# Total ramification $p = \epsilon \pi^n$ implies $F = K(\pi)$ and minimal polynomial of $\pi$ Eisenstein

Before I state my question, let me give the set-up / "what I know".

Let $F / K$ be an extension of number fields of degree $n$. Let $v$ be a (discrete) valuation on $K$ and let $$\mathfrak{O}_{v} = \{ \alpha \in K : v(\alpha) \geq 0 \}$$ be the valuation ring of $v$. Up to associates, $\mathfrak{O}_{v}$ has a unique irreducible element $p$ and $v(p) = 1$.

Let $v_1, \ldots, v_m$ be the extensions of $v$ to the field $F$, let $$\mathfrak{O}_{v_i} = \{ \alpha \in F : v_i(\alpha) \geq 0 \}$$ be the valuation ring of $v_i$ $(i = 1, \ldots, m)$, and let $\pi_i$ be the unique (up to associates) irreducible element of $\mathfrak{O}_{v_i}$ $(i = 1, \ldots, m)$. Then the integral closure of $\mathfrak{O}_{v}$ in $F$ is $$\overline{\mathfrak{O}}_{v} = \cap_{i=1}^{m} \mathfrak{O}_{v_i},$$ the elements $\pi_1, \ldots, p_m$ constitute all the nonassociate irreducibles in $\overline{\mathfrak{O}}_{v}$, and $p$ factors in $\overline{\mathfrak{O}}_{v}$ as $$p = \epsilon \pi_{1}^{e_1} \cdots \pi_{m}^{e_m}$$ where $\epsilon$ is a unit of $\overline{\mathfrak{O}}_{v}$, and where $e_i$ is the ramification index of $v_i$ with respect to $v$ (i.e., $e_i$ is the positive integer such that $v(\alpha) = v_i(\alpha)/e_i$ for all $\alpha \in K^{\times}$, i.e., $e_i = v_i(p)$).

The residue class field of $v$ is $\Sigma_{v} = \mathfrak{O}_v / m_v$, where $$m_v = (p) = \{ \alpha \in K : v(\alpha) > 0 \}$$ is the unique maximal ideal of $\mathfrak{O}_v$. Similarly, the residue class field of $v_i$ is $\Sigma_{v_i} = \mathfrak{O}_{v_i} / m_{v_i}$, where $$m_{v_i} = (\pi_i) = \{ \alpha \in F : v_i(\alpha) > 0 \}$$ is the unique maximal ideal of $\mathfrak{O}_{v_i}$. The residual degree of $v_i$ over $v$ is $f_i = [ \Sigma_{v_i} : \Sigma_{v} ]$

The following formula relating the ramification indices $e_i$, the residual degrees $f_i$, the number $m$ of extensions of $v$ to $F$, and the degree $n$ of $F / K$ holds: $$\sum_{i=1}^{m} e_i f_i = n$$

Now the question: Suppose that $p$ factors in $\overline{\mathfrak{O}}_{v}$ as $$p = \epsilon \pi^n$$ where $\epsilon$ is unit in $\overline{\mathfrak{O}}_{v}$ and $\pi$ is an irreducible element in $\overline{\mathfrak{O}}_{v}$.

Part 1: I want to show that $F = K(\pi)$ (that is, $\pi$ is a primitive element for $F/K$). To do this, it suffices to show the minimal polynomial of $\pi$ over $K$, $$\phi(X) = X^k + a_{k-1}X^{k-1} + \cdots + a_0 \qquad (a_i \in K),$$ has degree $n$. Of course, $k \leq n$.

