Let $p$ a prime number, is the polynomial
$$1+x+x^2+\cdots+x^{p-1}$$ irreducible in $\mathbb{Z}[x]$ ? Thanks in advance.
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What else can you write it as? After that, can you shift $x$ and make the polynomial Eisenstein at $p$? – anon May 31 '13 at 22:54
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See also: Irreducibility of $X^{p-1} + \ldots + X+1$ or Irreducibility issue or Eisenstein Criterion with a twist – Martin Sleziak Jun 01 '13 at 04:40
2 Answers
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Mr. Eisenstein certainly thinks so!
The polynomial can be rewritten as $\frac{x^p-1}{x-1}$. Setting $x=(y+1)$ has no effect on (ir)reducibility of the polynomial; doing so and applying the binomial theorem yields: $$\frac{(y+1)^p-1}{(y+1)-1}=\frac{1}{y}\sum_{k=1}^{p}\binom{p}{k}y^{k}=y^{p-1}+\binom{p}{p-1}y^{p-2}+\ldots+\binom{p}{2}y^1+\binom{p}{1}y^0$$
Since all the coefficients apart from the leading one are divisible by $p$ and the constant term is too small to be divisible by $p^2$, we can apply the criterion and conclude that the polynomial is irreducible over rationals (and thus also over integers).
Peter Košinár
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How do we know that $\sum_{k=0}^l {k \choose l}$ is divisible by p for $0\leq l \leq p-2$? I have brute-forced verified this for the special case $p=5$, but I fail to see a trick that will let me prove this for arbitrary prime $p$. – Pascal's Wager Mar 02 '18 at 16:08
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@Pascal'sWager Are you sure you are interested in the sum you specified (it doesn't include any mention of $p$ and is actually equal to $1$ since all the terms are zero apart from the one with $k=l$). When it comes to the sum in my answer, it's each binomial coefficient on its own being divisible by $p$ already: Since $\binom{p}{k}=\frac{p!}{k!(p-k)!}$ and the numerator is divisible by $p$ while all the multiplied terms in the denominator are strictly smaller than $p$, the whole binomial coefficient is divisible by $p$ too. – Peter Košinár Mar 02 '18 at 17:00
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I meant $\sum_{k=0}^{p-1} {k \choose l}$. Sorry for that mistake. – Pascal's Wager Mar 02 '18 at 17:18
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$1+x+x^2+\ldots +x^{p-1}=\frac{x^p-1}{x-1}$. So if we set $x=y+1$, then our polynomial becomes $\frac{(y+1)^p-1}{y}=y^{p-1}+py^{p-2}+\frac{p(p-1)}{2}y^{p-3}+\ldots+p$, which is irreducible by Eisenstein. But since $x \mapsto y+1$ is an isomomorphism from $\mathbb{Z}[x]$ to $\mathbb{Z}[y]$, our original polynomial is irreducible as well.
Chris Eagle
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