$ef$ theorem for inertial degree and ramification index

Question

$L/K$ is a finite algebraic extension of fields. $O_{P}$ and $O_{\mathcal{P}}$ are DVRs with $P$ and $\mathcal{P}$ maximal ideals respectively, their quotient fields are $K$ and $L$ respectively. Let $e$ be the ramification index of $\mathcal{P} / P$ and $f$ be the inertial degree of $\mathcal{P}/P$.

I am working on showing $ef \leq n=[L: K]$ (without separable condition). Not quite understand several details in the proof.

The proof in the book is: Let $\Pi$ be a generator of $\mathcal{P}$ and choose $\omega_{1},\ldots,\omega_{m}$ such that their modulo $\mathcal{P}$ are linearly independent over $O_{P}/P$. Then show $\omega_{i}\Pi^{j}$ are linearly independent over $K$.

Take a relation $$\sum_{j=0}^{e-1}\sum_{i=1}^{m}a_{ij}\omega_{i}\Pi^{j}=0.$$ If $a_{ij}$ are not all zero, we can assume the are all in $O_{P}$ and at least one of them is not in $P$ (I understand they could be put in $O_{P}$ since K is the quotient field, but why at least one not in $P$?).

Then consider the elements $$A_{j}=\sum_{i=1}^{m}a_{ij}\omega_{i}.$$ If some $a_{ij}\not \in P$, then $A_{j}$ is a unit in $O_{\mathcal{P}}$. Otherwise, $A_{j}$ is divisible by $\pi$, the generator of $P$ and so $\text{ord}_{\mathcal{P}}(A_{j}) \geq e$.

Thus, $\text{ord}_{\mathcal{P}}(\sum_{j=0}^{e-1}A_{j}\Pi^{j}) =j_{0}$ for some $j_{0} < e$ (how to get this conclusion? Just becasuce you have some $A_{j}$ with $0$ order and then multiply with $\Pi^{j}$?). Finally this is a contradiction since $\sum_{j=0}^{e-1}A_{j}\Pi^{j}=0$ (why is it a contradiction? Isn't $0$ has $\mathcal{P}$ order less than e?).

Thanks for any kind help you offer!

Answer 1

First question: you have a relation

$$\sum\limits_{i,j} a_{ij} \omega_i \Pi^j =0$$

for some $a_{ij} \in K$, not all zero, and you want to justify the claim that you can multiply this equation by some $c \in K$ so that all of the $ca_{ij}$ are in $\mathcal O_P$ with at least one of these not in $P$. The only way this would not be possible is if all the nonzero $a_{ij}$ had the same valuation. That would mean you could choose a $c \in K$ such that all the nonzero terms $ca_{ij}$ were units in $\mathcal O_P$.

Taking the equation $$\sum\limits_{i=1}^m \sum\limits_{j=0}^{e-1} ca_{ij} \omega_i \Pi^j = 0 \tag{1}$$ modulo $\mathcal P$ yields the equation $\sum\limits_{i=1}^{m} ca_{i0} \omega_i \equiv 0 \pmod{\mathcal P}$. The linear independence of the cosets $\omega_i + \mathcal P$ over the field $\mathcal O_P/P$ implies that all the $ca_{i0}$ are in $P$. Since the nonzero terms $ca_{i0}$ are units in $\mathcal O_P$, this means that all the $ca_{i0}, 1 \leq i \leq m$ must be $0$.

Equation (1) is then

$$\sum\limits_{i=1}^m \sum\limits_{j=1}^{e-1} ca_{ij} \omega_i \Pi^j = 0.$$

Divide this equation by $\Pi$ and reduce modulo $P$ again. The same argument as in the previous paragraph implies that all the terms $ca_{i1}, 1 \leq i \leq m$ are all zero. Iterating this argument shows that all the terms $ca_{ij}$ have to be zero. This contradicts the assumption that not all the $a_{ij}$ were zero.

Second question:

For each $A_j, 0 \leq j \leq e-1$, we have either $\operatorname{ord}_{\mathcal P}(A_j) =0$ or $\operatorname{ord}_{\mathcal P}(A_j) \geq e$ as they explain. Since there is at least one $a_{ij}$ which is not in $P$, this means that there is at least one $j$ for which $\operatorname{ord}_{\mathcal P}(A_j) = 0$.

The usual rule

$$\operatorname{ord}_{\mathcal P}(x+y) \geq \operatorname{Min}\{ \operatorname{ord}_{\mathcal P}(x), \operatorname{ord}_{\mathcal P}(y)\}$$

is written for nonzero $x, y \in L$, but it continues to make sense if we make the convention that $\operatorname{ord}_{\mathcal P}(0) = \infty$. Remember this is an exact equality if $\operatorname{ord}_{\mathcal P}(x) \neq \operatorname{ord}_{\mathcal P}(y)$. Let $S$ be the set of $0 \leq j \leq e-1$ such that $\operatorname{ord}_{\mathcal P}(A_j) = 0$ (we know it is nonempty), and let $T$ be the set of $0 \leq j \leq e-1$ such that $\operatorname{ord}_{\mathcal P}(A_j) \geq e$. If we set

$$X_S = \sum\limits_{j \in S} A_j \Pi^j, X_T = \sum\limits_{j \in T} A_j \Pi^j$$ then we have $$\sum\limits_{j=0}^{e-1} A_j \Pi^j = X_S + X_T = 0.$$

Since all the $\operatorname{ord}_{\mathcal P}(A_j)$ for $j \in S$ are $0$, we see that $\operatorname{ord}_{\mathcal P}(A_j \Pi^j) = j$ for all $j \in S$. Thus all the terms $A_j \Pi^j$ for $j \in S$ have distinct valuations, and so

$$\operatorname{ord}_{\mathcal P}(X_S)= j_0$$ where $j_0$ is the smallest number in $S \subseteq \{0,1, ... , e-1\}$.

Since all the terms $A_j \Pi^j : j \in T$ have order $\geq e$, the element $X_T$ has order $\geq e$. This implies

$$\operatorname{ord}_{\mathcal P}(X_S + X_T) = j_0 < e.$$

In particular, $\sum\limits_{j=0}^{e-1} A_j \Pi^j$ does not have infinite valuation, so it cannot be zero.

Answer 2

$L,K$ are discretely valued field, the valuation $v$ is surjective $L\to \Bbb{Z}\cup \infty$ and $K\to e\Bbb{Z}\cup \infty$.

$\Pi$ is chosen such that $v(\Pi)=1$.

$\omega_i$ are taken in $O_\mathcal{P}$ to be a $O_P/P$ basis of $O_\mathcal{P}/\mathcal{P}$, ie. $m=f$.

The $O_P/P$ independence of the $\omega_i$ gives that $$v(\sum\limits_{i=1}^m a_{ij} \omega_i) = \inf_i v(a_{ij})$$

Thus $v(\sum\limits_{i=1}^m a_{ij} \omega_i \Pi^j) \in j+ e\Bbb{Z}\cup \infty$.

For $v(b)\ne v(c)$, $v(b+c)=\inf( v(b),v(c))$. Thus $$v(\sum_{j=0}^{e-1} \Pi^j\sum\limits_{i=1}^m a_{ij} \omega_i )=\inf_j v(\Pi^j\sum\limits_{i=1}^m a_{ij} \omega_i)=\inf_j (j+\inf_i v(a_{ij}))$$ That it is $\infty$ means that each $v(a_{ij})=\infty$ ie. each $a_{ij}=0$.