Alex Bartel has offered a solution below, but I don't understand some of the details. Here is my understanding of his argument. By definition, $$0 = \phi(\pi) = \pi^k + a_{k-1}\pi^{k-1} + \cdots + a_0$$ On the left, for every $i$, $v_i(\phi(\pi)) = v_i(0) = \infty$. On the right, $v_i(\pi^k) = k$, $v_i(a_{j}\pi^{j}) = v_i(a_{j}) + j = nv(a_j) + j$ for $j=0, \ldots, k-1$, and so $$v_i(\phi(\pi)) \geq \min(v_i(\pi^k), v_i(a_{j}\pi^{j})) = \min(k, nv(a_j) + j)$$ where $j$ ranges from $0$ to $k-1$. If $k \neq nv(a_j) + j \neq nv(a_l) + l$ for all $j \neq l$, then equality holds in the last displayed equation, which contradicts that the left-hand side is infinite. Alex asserts that $k \neq nv(a_j) + j \neq nv(a_l) + l$ for all $j \neq l$ holds if $k < n$, meaning $k < n$ is impossible. I do not understand why this assertion should be ture

Part 2: I also want to show that the minimal polynomial of $\pi$ over $K$, $$\phi(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_0 \qquad (a_i \in K),$$ is Eisenstein with respect to $v$ (that is, $v(a_{i}) > 0$ for $i = 1, \ldots, n-1$ and $v(a_0) = 1$ [I think this is what Eisenstein with respect to $v$ means]).

Based on the formula $\sum_{i=1}^{m} e_i f_i = n$, I know that $v$ has only one extension to $F$, $v_1$, with $e_1 = n$, $f_1 = 1$, and irreducible element $\pi_1 = \pi$. So $\pi$ is the unique irreducible element of $\overline{\mathfrak{O}}_{v}$ and every nonzero element $\alpha$ in $\overline{\mathfrak{O}}_{v}$ factors as $$\alpha = \epsilon \pi^{a}$$ for some unit $\epsilon$ in $\overline{\mathfrak{O}}_{v}$ and some $a \geq 0$. Since the quotient field of $\overline{\mathfrak{O}}_{v}$ is $F$, any nonzero $\alpha$ in $F$ can be expressed as above if we allow $a$ to range through the integers.

That is where I am stuck.

I would like to solve this problem in the language of valuations as I am studying from Borevich and Shafarevich, and this is the language they use.

I would appreciate some help with this.

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See math.uconn.edu/~kconrad/blurbs/gradnumthy.totram.pdf. In your setting you will need to use an analogue of the ideal norm, which is developed in Borevich and Shafarevich's book (their function $N$ on divisors). –  KCd Mar 31 '11 at 8:45

First, to show that $F=K(\pi)$, it is enough to show that $K(\pi)$ has degree $n$ over $K$, since one inclusion is clear. In fact, you only need to show that $K(\pi)$ has degree at least $n$. Let $v$ be the normalised valuation on $K$ corresponding to $p$ (which by the way means that $v(p)=1$, not 0 as you wrote), and extend $v$ to $F$, so that $v(\pi)=1/n$. Let $f=\sum_{i=1}^m a_mX^m$ be the minimal polynomial of $\pi$ over $K$. If $m<n$, then the valuations of all the monomials in $f(\pi)$ are distinct, since they are distinct modulo $\mathbb{Z}$, so $f(\pi)\neq 0$. You deduce that we must have $m=n$ and this settles your first question.
To show that $f$ is Eisenstein, you again consider valuations of the monomials. As we have already observed, if some monomial in $f(\pi)$ has valuation strictly smaller than all the others, then $f(\pi)\neq 0$. Since $v(\pi^n) = 1$, you deduce that $v(a_i)\geq 1$ for all $i$. But if $v(a_0)>1$, then the top power term has the lowest valuation. which we agreed couldn't be the case, so you are done.
I don't understand why the monomials of $f(\pi)$ have distinct valuations. What do you mean by modulo $\mathbb{Z}$? –  admchrch Apr 1 '11 at 6:30
@admchrch The valuation of $\pi^i$ is $i/n$, the valuations of the coefficients $a_i$ are integers (since they are in the base field). So if $m < n$, then the valuations of the monomials have different fractional parts. That's what I mean by "modulo $\mathbb{Z}$". –  Alex B. Apr 1 '11 at 11:41