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The question is that $e^{2\pi i/6r}$ is a root of the polynomial $X^{2r}-X^{r}+1 \in \mathbb Q[X]$ , we want to prove that $X^{2r}-X^{r}+1$ is irreducible if and only if $r$ is of the form $2^{a}3^{b}(a,b\geq 0)$, and the following question $X^{2r}+X^{r}+1$ is irreducible if and only if $r$ is of the form a power of $3$.

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    What precisely do you mean by "$(a,b)\ge0$"? You needn't specify that the greatest common divisor of $a,b$ is nonnegative, as this is necessarily true. What, then, do you mean?2012-12-27
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    Thanks for your command, I only want to say that they are both greater than 0.2012-12-27
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    This link deals with the question of irreducibility of $\Phi_k(x^n)$ http://m-hikari.com/imf-2011/29-32-2011/damianouIMF29-32-2011.pdf2012-12-27
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    Cool, this is a very useful link.2012-12-27

3 Answers 3

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Note that

$$X^{2r}-X^r+1=\frac{X^{3r}+1}{X^r+1}=\frac{X^{6r}-1}{(X^r+1)(X^{3r}-1)}$$

$\Rightarrow$

Assume by contradiction that $r$ has a prime factor $p \geq 5$. Let $r=pq$. Then $e^{\frac{2 \pi i }{6q}}$ is a root of $X^{6r}-1$ but not of $(X^r+1)(X^{3r}-1)$.

This shows that the minimal polynomial of $e^{\frac{2 \pi i }{6q}}$ divides $X^{2r}-X^r+1$. Since the degree of the minimal polynomial is $\phi(6q) \neq 2r$, it follows that the minimal polynomial of $e^{\frac{2 \pi i }{6q}}$ is a proper divisor of $X^{2r}-X^r+1$.

$\Leftarrow$

If $r=2^a3^b$. It should be easy to prove from

$$X^{2r}-X^r+1=\frac{X^{6r}-1}{(X^r+1)(X^{3r}-1)}$$ that $X^{2r}-X^r+1$ is a cyclotomic polynomial. Note that

$$\phi(6r)=2^a\cdot 3^b \cdot 2=2r=6r-r-3r=\deg(X^{6r}-1)-\deg(X^r+1))-\deg(X^{3r}-1)$$

The second part should follow by the same idea, using

$$X^{2r}+X^r+1=\frac{X^{3r}-1}{X^r-1}$$

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    Thanks so much for such detailed answer, I appreciate that. :)2012-12-27
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In general, the degree of the minimal polynomial for $e^{\frac{2\pi i}n}$ is $\phi(n)$. So if $x^{2r}-x^r+1$ is irreducible, then $2r=\phi(6r)$. Write $r=2^a3^bD$ with $(6,D)=1$. Then $$2r = \phi(6r)=\phi(2^{a+1})\phi(3^{b+1})\phi(D)=2^a(2-1)3^b(3-1)\phi(D) = 2r\frac{\phi(D)}D$$

So $\phi(D)=D$, so $D=1$.

For the second case, $e^{\frac{2\pi i}{3r}}$ is a root, so for $x^{2r}+x^r+1$ to be irreducible, $2r=\phi(3r)$. Write $r=3^aD$ with $(D,3)=1$.

Then $2r = \phi(3r)=\phi(3^{a+1}D)=(3-1)3^a\phi(D) = 2r\frac{\phi(D)}D$, and again $D=1$.

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    Thanks a lot, this is an elegant proof, this one and the first answer are both helpful to me. :)2012-12-27
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    I wasn't sure if you could use that the degree of the minimal polynomial had to be $\phi(n)$ for the $n$th root of unity, since in some sense, it is a more general result than what you are trying to prove. Possibly, as an exercise, it was an attempt to get you to prove these special cases before motivating this more general result...2012-12-27
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    yes , I can use that, we have learnt the Euler's function and learnt cyclotomic polinomial2012-12-27
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This solution requires familiarity with Cyclotomic polynomials, specfically their irreducibility. I will factor $x^{2r}-x^{r}+1$ completely.

