# Variants of Eisenstein irreducibility

In his article where he stated what we know as Eisenstein's irreducibility criterion (which actually was first proved by Schönemann, as was Scholz's reciprocity law and Hensel's Lemma), he claimed that the result also holds in the following case: assume that $$f(x) = x^m + a_{m-1}x^{m-1} + \ldots + a_0,$$ where the $a_i$ are integers divisible by $p$, and where $p^2 \nmid a_j$ for some $0 \le j \le m-1$. There is an obvious counterexample provided by the quadratic polynomial $f(x) = (x-p)^2 = x^2 -2px + p^2$, so that could be the end of that story. But it isn't: I've read somewhere that some form of this criterion holds for polynomials of degree $\ge 3$, and that the degrees of the possible factors of counterexamples can be predicted in terms of this index $j$. Unfortunately, I don't remember the exact statement (those who are familiar with Newton polygons will probably be able to figure out a correct version) or where I've seen this. Can anyone help?

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If you have a kink in the NP (i.e. if it's not just one line) then surely an arbitrarily small perturbation of your polynomial will factor over Q? Here's an example of something that is true: if f is monic of odd degree m, and p^2 divides all the a_i for 0<=i<=m-1, but p^3 doesn't divide a_0, then f will be irreducible, because the local extension will contain things of valuation 2/m and hence things of valuation 1/m. –  Kevin Buzzard Feb 16 '10 at 15:28
On the other hand if p exactly divides a_j for j>0 but the a_i for 0<=i<j are allowed to be divisible by arbitrary powers of p then you can't possibly deduce irreducibility: consider (x^(m-j)-p)*(x^j-p^1000). –  Kevin Buzzard Feb 16 '10 at 15:29

Basically all such criteria boil down to some argument involving the Newton polygon as Kevin Buzzard mentions in the comments. While something as general as your statement has trivial counter examples the following generalization holds:

Let $R$ be a unique factorization domain and $f(x) =a_nx^n+\cdots +a_0\in R[x]$ with $a_0a_n\neq 0$. If the Newton polygon of $f$ with respect to some prime $p\in R$ consists of the only line segment from $(0,m)$ to $(n, 0)$ and if $gcd(n,m) = 1$ then $f$ is irreducible in $R[X]$.

I've heard this called the Eisenstein-Dumas criterion of irreducibility (it also proves the example given in the comments). Another generalization of Eisenstein's criterion is the following:

If $p|a_0,a_1,\dots,a_k$ but $p^2\not | a_0$ then $f(x)$ has an irreducible factor of degree $\geq k+1$

(This is how you prove for example, that a polynomial like $x^n+5x^{n-1}+3$ is irreducible, after checking that it has no linear factors.) If not answering your question, at least I hope that this refreshes your memory of the statement you claim above. :)

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Interesting. I've only ever seen the problem you mention solved using Perron's criterion, but that's a nice solution. –  Qiaochu Yuan Jul 12 '10 at 6:18
That example comes from some olympiad and while the intended (proposed) solution was with Perron's criterion, everyone in the competition solved it using Eisenstein-like criteria. –  Gjergji Zaimi Jul 12 '10 at 6:34

Such a generalization (Dumas' theorem) was discussed here: Is a polynomial with 1 very large coefficient irreducible?

A good source to learn about it is Prasolov's book on polynomials: http://tinyurl.com/prasolov - see page 53, Dumas' theorem (and a bit before this theorem).

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Yeah, I had your answer from the other question in mind when mentioning Dumas' theorem above :) –  Gjergji Zaimi Feb 17 '10 at 11:07
Good to know :-) –  Vladimir Dotsenko Feb 17 '10 at 11:26
Thanks for the reference to Prasolov's book. –  Franz Lemmermeyer Feb 17 '10 at 14:21

Franz, if memory serves correct there is extensive discussion of variants of Eisenstein's criterion and relations with Newton polygon's etc in some of Filaseta's work, e.g. in his interesting book (draft) "The theory of irreducible polynomials". It was previously available at [1] but you may now need to write him for a password to access it. He also has some software available, e.g. Java applets for computing Newton polygons, etc.

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There, Netto proved the following: A polynomial $$f(x) = x^n + a_{n-1} p x^{n-1} + \ldots + a_{k+1} p z^{k+1} + a_k p^2 z^k + \ldots + a_0 p^2$$ with degree $n > 2k$, in which the $a_j$ are integers such that $p \nmid a_0$, does not have a factor of degree less than $k+1$. This is similar to Gjergji's second example, but allows the divisibility by $p^2$ that I had had in mind.