October 17th
Today I learned a little bit about Eisenstein's criterion, in hopes of finding a polynomial for which Eisenstein's criterion fails for any transformation. Namely, Keith Conrad shows here that if the minimal polynomial for $\alpha$ satisfies Eisenstein's criterion with $p$ ("is $p$-Eisenstein''), then $p$ totally ramifies in $\QQ(\alpha).$ The converse is also true—if $p$ totally ramifies in $K,$ then there exists an $\alpha$ with $K=\QQ(\alpha)$ such that the minimal polynomial for $\alpha$ is $p$-Eisenstein—but we don't show that here.
Indeed, fix a root $\alpha$ of a monic polynomial $f\in\ZZ[x]$ of degree $n$ which is $p$-Eisenstein and therefore irreducible. (If the polynomial wasn't monic, we could multiply it by the leading coefficient enough times until it was without changing the $p$-Eisenstein condition or the generated field $\QQ(\alpha).$) Explicitly, we know that with\[f(x)=\sum_{k=0}^na_kx^k,\]we have $a_n=1,$ each $a_k$ is divisible by $p$ for $k\in[0,n),$ and $p^2\nmid a_0.$
We're going to show that $p$ is totally ramified in $K=\QQ(\alpha).$ We suspect there is only one prime above $p,$ so let $\mf p$ be a prime above $p.$ We want to show $(p)=\mf p^n,$ but for now let $(p)=\mf p^eI$ for some $\mathcal O_K$-ideal $I.$ The hard part of this argument is showing $e\ge n.$
Now we return to talking about $f(x).$ The main idea is to reduce\[\alpha^n+\sum_{k=1}^na_k\alpha^k=f(\alpha)\equiv0\pmod{\mf p^\bullet}\]for some large power $\mf p^\bullet$; it turns out most of these terms are already divisible by a large power. Being $p$-Eisenstein tells us that $a_k\in(p)$ for $k\in[0,n),$ which gives $a_k\in\mf p^e$ for free. So, reducing gives that\[\alpha^n\equiv0\pmod{\mf p^e},\]and in particular $\alpha\in\mf p.$ But then we can push this to $a_k\alpha^k\in\mf p^{e+1}$ for $k\in[1,n),$ so we know\[\alpha^n+a_0\equiv0\pmod{\mf p^{e+1}}.\]
Quickly, the other $p$-Eisenstein condition is that $p^2\nmid a_0,$ which will tell us that $a_0\notin\mf p^{e+1}.$ Indeed, this is equivalent to $\mf p^{e+1}\nmid(a_0),$ so it's enough to show that $\nu_\mf p(a_0\mathcal O_K)=e.$ We may let $a_0=pa$ where $p\nmid a,$ and then expanding this as a unique prime factorization tells us\[a_0\mathcal O_K=p\mathcal O_K\cdot a\mathcal O_K=\mf p^eI\cdot a\mathcal O_K.\]The above has successfully factored $\mf p^e$ out of $a_0\mathcal O_K,$ so we want to know that the remaining factors have no $\mf p.$ Well, $I$ is coprime by definition, and $\mf p\mid a\mathcal O_K$ is equivalent to $a\in\mf p$ implies $a\in\mf p\cap\ZZ=p\ZZ,$ contradicting the definition of $a.$
Returning to the argument, we see $\alpha^n\equiv-a_0\pmod{\mf p^{e+1}},$ so $a_0\notin\mf p^{e+1}$ implies that $\alpha^n\notin\mf p^{e+1}.$ This means $\mf p^{e+1}\nmid(\alpha)^n,$ but $(\alpha)$ is divisible by at least one $\mf p\ni\alpha,$ so we get $\mf p^{e+1}\nmid\mf p^n.$ It follows $e+1 \gt n,$ or $e\ge n.$ This finishes the hard part of the argument.
To finish, we use the fundamental identity. Indeed, note\[n=\sum_{\mf q/p}e(\mf q/p)f(\mf q/p)\ge nf(\mf p/p)+\sum_{\substack{\mf q/p\\\mf q\ne \mf p}}e(\mf q/p)f(\mf q/p)\ge n\]forces $f(\mf p/p)=1$ and removes the possibility of any other primes above $p.$ Thus, $p\mathcal O_K=\mf p^n,$ which is total ramification.