As N.S. observed, $$x^{2r}-x^{r}+1 = \frac{x^{3r}+1}{x^r+1}=\frac{x^{6r}-1}{x^{3r}-1}\frac{x^r-1}{x^{2r}-1}$$ Remember that $\prod_{d|n} \phi_d(x) = x^n-1$, thus: $$x^{2r}-x^{r}+1 = \frac{\prod_{d|6r} \phi_d(x)}{\prod_{d|3r} \phi_d(x)}\frac{\prod_{d|r} \phi_d(x)}{\prod_{d|2r} \phi_d(x)}$$ Most of the terms cancel - $$\frac{\prod_{d|6r} \phi_d(x)}{\prod_{d|3r} \phi_d(x)} = \prod_{d|6r, d\nmid 3r} \phi_{d}$$ $$\frac{\prod_{d|r} \phi_d(x)}{\prod_{d|2r} \phi_d(x)} = \frac{1}{\prod_{d|2r, d\nmid r} \phi_{d}}$$

So $$x^{2r}-x^{r}+1 = \prod_{d|6r,d\nmid 3r,(d|r \text{ or } d\nmid 2r)} \phi_{d}(x)$$

When $r=3^a 2^b$, the unique positive integer satisfying the three conditions $d|6r,d\nmid 3r,(d|r \text{ or } d\nmid 2r)$ is $d=6r$ itself - the condition $d|6r,d\nmid 3r$ requires that $d$ is divisible by the maximal power of $2$ dividing $6r$, so $d=2^{b+1}3^{k \le a+1}$. $d$ can't divide $r$, so it must not divide $2d$, provided $k\ge a+1$, i.e. $k=a+1$. Thus $x^{2r}-x^r+1 = \phi_{6r}(x)$.

When $r$ is not a product of a power of 2 and a power of 3, $d=6r$ appears in the product, but also any $\frac{6r}{q}$ where $q$ divides $r$ but $(q,2)=(q,3)=1$: $$x^{2r}-x^{r}+1 = \prod_{d=6r/q,q|r,(q,6)=1} \phi_{d}(x)$$

(this formula is always valid.) If we write $r=2^{a}3^{b}r', (r',6)=1$, we find that the number of factors in the factorization of the polynomials equals the number of divisors of $r'$.

EDIT: In general, by using $\phi_{n} (x) = \prod_{d|n} (x^d - 1)^{ \mu (n/d) }$, one can show: $$\phi_{n}(x^m) = \prod_{d|m,(d,n)=1} \phi_{nm/d}(x) $$ So $\phi_{n}(x^m)$ is irreducible iff $p|m \implies p |n$, in which case $\phi_{n}(x^m) = \phi_{nm}(x)$.

To prove the last statement (without using my identity), notice that $\phi_{n}(x^m)$ must be divisible by $\phi_{nm}(x)$, and by comparing degrees as in the other solutions, to have equality we must have $m\phi(n)=\phi(nm)$, i.e. $nm \prod_{p|n}(1-\frac{1}{p}) = nm\prod_{p|nm}(1-\frac{1}{p})$, or $\prod_{p|nm, p \nmid n} (1-\frac{1}{p}) = 1$, which is equivalent to $p|m \implies p|n$.

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    thanks so much for giving me so detailed answer, I appreciate your effort. :)2012-12-27
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    @user53800 - In general, I proved $\phi_n(x^m) = \prod_{d|m, (d,n)=1} \phi_{nm/d}(x)$, using the explicit formula for $\phi_n$ (the one that uses the Mobius functions). Thus, $\phi_n(x^m)$ is irreducible iff each prime dividing $m$ also divides $n$, in which case $\phi_n(x^m)=\phi_{nm}(x)$.2012-12-31