# Am I the only one constantly forgetting the Eisenstein criterion? [closed]

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.

• Maybe memorizing a few specific examples (and non-examples) can help. Commented Nov 15, 2018 at 17:33
• Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this. Commented Nov 15, 2018 at 17:57
• 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. Commented Nov 15, 2018 at 18:23
• 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? Commented Nov 15, 2018 at 20:56

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.

• Why does $i,j\ge 1$ imply that $p$ divides both $g(0)$ and $h(0)$? Commented Nov 15, 2018 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 $\Bbb F_p = \Bbb Z/p)$ Commented Nov 15, 2018 at 19:00
• Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma. Commented Nov 15, 2018 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 $x^n$? Commented Nov 17, 2018 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). Commented Nov 17, 2018 at 16:19

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.