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I could use some help with proving the following irreducibility criterion. (It came up in class and got me interested.)

Let p be a prime. For an integer $n = p^k n_0$, where p doesn't divide $n_0$, set: $e_p(n) = k$. Let $f(x) = a_n x^n + \cdots + a_1 x + a_0$ be a polynomial with integer coefficients. If:

  1. $e_p(a_n) = 0$,
  2. $e_p(a_i) \geq n - i$, where $i = 1, 2, \ldots, n-1$,
  3. $e_p(a_0) = n - 1$,

then f is irreducible over the rationals.

Reducing mod p and mimicking the proof of Eisenstein's criterion doesn't cut it (I think). I also tried playing with reduction mod $p^k$, but got stuck since $Z_{p^k}[X]$ is not a UFD.

Also, does this criterion has a name?

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    @ArturoMagidin Thank you very much! Your explanation is perfect and helped me understand this problem!2014-09-26

2 Answers 2

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Apply Eisenstein's criterion to ${1 \over p^{n-1}}x^nf({p \over x})$.

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    OK, I get in now. Thanks a lot! I'm accepting this one since I understand it better.2010-12-29
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One way to prove the irreducibility seems to be to use the Newton Polygon. The condition on the coefficients of $f$ means that the Newton Polygon has a side of slope $\dfrac{1}{n}-1$ and hence that $f$ has a root $\alpha$ in some algebraic closure $F$ of $\mathbf{Q}_p$ having valuation $1 - \dfrac{1}{n}$ (there is a unique way to prolong $e_p$ to a valuation of $F$).

But then the extension $\mathbf{Q}_p \subset \mathbf{Q}_p(\alpha)$ is totally ramified of degree $n$ and $f$ must be irreducible over $\mathbf{Q}_p$ hence a fortiori irreducible over $\mathbf{Q}$.

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    Great! I'll definitely take a look. Thank you once more!2010-12-29