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



Normal crossings on a surface and ordinary double points
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. 


