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How do I show that $f(x)=1+2x+\cdots+(p-1){x}^{p-2}$ is not reducible on $\mathbb{Q}$, where $p$ is prime.

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up vote 6 down vote accepted

By Gauss's lemma, we only need to prove that $f$ is not irreducible over $\mathbb{Z}$. We have $$g(x):=(x-1)^2f(x)=(p-1)x^p-px^{p-1}+1.$$

Consider $g(x)$ in the field $\mathbb{Z}/p\mathbb{Z}$, where $g(x)=(p-1)(x-1)^p$. Therefore, we have $f(x)=(p-1)(x-1)^{p-2}$ in $\mathbb{Z}/p\mathbb{Z}$. Hence, $f(x+1)=(p-1)x^{p-2}$ over $\mathbb{F}_p[x]$. The constant term of $f(x+1)$ is $(p-1)C_2^p-pC_2^{p-1}$, which is divisible by $p$ but not by $p^2$. Then by Einsenstein's criterion, $f(x+1)$ is irreducible. As a result, $f(x)$ is irreducible.

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can you show me in elementary method. – Leitingok Nov 25 '12 at 13:03
Which step/steps is/are not elementary? – Did Nov 25 '12 at 13:15
I do not know if this lemma can lead to a possible "analytic" proof of irreducibility, but it is not difficult to prove that the polynomial $x^p-px+(p-1)$ has only 3 real roots, one simple root less than $-1$ and a double root in $x=1$, so there are $p-3$ complex conjugated roots. Moreover, all the roots $\xi$ of the polynomial lie in the thin annulus $$ 1-\frac{2}{p}\leq|\xi|\leq 1+\frac{2}{\sqrt{p-1}}.$$ – Jack D'Aurizio Nov 27 '12 at 16:12

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