Local fields and infinite extensions, basic questions Notation throughout: Let $K$ be a discrete valuation field and $L/K$ an infinite (not necessarily Galois) extension of $K$.
1) How can/does one define a ramification index $e(L/K)$ for $L/K$? It seems that in order to first define an inertia group for $L/K$ one needs that $K$ be at least henselian and that $L/K$ be Galois (or else take a Galois closure in $L$); assuming these hypothesis, what is the definition of the ramification index $e(L/K)$? - is finiteness easy to see? aside: if $\pi_K$ is a uniformizer of $K$, (when) can we explicitly write down the uniformizer of $L$ in terms of $\pi_K$? 
2) Separate question: let $\pi$ be a uniformizer of $K$ (not-necessarily henselian). We know that $1+\pi$ is a unit of $K$; how can one explicitly write down the inverse of $1+\pi$ in $K$? (we can't 'expand' as $K$ is not assumed, e.g., to be complete)
 A: 1) You cannot, in general. The valuation on $L$ need no longer be discrete. For example, consider the extension $\mathbf Q_p(\zeta_{p^\infty})$ obtained by adding all $p$-power roots of $p$ to $\mathbf Q_p$. The only reasonable value for the ramification index of this extension is $e=+\infty$. As for writing down a uniformizer of $L$ in terms of $\pi_K$ : once again, this makes sense only if the valuation on $L$ is discrete. Assuming that it is, one way to do this would be to look at a finite sub-extension $L'/K$ which has the same ramification as the infinite extension $L$, and to find a uniformizer there (assuming you can solve the problem for a finite extension). (However, it is possible to define the ramification group of an infinite Galois extension.)
2) What kind of explicit expression are you looking for? Suppose $K = \mathbf C(z)$ with the discrete valuation at the origin. Then a uniformizer is $\pi = z$, and the inverse of $1+z$ is... $\frac{1}{1+z}$! How would you like to write $\frac{1}{1+z}$ more explicitly than that?
A: Regarding (1), positive examples can be given. For example, in the following situation:
Let $k$ be an imperfect field. For simplicity, let's take $K = k((t))$ so $K$ is a complete discrete valuation field with uniformizer $t$. Now let $k'$ denote the perfect closure of $k$ in (a fixed) $k^{\text{sep}}$ and let $K' = k'((t))$. Then there are finite extensions $L/K'$ with $e(L/K') = e(L/K) <\infty$, although $L/K$ is an infinite extension since $k'/k$ is infinite.
As a concrete example, suppose $k = \mathbb{F}_p((z))$ for $z$ an indeterminate, and take $L/K'$ to be the Artin-Schreier extension given by solving $T^p - T= 1/t$. Then $p = e(L/K') = e(L/K).$ Note that $L$ is defined over $K$. But now consider the element $z':= z^{1/p^N}$ in $k' = \mathbb{F}_p((z^{1/p^{\infty}})) $. Then if $L/K'$ is defined by $T^p - T = z'/t^p$, we still have $p = e(L/K')$ but we encounter problems in trying to define $e(L/K)$ since $L$ is no longer defined over $K$...
