This is an arithmetic follow-up to my previous question [Does there exist a non-trivial semi-stable curve of genus >1 with only 4 singular fibres](https://mathoverflow.net/questions/64112/does-there-exist-a-non-trivial-semi-stable-curve-of-genus-1-with-only-4-singular) 


Let $k$ be an algebraically closed field and let $f:X\longrightarrow \mathbf{P}^1\_{k}$ be a semi-stable curve. Let $s$ denote the number of singular fibres. If $X$ is non-isotrivial and of positive genus, we have that $s>2$ (Beauville and Szpiro). As Angelo stated in my previous question, for genus >1 and $k=\mathbf{C}$, Sheng-Li Tan has shown that $s>4$.


Now, let $S=\textrm{Spec} \mathbf{Z}$ and let $X\longrightarrow S$ be a (regular) semi-stable arithmetic surface. Let $s$ be the number of singular fibres. Fontaine has shown that $s>0$ if $X$ is of positive genus. 


**Question.** Let $g>0$ be an integer. Does there exist a semi-stable arithmetic surface $X\longrightarrow S$ of genus $g$ with precisely one singular fibre?


I expect the answer to be yes for $g=1$ but no for $g>1$.


**Example.** The modular curve $X\_1(\ell)$ ($\ell$ big enough) has semi-stable reduction over Spec $\mathbf{Z}[\zeta\_{l}]$. This model has precisely one singular fibre. Note that the modular curve $X\_1(l)$ does not have semi-stable reduction over $\mathbf{Z}$.



The number of singular fibres of a semi-stable arithmetic surface over \Z
I have the opposite intuition -- I would think the answer would be yes for all g. In genus 1, you are asking (I think) whether there are elliptic curves with prime conductor. There are a lot of elliptic curves of prime conductor; I believe the question of whether there are infinitely many is open, and considered hard. 


In higher genus, I would still expect a lot of semistable curves with only one singular fiber; after all, the singular fibers in your question are CLOSED fibers, not GEOMETRIC fibers as in the case of k=C. A better analogy would be curves over P^1\_k where k is a finite field; it is much harder to find a curve whose bad fibers form a subscheme of degree 1 than it is to find a curve whose bad fibers form a subscheme with 1 irreducible component. You are asking for the latter.


Loosely speaking, I think if you write down a hyperelliptic curve y^2 = f(x) and the discriminant of f(x) is prime, you're almost there, maybe with some problems at 2. Are there infinitely many polynomials of a given degree with prime discriminant? Surely yes, though again this is a "Schinzel-type" statement which would be very hard to prove.


