I am trying to understand two portions of the proof of Theorem 2.4 here.
Let $L/K$ be an extension of nonarchimedean fields. Say the valuation on $K$ is $v$ extending to $w$ on $L$. A polynomial $f = a_nx^n + … + a_1x + a_0 \in K[x]$ is called Eisenstein if $v(a_0) = 0$, $v(a_i) \geq 1$ for all $1 \leq i < n$, and $v(a_n) = 1$.
Let $\pi_L$ be a uniformizer of $L$. First, why does every root $\alpha_i$ of a degree $n$ Eisenstein polynomial over $K$ having extended valuation $w(\alpha_i) = 1/n$ force the ramification index $e = (w(L^*):v(K^*)) = v(\pi_L)$ to be at least $n$?
Second, suppose you have $\alpha \in L$ such that $w(\alpha) = 1$ and that $L/K$ is totally ramified; that is $e := v(\pi_L) = [L:K]$. It follows that the $1,\alpha,…,\alpha^{n-1}$ form a $K$-linear independent set, so
$$a_0\alpha^n + a_1\alpha^{n-1} … + a_{n-1}\alpha + a_n = 0$$
The proof then claims that it must be the case that there exist $i,j$ such that $w(a_i\alpha^i) = w(a_j\alpha^j)$ are of maximal valuation. Why? And why does this then force $f = \sum a_ix^i$ to be Eisenstein?
Best Answer
It can be a bit confusing with various different normalizations of the valuation, but let's agree to say that $v(\pi_K) = 1$.
If $L/K$ is a finite extension, and $\pi_L$ is a uniformizer of $L$, then since $\pi_K \in L$ you can write
$$\pi_K = u \cdot \pi^e_L$$
for some unit $u \in \mathcal{O}^{\times}_L$. The number $e$ here the ramification index. The valuation $v$ extends to $L$ and you see you have the formula
$$1 = v(\pi_K) = e v(\pi_L)$$
so $v(\pi_L) = 1/e$.
Now suppose that $L/K$ is Eisenstein, so $L = K(\alpha)$. From the Newton Polygon, you can deduce that $v(\alpha) = 1/n$. But you can also write $\alpha = u \pi^r_L$ in $L$ for some $r$, and then you see that
$$1/n = v(\alpha) = r v(\pi_L) = r/e,$$
so $e = nr$. But you also know that $e \le n$, because otherwise $\pi_L$ could not satisfy a degree $\le n$ polynomial over $K$. So you deduce not only that $e = n$ but also that $r = 1$, that is, that $\alpha$ is a uniformizer of $L$.
For the second question, the $p$-adic valuation has the following behavior: if $\alpha$ and $\beta$ are elements of $L$, then $v(\alpha + \beta) \ge \min\{v(\alpha),v(\beta)\}$. But it actually satisfies a stronger property: if $v(\alpha)$ and $v(\beta)$ are different, then $v(\alpha+\beta) = \min\{v(\alpha),v(\beta)\}$. By induction, that means that if you add together $p$-adic numbers all with different valuation then the valuation of the sum is the smallest valuation of any term. But the sum in this case is $0$, which has infinite valuation, so this can't happen, that is, there must be two terms with the same minimal (not maximal) valuation.
Now back to the particular sum. First of all, we want to make sure we don't get confused about valuations. The implicit normalization here is that $\alpha$ is a uniformizer of $L$ and $v(\alpha)=v(\pi_L)=1$. Then this implies that $e=n=[L:K]$ and then $v(\pi_K)=n$. That is, the valuation of any element of $K$ is divisible by $n$. But now look at the sum.
One can certainly assume that that $a_i$ are all in $\mathcal{O}_K$ and are not all divisible by $\pi_K$. That means they all have valuations which are divisible by $n$. Moreover, there is at least on $k$ with $a_k$ a unit. Consider the two terms of minimal valuation: since $v(\alpha) = 1$ and $v(a_i) \in n \mathbf{Z}$, the valuation of the $k$th term is $k \bmod n$. So the only way that two terms can have equal valuation is if they are the first and last terms. Moreover, these are the terms of smallest valuation. The valuation of $a_0 \alpha^n$ is at least $n$ since $a_0$ is assumed to be in $\mathcal{O}_K$. But we are assuming that there is a $k$ with $a_k$ a unit; the corresponding term must have valuation exactly $v(a_k \alpha^k) = k \le n$. So the only way the first and last terms can have the smallest equal valuation and the
These together say exactly that the polynomial is Eisenstein.