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Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.

roi_saumon
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    Maybe memorizing a few specific examples (and non-examples) can help. – Berci Nov 15 '18 at 17:33
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    Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this. – Viktor Vaughn Nov 15 '18 at 17:57
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    I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic. – rschwieb Nov 15 '18 at 18:23
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    How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right? – Jyrki Lahtonen Nov 15 '18 at 20:56

2 Answers2

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It's memorable if you view the proof as a consequence of the unique factorization of prime products.

The first part of the hypothesis says: $\bmod p\!:\ f \equiv a x^n\,$ is (an associate of) a prime power $x^n$

By uniqueness a $\rm\color{#c00}{proper}$ factorization has form $\, gh\equiv (b x^i) (c x^j)\,$ for $\,\color{#c00}{i,j \ge 1}$

But $\,\color{#c00}{i,j \ge 1}\,\Rightarrow\, p\mid g(0),h(0)\,\Rightarrow\, p^2\mid g(0)h(0)\!=\!f(0),\,$ contra hypothesis.

Said in ideal language it is $\,(p,x)^2\equiv (p^2)\,\pmod{\! x}$

Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $x\mapsto x+c$ that are Eisenstein, e.g. see this answer.

Bill Dubuque
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  • Why does $i,j\ge 1$ imply that $p$ divides both $g(0)$ and $h(0)$? – Alex Ortiz Nov 15 '18 at 18:52
  • @AOrtiz $\bmod p!:\ g\equiv b x^{\large i},\ i\ge 1,\Rightarrow g(0)\equiv b\cdot 0^{\large i}\equiv 0\ $ so $\ p\mid g(0),\ $ i.e. a nonconstant monomial $,bx^{\large i}$ has zero constant term (in $\Bbb F_p = \Bbb Z/p)$ – Bill Dubuque Nov 15 '18 at 19:00
  • Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma. – Bill Dubuque Nov 15 '18 at 19:07
  • @BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$? – roi_saumon Nov 17 '18 at 14:58
  • @roi Yes, $D = \Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^{n-i},$ (up to unit factors). – Bill Dubuque Nov 17 '18 at 16:19
  • @BillDubuque, I cannot open the link in "this" link in the comment. I would be interested to explore it – edamondo Jun 11 '25 at 14:55
  • @edamondo That question was deleted, see instead here. $\ \ $ – Bill Dubuque Jun 11 '25 at 15:59
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The way I always remember it is via the Newton polygon.

The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.

Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.

Alex J Best
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