Hi everyone,

I am dealing with a family of curves like f:X---->S, with X and S (and hence f) projective. In fact S, is the projective line but I don't know this helps or not. The problem is that, X is not a smooth surface. It has only one singular point which lies on a singular fiber of f. Now, can I take a semi-stable reduction for this family without desingularizating X? If not, and if g:Y--->X is a resolution of singularities of X, what happens to the singular fibers, especially the one which contains the surface singularity? in particular, does it become semi-stable or not? and more importantly, does the number of singular fibers change?

  • 3
    $\begingroup$ In general, you may need to do a base change to obtain a semistable reduction, so the number of singular fibres can change. Also, in a semistable reduction, you would have to modify the singular fibres anyway, so you shouldn't have to desingularize first. One last comment, please spell check "redcution". $\endgroup$ Apr 6, 2011 at 22:52
  • $\begingroup$ Thank you for your answer. Actually, in my case, I have set the scene such that, the number of singular fibers does not change. But, I asked this question, because in the book [compact complex surfaces, Barth, Peters, Van de Ven , p.95] the stable reduction theorem that they state has the condition that the surface X is smooth. So I was curious to know whether one can still use this theorem in the case X is singular and as I understood from your answer, I can do that. $\endgroup$
    – Cyrus
    Apr 7, 2011 at 20:08

1 Answer 1


At least in the contexts that I am familiar with, a family $X \to S$ with $S$ a smooth curve is called semistable at a closed point $s$ if it has a local model (local in the analytic topology if we are working over $\mathbb C$, or local in the etale topology in general) of the form Spec $k[X_1,\ldots,X_n,t]/(X_1\ldots X_m -t)$ for some $m \leq n,$ where $t$ is a uniformizing paramater of the curves $S$ at $s$.

Note that such a local model is regular, and so if $X$ is semistable at every point of $S$ then it is smooth.

I believe that semistability in this sense can also be characterized as follows: (i) $X$ is regular; (ii) the fibres of $X$ over points of $S$ are reduced normal crossings divisors.

This is presumably why it is required that $X$ is smooth in your context; if $X$ were not smooth, and $Y$ was a "semistable reduction" for $X$ which didn't resolve the singularities of $X$, then $Y$ could not actually be semistable in the usual sense.

A good example to think about, in the context of a family of curves, is a singularity of the following kind: Spec $k[x,y,t]/(x y = t^{n+1})$. This has a reduced normal crossings divisor over $t = 0$, but is not regular if $n > 0$. It is easy to resolve the singularity, though: just repeatedly blow-up the singular point. After you do this $n$ times, you will resolve the singularity, and you will get a reduced normal crossings divisor with $n + 2$ components: the original two components, plus $n$ components (each a $\mathbb P^1$) obtained from the blow-ups.

[Added to answer a question in the comments: If $X$ is singular, say with isolated singularities, with all the singular points lying in the singular fibres, then when we resolve the singularities of $X$, $X$ does not change in the complement of its singular points, and the singular points get replaced by various exceptional divisors. So no singular fibres will be added, but the component structure of the singular fibres will be changed.]

  • $\begingroup$ Thank you very much. Actually, when I take the semistable reduction, then I will have to normalize the fiber product (the pullback of the map f by the finite base change) and then desingularize that, so ultimately, the Y that I get is smooth. Just one more question which is probably very naive. If I resolve the singularities of X, could it change the number of the singular fibers?(Remember that the singularity of the surface X lies in a singular fiber of f and is in fact in the singular point of that fiber ) $\endgroup$
    – Cyrus
    Apr 8, 2011 at 12:12
  • $\begingroup$ Dear Cyrus, The number of singular fibres won't change (because $X$ doesn't change outside its singular locus when you resolve its singularities), but the component structure of the singular fibres will change. (I added a paragraph to this effect in my answer.) Regards, Matthew $\endgroup$
    – Emerton
    Apr 8, 2011 at 12:35
  • $\begingroup$ Thank you very much again. Your answer was very illuminating for me. $\endgroup$
    – Cyrus
    Apr 8, 2011 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.