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I have been thinking about this for quite sometime.

  • Eisentein Criterion for Irreducibility: Let $f$ be a primitive polynomial over a commutative unique factorization domain $R$, say $$f(x)=a_0 + a_1x + a_2x^2 + \ldots + a_nx^n \;.$$ If $R$ has an irreducible element $p$ such that $$p\mid a_m \qquad 0\le m\le n-1$$ $$p^2 \nmid a_0$$ $$p \nmid a_n$$ then $f$ is irreducible.

Can anyone give me an explanation of how one might have conjectured this problem? Thinking along, the same lines the first polynomial which came to my mind was $x^{2}+1 \in \mathbb{R}[x]$ which is irreducible. But there are lot's of polynomials and it's very difficult to think of a condition, which would make them irreducible.

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See David Cox's paper "Why Eisenstein proved the Eisenstein criterion and why Schonemann discovered it first" cs.amherst.edu/~dac/normat.pdf –  Weaam Feb 26 '11 at 18:27

3 Answers 3

up vote 13 down vote accepted

As Qiaochu hinted, Eisenstein's Criterion arises from exploiting factorization properties in simpler image rings. Here the property exploited is simply that the image reduces to a prime power, and prime products always factor uniquely in domains. With that in mind one can easily generalize Eisenstein's criterion to other polynomials of similar shape. However, there is a limit to how far one can go this way since the criterion corresponds to totally ramified primes, but there are number fields without such primes.

In order to find fruitful generalizations one may apply valuation-theoretic ideas. The Eisenstein and related irreducibility tests are all special cases of much more general techniques exploiting Newton polygons. They yield the master theorem behind all these related results. An excellent comprehensive introduction can be found in Filaseta's lecture notes - see the links below. See also this MathOverflow question.

[1] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/latexbook/

[2] http://www.math.sc.edu/~filaseta/gradcourses/Math788F/NewtonPolygonsTalk.pdf

[3] Newton Polygon Applet http://www.math.sc.edu/~filaseta/newton/newton.html

[4] Abhyankar, Shreeram S.
Historical ramblings in algebraic geometry and related algebra.
Amer. Math. Monthly 83 (1976), no. 6, 409-448.
http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=2964&pf=1

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The point is this: if $R \to S$ is a homomorphism and the image of $f$ is irreducible in $S[x]$, then $f$ must also be irreducible in $R[x]$ (since any factorization of $f$ in $R[x]$ maps to a factorization of $f$ in $S[x]$). This idea gets to the heart of what the point of homomorphisms is: they are maps that send algebraic statements ($f$ has a factorization) to other algebraic statements.

I guess the best motivated way to approach this is to make two choices for $S$. First we choose $S = \mathbb{Z}/p\mathbb{Z}$. This is a surprisingly useful choice for specific polynomials, especially of low degree (it works for some prime $p$ if and only if the Galois group of the polynomial $f$ contains a $\deg f$-cycle, which always happens in degree $2, 3$ and is still very likely in higher degrees). For example the polynomial $x^3 - x + 1$ must be irreducible because it has no roots $\bmod 2$.

At some point we might encounter a polynomial like $x^3 - 2x - 2$. Now this polynomial is reducible $\bmod 2$ since it factors as $x^3 = x \cdot x \cdot x$, but this implies that if $x^3 - 2x - 2$ had an actual factorization $fg$ in the integers then $f, g$ would both have to have all their non-leading terms divisible by $2$, hence the constant term of $fg$ is divisible by $4$; contradiction. Eisenstein's criterion is a simple generalization of this example; it corresponds to choosing $S = \mathbb{Z}/p^2\mathbb{Z}$ or $S = \mathbb{Z}_p$.

Nowadays it is understood that Eisenstein's criterion is really a statement about total ramification, but my guess is that this was not known when it was first written down. My guess is that Eisenstein wrote down this criterion for the sole purpose of proving that the cyclotomic polynomials $\Phi_p(x) = x^{p-1} + ... + 1$ are irreducible, and here it is very natural to think about what happens $\bmod p$ since $(x - 1) \Phi_p(x) = x^p - 1$ splits into linear factors $\bmod p$ by Fermat's little theorem. It is not a big leap from here to the general Eisenstein criterion.

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In the third paragraph, "...f,g would both have to have all their non-leading terms divisible by 2," I don't understand why this is true. Is there a theorem? –  Mark Aug 19 '11 at 3:19
    
@Mark: the only possible factorizations of $x^3$ in $\mathbb{F}_2[x]$ have factors of the form $1, x, x^2, x^3$. It follows that the image of any factor over $\mathbb{Z}$ is congruent to $1, x, x^2$ or $x^3 \bmod 2$. –  Qiaochu Yuan Aug 19 '11 at 3:38
    
Ok that explains why the leading term isn't divisible by 2. But how does it follow that the rest of the terms of both f and g have to be divisible by 2? That's where I am stuck. –  Mark Aug 19 '11 at 3:48
    
@Mark: suppose $f$ is an integer polynomial which is a factor of $x^3 - 2x - 2$. The argument above shows that $f(x) \equiv 1, x, x^2, x^3 \bmod 2$. This is equivalent to the statement that the non-leading terms of $f$ are divisible by $2$. –  Qiaochu Yuan Aug 19 '11 at 3:51
    
Oh Fermat's theorem. Shoot. Thanks for the clarification. –  Mark Aug 19 '11 at 3:58

If you don't know about ramification or valuations and things, there is still a great easy way to "see" Eisenstein from the $\mathbb{Z}[x]$ case. This is presented in Integers, Polynomials, and Rings by Ron Irving as a series of exercises to get the student to guess Eisenstein on their own.

First examine the case $x^n-p$ where $p$ is a prime. If this factors, then there are $f(x)=\sum_{k=0}^m a_kx^k$ and $g(x)=\sum_{k=0}^r b_kx^k$ for which $f(x)g(x)=x^n-p$. First, you get that $a_0b_0=p$ and hence without loss of generality $a_0=p$ and $p$ does not divide $b_0$ (we'll merely use this fact rather than $b_0=1$).

Now since $a_0b_1+a_1b_0=0$ we get that $p|(a_0b_1+a_1b_0)$ and $p|a_0$ so $p|a_1b_0$. This means $p|a_1$ or $p|b_0$, but $p$ does not divide $b_0$, so $p|a_1$. You can keep bootstrapping this argument up all the coefficients $a_k$ to get that $p|a_m$ the top coefficient, which is a contradiction since $a_mb_r=1$.

But what was really used here? Not much. So you could repeat the exact same argument with $x^n-pm$ where $(p,m)=1$. But that still isn't as general as you could go. You didn't need that $\sum_{j+k=l} a_jb_k=0$, you only needed that $p$ divided that sum to make the argument work, which is the same as checking that the polynomial $x^n+c_{n-1}x^{n-1}+\cdots +c_0$ has all $c_i$ divisible by $p$ (and $c_0$ not divisible by $p^2$).

Yet again, this still isn't as general as the argument allows you to go, because the contradiction only came from the fact that $p$ did not divide the leading coefficient, it wasn't that it was $1$.

So merely extrapolating what made the proof that $x^n-p$ is irreducible in $\mathbb{Z}[x]$ work gives you the full Eisenstein in $\mathbb{Z}[x]$. Lastly, if you want to go to $R[x]$, you need to figure out which facts about division you needed about $\mathbb{Z}$. But Eisenstein really is a criterion invented artificially in $\mathbb{Z}$ that could be guessed from examining what the key properties of one proof were.

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