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I want to prove that $f(x)=1+x+x^2+ \dots +x^6 $ is irreducible over $\mathbb{Q}[x]$.

My approach, by contradiction:

First, I define $\alpha = \dfrac{m}{n}$ could be a rational root of $f$. Now, if this is satisfied, I observed that $m|1$ and $n|1$. In consecquence, $m= \pm 1$ and $n = \pm 1$. Thus, the possible rational roots of $f$ are $\alpha = \dfrac{m}{n} = \pm 1$.

Nonetheless, $$ f(-1)= 1 \not=0 \quad \text{and} \quad f(1) = 7 \not=0$$ So $f$ does not have rational roots over.

Later on, $x^7-1 = (x-1)(x^6+x^5+\dots+1)$ which means that $f$ roots are the same as $(x^7-1)$. Also, by using binomial theorem, $(x-1)^7 = \sum_{i=0}^7 \binom{7}{i} x^i (-1)^{7-i} $ and I got stuck in here.

I appreciate any kind of help or hint. Thank you.

Rodrigo de Azevedo
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Davshock
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    This is a standard application of Eisenstein's criterion (applicable for any prime $p$, not just $p=7$) – lulu Apr 16 '23 at 16:07
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    Here is a standard proof. Here is a "non-standard" proof. – durianice Apr 16 '23 at 16:11
  • "which means that roots are the same as (^7−1)". No. It means that the set of roots of f union the set {1} is equal to the set of roots of (x^7 - 1). – Dan Asimov Apr 16 '23 at 20:50
  • It’s worth noting that trying to prove that a polynomial is irreducible by making some argument about the roots is probably not going to work if the polynomial is of degree greater than 3. Even if you get that none of the roots are in $\mathbb Q$, that doesn’t rule out the possibility of the polynomial factoring further. E.g. $x^4 -2x^2 -3 = (x^2 - 3)(x^2 + 1)$ has no rational roots but is not irreducible. – Kenanski Bowspleefi Apr 17 '23 at 16:23

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