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Let $R$ be a discrete valuation ring with residue field $F$, and let $X/R$ be a regular projective relative curve with closed fiber $X_0$. Then $X_0/F$ is a projective curve. We know that $X$ may be transformed by blow-up into a regular proj. rel. curve for which $X_0$ has normal crossings. As I understand it, the reduced structure $C=(X_0)_{red}$ has regular irreducible components, at most two of which meet at any point, and when they meet, they meet transversally. My questions: (1) are the singular points of $C$ necessarily ordinary double points? (2) If not, does the answer change if $F$ is perfect? I'm not an algebraic/arithmetic geometer, and I'm trying to read a paper by Saito (Class field theory for curves over local fields, JNT 21 (1985), 44-80). Thank you in advance for any comments/references.

EricB

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    $\begingroup$ Now all we need is for Rakim to show up, then this party'll be paid in full; ya know what I'm sayin'? $\endgroup$
    – Harry Gindi
    Commented Feb 22, 2011 at 2:46

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For the question (1), it depends on the definition of ordinary double point. If you use that of Deligne-Mumford or Bosch-Lütkebohmert-Raynaud ($C$ at the singular point is isomorphic to $\mathrm{Spec}(F[x,y]/(xy))$ for the étale topology) then it implies that the residue field at the singular point is separable over $F$. In general embedded resolution of singularities gives you regular (not necessary smooth) irreducible component, and the residue field at the singular points of $C$ may be inseparable over $F$, even when the generic fiber is smooth over $\mathrm{Frac}(R)$.

Example: take $R=F[[t]]$ and use a local equation $y^2=x^p-\lambda+tx$ with $p=\mathrm{char}(F)>0$ and $\lambda\in F$ is not a $p$-th power in $F$. You get an irreducible component which is regular but not smooth. Then you can blow-up as many times as you want, the stricit transform of this component will be finite and birational (hence isomorphic) to the initial one, so you never get a smooth component. Similar constructions can be done for singular points in $C$. If you have a singular point in $C$ with inseparable residue field, then blowing-up this point still gives inseparable points.

(2) If $F$ is perfect, then yes, everything agree.

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Normal crossing singularities of a reduced curve are ordinary double points by definition. If you allow base change, then you can even ask that $X_0$ be reduced.

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That would be great, and thanks for the quick reply. Let me make sure I understand: Say $X_0$ is reduced, and $x$ is a singular point. The definition I have for normal crossings would say that $X_0$ would be cut out at $x$ by a regular system of parameters $\{f,g\}$ for the local ring $O_{X,x}$. If $x$ is an ordinary double point for $X_0/F$, then (by the definition I have) the completion at $x$ looks like $F[[\bar f,\bar g]]/(\bar f\bar g)$. How do you get from one to the other? For example if a uniformizer $t$ for $R$ is not prime in $O_{X,x}$, the structure theorem for complete regular local rings is not very explicit.

I thought it was an interesting question because the normal crossings definition seems to rely on the ambient scheme, whereas "double point" seems more intrinsic to the curve.

EricB

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  • $\begingroup$ EricB, in terms of the definition for normal crossings, I often see simple normal crossings for the definition you gave, if it holds at all points, to avoid this ambiguity. In particular, the definition you gave rules out the nodal curve (in a smooth ambient space) having normal crossings at the node point. Although locally analytically, it certainly does have normal crossings (via your definition). $\endgroup$
    – Karl Schwede
    Commented Feb 22, 2011 at 2:20
  • $\begingroup$ I think it depends on how you define normal crossings though. Another definition I've seen is that two curves meeting at a point $x \in X$ having normal crossings means that the two curves are smooth at $x$ and cross at $x$ transversally, and that point $x$ is a regular point (or maybe smooth point depending on the context) of $X$. Of course, if $X$ is not smooth at a point $x$, you could still have two smooth curves crossing transversally at that point (for example, the two rulings of a quadric cone), but it wouldn't be called normal crossings. $\endgroup$
    – Karl Schwede
    Commented Feb 22, 2011 at 2:23
  • $\begingroup$ Thanks for the comment Karl. I'm using a definition in Grothendieck-Murre (Tamely ramified covers...), which is essentially compatible with one in the AG book by Q. Liu. It does not speak of smooth, I believe, only regular. I guess the point I'm interested in is whether Lipman's embedded resolution theorem can produce a relative curve such that $(X_0)_red$ has only ordinary double points for singularities, hang the normal crossings! Especially About your other comment, I think I've seen that definition, but isn't there the same problem in either case to translate to an ordinary double point? $\endgroup$
    – EricB
    Commented Feb 22, 2011 at 2:27
  • $\begingroup$ Eric, I almost made the same remark as Karl regarding normal crossings versus simple normal crossings. The difference is whether you allow self-intersections of the irreducible components or not. As far as the singularities are concerned, they are ordinary double points in each case. In other words, normal crossings only asks for a certain behaviour in an analytic or formal neighbourhood while simple normal crossings ask for that in a Zarsiki neighbourhood. As far as ordinary double points are concerned, there is no difference. $\endgroup$
    – Sándor Kovács
    Commented Feb 22, 2011 at 3:03
  • $\begingroup$ The last statement in my previous comment follows from the fact that if the completion of a local ring is isomorphic to the completion of the local ring of an ordinary double point of a curve, then the original local ring was the local ring of an ordinary double point. This is an exercise somewhere near the end of the first chapter of Hartshorne. $\endgroup$
    – Sándor Kovács
    Commented Feb 22, 2011 at 3:05

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