Primes of ramification index 1 with inseparable residue field extension I've been reading through Neukirch's Algebraic Number Theory, and I'm a little puzzled about a possibility with ramification of primes.
As usual, let $\mathcal{O}_K$ be a Dedekind domain with field of fractions $K$, let $L/K$ be a finite algebraic extension, and let $\mathcal{O}_L$ be the algebraic closure of $\mathcal{O}_K$ in $L$. Let $\mathfrak{p}$ be a prime of $\mathcal{O}_K$, and let $\mathfrak{p} = \prod_{i=1}^{r} \mathfrak{P}_i^{e_i}$ be its prime factorization in $\mathcal{O}_L$.
Neukirch defines $\mathfrak{P}_i$ to be unramified over $\mathcal{O}_K$ provided that $e_i = 1$ and the residue field $\mathcal{O}_L/\mathfrak{P}_i$ is separable over $\mathcal{O}_K/\mathfrak{p}$. It's the latter condition that has me wondering: What if $e_i = 1$, but the residue field extension is inseparable (perhaps even purely inseparable)? In other words, can a prime be ramified despite having ramification index 1? The definition suggests it can, but no examples readily come to mind, probably because most of the examples I've worked with are perfect fields.
Also, is there some good geometric way of thinking about what it means for a point to be ramified with ramification index 1? I usually think of ramification points as "points with multiplicity" or something along those lines, but that doesn't make sense if $e_i = 1$. Is there an example — ideally, a reasonably natural one — that illustrates this phenomenon?
 A: Let's work locally at the prime $\mathfrak p$ (or complete at $\mathfrak p$ if you prefer), and so assume $\mathcal O_K$ is a DVR with maximal ideal $\mathfrak p$.  Also, let's assume that $\mathcal O_L = \mathcal O_K[\alpha]$ for some $\alpha$.  (Even if we pass to the completion at $\mathfrak p$  this isn't always true in the inseparable residue field context, because it's not always true that inseparable field extensions are simple, but it makes things more transparent if we assume this.)
If $f(x) \in \mathcal O_K[x]$ is the minimal polynomial of $\alpha$, 
then we can write $\mathcal O_L = \mathcal O_K[x]/(f(x)) .$
Now unramified means that this extension should be etale in the sense of commutative algebra/algebraic geometry, i.e. that the derivative $f'(x)$ should be invertible in the quotient $\mathcal O_L$.  (Actually etale equals unramified plus flat, but flatness of $\mathcal O_L$ over $\mathcal O_K$ is automatic in our setting; so don't worry if this doesn't mean anything to you!) Now being invertible in a DVR just means that $f'(x)$ is non-zero in the residue field, i.e. that $k[X]/(f(X))$ is a product of separable field extensions if, where $k = \mathcal O_K/\mathfrak p$, i.e. that $\mathcal O_L/\mathfrak p$ is a product of separable field extensions of $k$.  In other  words, $\mathcal O_L$ is unramified over $\mathcal O_K$ precisely if $\mathfrak p$ factors into a product of primes in $\mathcal O_L$, each appearing with multiplicity one, and each having a residue field that is separable over $k$.

For a geometric perspective, note that a non-separable residue field extension in the context of curves over a field will turn into multiplcity if we extend scalars.
E.g. consider the field $k = \mathbb F_p(T)$, and the map $\phi: \mathbb A^1_{/k} \to \mathbb A^1_{/k} $ defined by $X \mapsto T^{-1}X^p.$  (Here $X$ is the coordinate on $\mathbb A^1$; and remember that $T$ is an element of the ground field $k$.)  This gives the extension of Dedekind domains (and $k$-algebras) 
$$\mathcal O_K :=k[Y] \subset k[Y][X]/(X^p - TY)= k[X] =: \mathcal O_L.$$
If we take $\mathfrak p = (Y)$, corresponding to the point $0$ in $\mathbb A^1$,
then it factors as $(Y) =(TY)$ (remember $T$ is a unit) $= (X)^p$, so the extension of residue fields is trivial (just $k$ over $k$), while $e = p$.
On the other hand, if we look over the point $(Y -1)$, corresponding to the    points $1$ in $\mathbb A^1$, there is a unique prime lying over it, namely $(X^p - T)$.  So $e = 1$, but the residue field extension is non-trivial, and non-separated (it is $K(T^{1/p})$ over $k$.) 
Now let's extend sclaras from $k$ to the algebraic closure of $k$ (or even just to $l := k(T^{1/p})$).  Then our extension of Dedekind domains becomes 
$$l[Y] \subset l[Y][X]/(X^p - TY) = l[Y][X] / (X - T^{1/p}Y)^p = l[X].$$ Above $0$ the fibre doesn't change, we still have $e = p$ and a trivial residue field extensions, but above $1$, we now have $(Y-1) = (X - T^{1/p})^p$, i.e. $e = p$ and a trivial extension of residue fields.
Our original map $\phi$ was not the Frobenius map, but was a twisted form of Frobenius (twisted by $T$).  After extending scalars to $l$, we could untwist, and our map just becomes the Frobenius (after a change of coordinates to $T^{1/p}X$ on the source).  So after this base-change, we have $e = p$ above every point/prime in the target; but before it, we didn't have $e = p$ in general --- rather we had $e = 1$ but an inseparable residue field extension of degree $p$.
So the notion of $e = 1$ is not stable under extension of scalars (when we are in the geometric context of curves over fields), but the notion of unramified   ($e = 1$ and separable residue field extension) is stable under extension of scalars.  Typically, properties that are separable under extension of scalars are better behaved --- of course, tautologically they behave better under extension of scalars (!), but this has a conceptual meaning: it means that they are capturing a true aspect of the underlying geometry, rather than some more transient phenomena that just has to do with us working over perhaps a  field of scalars that is too small to reveal all the geometric phenomena in play.
Here I am restricting my discussion of extension of scalars to the setting of DVRs that come from curves over a field, because in the number theory setting (for instance) it is not so obvious how to make extensions of scalars that keep us inside the realm of standard algebraic number theory (DVRs inside finite extensions of $\mathbb Q$ or $\mathbb Q_p$).  However, when defined properly, the concepts of unramified and etale make sense for any extension (or, more generally, morphism) of rings (or even more generally, morphism of schemes), and then both are stable under arbitrary base-change.  
(In the formulation I gave above, writing $\mathcal O_L := \mathcal O_K[x]/(f(x))$ with $f'(x)$ being a unit in $\mathcal O_L$, this perservation under arbitrary base-change $\mathcal O_K \to A$ is pretty clear.)
