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I am trying to prove the following assertion:

Given a prime number $p\in \mathbb{Z}$ , let $f =\sum_{i=0}^{2n+1}{a_ix^i}\in \mathbb{Z}[x]$ which is a polynomial of odd degree. Furthermore, we assume $p\nmid a_{2n+1}, p^2\mid a_{0}, \ldots, a_n, p\mid a_{n+1}, \ldots, a_{2n}$ and $p^3\nmid a_0$.

The aim is to prove $f$ is irreducible. Obviously, the Eisenstien Criterion cannot be used rightaway. What I tried was to make a linear change of variables. But it did not simplify matters. Any hints?

Thanks.

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    I think it is the hypothesis that $P^2\mid a_0$. Okay, thanks. I will try again.2012-03-18

1 Answers 1

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Suppose we have a decomposition of $f=P\cdot Q$ with $P(X)=b_k X^k+\ldots+b_0,Q(X)=c_m X^m+\ldots c_0\in\mathbb{Z}[X]$ and since the degrees add up to $2n+1$ we suppose without loss of generality that $m=\text{deg}(Q)\leq n<\text{deg}(P)$.

  1. Step: Reduce mod $p$ to show that $P$ and $Q$ must be Eisenstein polynomials in $p$.
  2. Step: Derive a contradiction from the fact that $p^2|a_m=c_m b_0+\ldots+c_0 b_m$ by showing that $p^2$ divides all the summands except $c_m b_0$.
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    Michalis, done.2012-03-18