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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?

swan
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  • How do you define $e_p(0)$? Presumably as $\infty$? Also, the fact that $\mathbb{Z}{p^k}[X]$ is not a UFD need not be an obstacle; if the polynomial is irreducible modulo $p^k$, then it is necessarily irreducible modulo $p^k$. The concept of irreducibility still makes sense in $\mathbb{Z}{p^k}[X]$. – Arturo Magidin Dec 21 '10 at 16:22
  • @Arturo: Oops, my bad! You're right, I should've added that. Fortunately this omission isn't too confusing. Also, I was trying to argue somewhere along these lines: reduce f mod $p^k$ (k as big as possible) to obtain a polynomial of the form $cx^n$. Now, had $Z_{p^k}[X]$ been a UFD, the only possible factorization would be into polynomials $dx^s$ and $ex^{k-s}$... – swan Dec 22 '10 at 12:38
  • @ArturoMagidin Thank you very much! Your explanation is perfect and helped me understand this problem! – Bman72 Sep 26 '14 at 13:57

2 Answers2

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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 $\mathbf{Q}$.

George McNinch
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  • Thanks for that! Gotta learn about the Newton Polygons. I wonder if there's a more elementary way of proving validity of the criterion though. I mean, Newton Polygons are not something I could've came up with myself. – swan Dec 22 '10 at 12:46
  • @swan: Right. Well, Zarrax's solution is more elementary. But the Newton polygon seems to work pretty "naturally" for this problem. There is a treatment of the NP in the 1st chapter of Dwork, Gerotto and Sullivan "An Intro to G-Functions". – George McNinch Dec 24 '10 at 00:09
  • Great! I'll definitely take a look. Thank you once more! – swan Dec 29 '10 at 12:09
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Apply Eisenstein's criterion to ${1 \over p^{n-1}}x^nf({p \over x})$.

Zarrax
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  • Why would irreducibility of $\frac{1}{p^{n-1}}x^nf(\frac{p}{x})$ imply irreducibility of f? Also I have to admit I don't really know how to interpret $f(\frac{p}{x})$. – swan Dec 22 '10 at 12:47
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    @swan: $x^nf(\frac{1}{x})$ is the polynomial you get by 'reversing' the powers. If you had $a_nx^n + a_{n-1}x^{n-1}+\cdots+a_0$ (with $a_n\neq 0$, then $x^nf(\frac{1}{x}) = a_n + a_{n-1}x + \cdots + a_1x^{n-1} + a_0x^n$. Now see what happens if you do $x^nf(\frac{p}{x})$ (write out the powers of $p$ in the $a_i$ explicitly, that will help), and then see if you can factor out $p^{n-1}$. Factorizations of $f(x)$ correspond to factorizations of $x^nf(\frac{1}{x})$ in the "obvious way" (just 'reverse the powers' in the factors). – Arturo Magidin Dec 23 '10 at 06:45
  • OK, I get in now. Thanks a lot! I'm accepting this one since I understand it better. – swan Dec 29 '10 at 12:08