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.
Best Answer
$R$ 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 $r,s\in v(P)$.