MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a general point of confusion for me, which is why the question is going to be a bit vague. Think of the main question as: What is the right way to think about this?

I will present the question in the context I was thinking of, but it may be that the ideas are more general.

Let's say we have a morphism $X \rightarrow Y$ of relative curves over a complete DVR, $R$. Let's also assume that $Y$ is a nodal curve (meaning its closed fiber is nodal). In fact, for simplicity, assume that $Y$'s closed fiber is made up of exactly two irreducible components meeting at a node. There are many ways of blowing that "point" up: if the node is given by $xy=t^n$, we may blow up at $(x,t)$, $(x,t^2)$, and so forth.

Now, it may be that $X$ is not nodal. What I would like to compare is blowing up at all the pre-images of the node in $X$, and blowing up at the node in $X$ and then normalizing.

share|cite|improve this question
Do you mean blow up at $(x,y,t)$, $(x,y,t^2)$, etc., or perhaps $(x,y,t)$, $(x,y,t)^2$, etc.? (Presumably your curves have smooth generic fiber?) – BCnrd Jun 12 '10 at 18:32
I believe I mean the former. – Makhalan Duff Jun 12 '10 at 19:07
Sorry to resurrect this, but I'm confused. I'm assuming that the DVR is k[[t]], and Y is maybe the family of curves that has a chart that looks like $xy = t^n$. When you say blow up $(x, t)$ you are intending to blow up one irreducible component of the closed fiber? More specifically, when you say "the point", can you say which point you mean? – Karl Schwede Jul 11 '10 at 19:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.