If p is a prime, prove that $$x^{p-1}-x^{p-2}+x^{p-3}-\cdots-x+1$$ is irreducible over $\Bbb Q$. Hope somebody can show me some details. Thanks
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1Look at $f(-x-1)$. Expand it out using the binomial theorem, and apply Eisenstein. – Prahlad Vaidyanathan Nov 10 '13 at 03:42
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1Or f(x+1) and use Eisenstein"s :) – Mr.Fry Nov 10 '13 at 04:16
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Duplicate: http://math.stackexchange.com/questions/87609/eisenstein-criterion-with-a-twist?lq=1 – Prism Nov 10 '13 at 04:44
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Another duplicate: http://math.stackexchange.com/questions/408072/is-polynomial-1xx2-cdotsxp-1-irreducible?lq=1 – Prism Nov 10 '13 at 04:45
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I'm assuming that $p$ is odd; otherwise the alternating signs don't match up. It's well known that $F(x)x^{p-1}+x^{p-2}+x^{p-3}+\cdots +x+1$ is irreducible, but I'll reproduce the proof here.
$F(x)$ is irreducible iff $F(x-1)$ is irreducible. But $$ F(x)=\frac{x^p-1}{x-1}=\frac{(y+1)^p-1}{y} $$ Where $y=x-1$. Expanding using the binomial theorem, this is $$ y^{p-1}\binom{p}{p}+y^{p-2}\binom{p}{p-1}+\cdots+y\binom{p}{2}+\binom{p}{1} $$ But by Eisenstein's criterion, this is irreducible, as $p\mid \binom{p}{k}$ for $1\le k \le p-1$, and $p^2\nmid \binom{p}{1}$.
Now the original polynomial, $x^{p-1}-x^{p-2}+\cdots -x+1$ is simply $F(-x)$, so it is irreducible as well.
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