# Eisenstein criterion and Newton polygon

Eisenstein criterion of irreducibility over $$\Bbb Q_p$$(so ofcourse over $$\Bbb Q$$) is also proved by using Newton polygon.

The proof goes like this, Let $$f(x)＝a_nx^n＋･･･＋a_1x＋a_0$$ be $$p$$-Eisenstein polynomial in $$\Bbb Z_p[x]$$. Then, Newton polygon of $$f(x)$$ consists of one segment of slope $$1/n$$. Thus, all roots of $$f(x)$$ in algebraic closure of $$\Bbb Q_p$$, say $$α_1、α_2､･･･,α_n$$, have the same $$p$$-adic value $$1/n$$.

Now, $$f(x)＝(x-α_1)(x-α_2)･･･(x-α_n)$$, and if $$f$$ is reducible over $$\Bbb Q_p$$, product of some proper subset of the roots, say $$β_1β_2･･β_m$$ must have integer valuation, but $$ord_p( β_1β_2･･β_m)＝m/n$$ is not integer. This contradicts the fact that $$ord_p$$ takes integer value.

(Note that above discussion proved Eisenstein criterion without using, what we call Gauss lemma)

My question:

But Eisenstein criterion also holds on fraction field of valuation ring, furthermore, over fraction field over any integral domain. Above argument uses the value group is discrete, so the proof cannot apply to this general case.

Can Newton polygon argument deduce generalized (over integral domain) Eisenstein criterion ?

Thank you in advance.

• What exactly is the formulation you are having in mind where we are dealing with a non-discrete valuation? Because the valuation in reuns' answer ($\mathfrak p$-adic value for a prime ideal $\mathfrak p$), which applies in particular in the number field case, looks like a discrete one to me ... Commented Dec 10, 2021 at 21:33
• For example, UFD which is not DVR is in mind. Newton polygon argument only applies to a field which has discrete valuation, so the newton polygon argument seem meaningless to rings which is not DVR. Is my understanding correct ? Commented Dec 11, 2021 at 10:53
• $\mathbb Z$ is not a DVR, yet the $p$-adic valuation on $\mathbb Z$ is discrete ... Commented Dec 11, 2021 at 16:34

$$R$$ is an integral domain, $$P$$ a prime ideal, $$f\in R[x]_{monic}$$, $$f\equiv x^n \bmod P$$,
If $$f=gh$$ with $$g,h\in R[x], \not \in R$$ then $$g\equiv u x^a \bmod P,h\equiv w x^b\bmod P$$ with $$u,w\in R/P^\times$$ and $$a+b=n,a\ge 1,b\ge 1$$. So $$g(0),h(0)\in P$$ which implies that $$f(0)\in P^2$$.
You can take a valuation over $$P$$ (that is a valuation on $$Frac(R)$$ such that for $$a\in R, v(a)\ge 0$$ and $$v(a)=0$$ iff $$a\not \in P$$) to rephrase it in term of valuations, that is $$f$$ reducible implies that $$v(f(0)) = r+s$$ for some $$r,s\in v(P)$$.
• In general it can happen that $P^2=P$, and even if we have a valuation, if it is not discrete (e.g. $R= \mathbb Z_p[\mu_{p^\infty}]$), $v(P)$ can be a divisible group and $v(f(0)) > 0$ might well be equivalent to $v(f(0)) = r+s$ for some $r, s \in v(P)$. Of course everything you write is still correct, I just highlight that to the OP regarding my clarification request about what an "Eisenstein criterion" in the case of non-discrete valuation would even be ... Commented Dec 13, 2021 at 18:26