$ is a cubic field. Thus, $K(\omega,\sqrt[3]{2})/K$ is a sextic field which is a contradiction (do you see it?)*

*)Fact from Field Theory: If $F(\alpha)/F$ is degree $n$ and $F(\beta)/F$ is degree $m$ with $\gcd(n,m)=1$ then $F(\alpha,\beta)/F$ is degree $n\cdot m$.

share|cite|improve this answer

Hint for the second part: Note that the splitting field of $x^3-2$ is exactly ${\mathbb Q}(\sqrt[3]{2},\omega)$. If you can show that $[{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}] = 6$, you should be done (why?).

For convinience consider: $ [{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}] = [{\mathbb Q}(\sqrt[3]{2}):{\mathbb Q}] \times [{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}(\sqrt[3]{2})] $ and note that $\omega\not\in {\mathbb Q}(\sqrt[3]{2})$ (why?). :)

EDIT: Without Galois Theory, the first part looks very messy.

Here goes an idea: We are done, in view of the previous hint, if we can show ${\mathbb Q}(\alpha) = {\mathbb Q}(\sqrt[3]{2},\omega)$ (in fact, this should prove the second part directly).

Note that if $\omega\in {\mathbb Q}(\alpha)$, the so is $\sqrt[3]{2}$, and similarly if $\sqrt[3]{2}\in {\mathbb Q}(\alpha)$ then $\omega\in {\mathbb Q}(\alpha)$. Thus we will assume that $\omega\not\in {\mathbb Q}(\alpha)$ and $\sqrt[3]{2}\not\in {\mathbb Q}(\alpha)$, else we would be done.

We have, since ${\mathbb Q}(\alpha,\omega) = {\mathbb Q}(\sqrt[3]{2},\omega)$ (why?) : $$6 = [{\mathbb Q}(\alpha,\omega):{\mathbb Q}] =[{\mathbb Q}(\alpha,\omega):{\mathbb Q}(\alpha)]\times [{\mathbb Q}(\alpha):{\mathbb Q}]$$

Now use this and $\omega\not \in {\mathbb Q}(\alpha)$ to conclude that $[{\mathbb Q}(\alpha):{\mathbb Q}]=3$. Use this to prove that $2=[{\mathbb Q}(\alpha,\sqrt[3]{2}):{\mathbb Q}(\alpha)]$.

Now the minimal polynomial $m(x)$ for $\sqrt[3]{2}$ over ${\mathbb Q}(\alpha)[x]$ must divide $x^3-2 \in {\mathbb Q}(\alpha)[x]$ and $\deg m(x) = 2$.

Dicuss then according to what the 2 roots of $m(x)$ should be, bearing in mind that $m(x) | (x^3 - 2)$.

Finally, prove that if a possible choices produced a polynomial in ${\mathbb Q}(\alpha)[x]$, then either $\omega\in {\mathbb Q}(\alpha)$ or $\sqrt[3]{2}\in {\mathbb Q}(\alpha)$ contradicting our assumption.

share|cite|improve this answer
At least two students from the same class asking the same question:… – Alex B. Oct 23 '13 at 14:18
Thank you! However, we have NOT had any Galois Theory. Is is possible to answer the question without Galois Theory?? – incompleteness Oct 23 '13 at 15:05
I am trying to prove the part about the irreducible polynomial, but withouth Galois Theory it is getting very messy. I might edit my post soon if I get somewhere. – Pablo Rotondo Oct 23 '13 at 15:23
@incompleteness Edited the post. Very messy though, but this seems to prove (if correct) the second part directly. – Pablo Rotondo Oct 23 '13 at 15:29

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