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?

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    $\begingroup$ Daniel, yes. You certainly can have $e_i=1$ and an inseparable extension of residue fields--an example can be found here, written very nicely by Brian Conrad: math.stanford.edu/~conrad/248APage/handouts/weirdfield.pdf Now, as for the geometric intuition about the case $e_i=1$, I am not entirely sure there is one--at least not one that will satisfy you. Indeed, in the actual geometric case all residue fields are perfect, and so you can't actually "visualize" it. Are you interested in something analgous to "the existence of positive $f$ means that points are stuck together"? $\endgroup$ Aug 9, 2013 at 4:18
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    $\begingroup$ Also, I forget--does Neukirch explain "why" this is the correct definition of ramification (so that the classificaiton of ramified primes works out nicely)? $\endgroup$ Aug 9, 2013 at 4:34
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    $\begingroup$ Ok, I just think we need to stay calm...and wait for Matthew Emerton. $\endgroup$ Aug 9, 2013 at 4:43
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    $\begingroup$ We need to get him like a Batman signal. $\endgroup$ Aug 9, 2013 at 4:47
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    $\begingroup$ @AlexYoucis: Dear Alex, I saw your comment and posted an answer. Hopefully it helps. Regards, $\endgroup$
    – Matt E
    Aug 9, 2013 at 22:27

1 Answer 1


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.)

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    $\begingroup$ Matt E, thank you for the fantastic response. I am not quite sure if this answers the OP's question, but it only answers half of mine. I already understood, in a much less sophisticated way, why needing both the inseparability of residue fields and trivial ramification index was necessary to "make things work correctly". $\endgroup$ Aug 10, 2013 at 1:19
  • $\begingroup$ ...I was curious if you had some kind of quick, probably inaccurate, way to think visually about ramification that encompasses both non-trivial ramification index or inseparability of residue fields? Clearly thinking of a projection map $\pi:C\to C'$ and some point $p\in C'$ having too few elements of its fiber (so that different strands come together) works to visualize the non-triviality of ramification degree, but doesn't do us anything for non-separable residue extensions. Perhaps it's caught up in the intuition of Etale, which I have none for yet. Regardless, thanks so much! $\endgroup$ Aug 10, 2013 at 1:19
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    $\begingroup$ Thank you very much for the answer; that's pretty much exactly what I was looking for! (The geometric perspective really helped.) Just to make sure my intuitive understanding is correct, does that mean that a ramification point with inseparable residue extension can generally be thought of as having a higher ramification index "hidden behind" the lack of roots in the base field? In other words, does the inseparability degree of the residue extension measure the extent to which the ramification index can increase under base change? $\endgroup$ Aug 10, 2013 at 2:35
  • $\begingroup$ @DanielHast: Dear Daniel, Yes, that is exactly the point. Regards, $\endgroup$
    – Matt E
    Aug 10, 2013 at 3:30
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    $\begingroup$ @AlexYoucis: Dear Alex, If you look at Daniel's comment just above, he does a pretty good job of summarizing my answer in one sentence: an inseparable residue field does correspond to higher ramification index, except that it is "hidden" because the roots of the inseparable polynomial that you need to reveal it aren't available in the base-field; it only manifests itself when you make a sufficiently large extension of the ground field. Regards, $\endgroup$
    – Matt E
    Aug 10, 2013 at 3:32

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