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Is this $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ irreducible in $\mathbb Q[x]$? Can you help please?

mathlover
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    It is the seventh cyclotomic polynomial, and they are all irreducible. Alternately, set $x \to x+1$ and see if you can Eisenstein. – Sarvesh Ravichandran Iyer Nov 25 '20 at 06:47
  • It would be useful if you could tell us what you have already tried. For example, a standard irreducibility test is the reduction mod $p$ test. If you try it (with a computer for factoring, perhaps) then you'll find it helps for certain small $p$. – KCd Nov 25 '20 at 06:51
  • @KCd that's a cool suggestion too, the mod $p$ test can take over when Eisenstein doesn't work! – Rachid Atmai Nov 25 '20 at 07:08
  • This polynomial is irreducible modulo a prime $p$ if and only if $p$ is a primitive root modulo $7$. That is, if and only if $p\equiv3,5\pmod7$. – Jyrki Lahtonen Nov 25 '20 at 08:42
  • This and this thread are related. – Jyrki Lahtonen Nov 25 '20 at 08:43
  • I do want to refer the asker to Wikipedia as well as our guide for new askers. Neither of these are easy enough to find. Absorbing the contents of the latter will make your future on this site much more pleasurable. More concretely, it will help you avoid the votes to close. The WP-link is hard to find unless you know the buzzword (which makes it show on top of most of the search engines). – Jyrki Lahtonen Nov 25 '20 at 08:46

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Let $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$. Define the new polynomial $f(x+1)=x^6+7x^5+21x^4+35x^2+21x+7$, using the binomial theorem expand $f(x+1)$. This polynomial is irreducible over $\mathbb{Q}$ by Eisenstein criterion, for $p=7$

If $f(x)$ were reducible over the rationals then let $p(x), q(x)\in \mathbb{Q}[x]$ such that $$f(x)=p(x)q(x)$$ where $\deg(p)< 6$ and $\deg(q)<6$ We would then have $$f(x+1)=p(x+1)q(x+1)$$ but that's impossible! Hence $x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Q}[x]$

Rachid Atmai
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