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I'd like to get my hands on some simple examples of families of stable curves. Ideally these would come in the form of a projective curve $C$ over a 1 dimensional base $B$, say $B = \mathbb{A}^1$. The generic fiber would be smooth and the special fiber would be nodal.

For genus 0 there is the nice example of the closure in $\mathbb{P}^2_{\mathbb{A}^1}$ of Spec $k[x,y,t]/(xy - t)$ where $t$ is the coordinate on $\mathbb{A}^1$.

For genus 1 there is the closure in $\mathbb{P}^2_{\mathbb{A}^1}$ of Spec $k[x,y,t]/(y^2 = x(x-t)(x+1))$.

I know for genus $2$ you can't expect to have an example in $\mathbb{P}^2_{\mathbb{A}^1}$ but what about something in $\mathbb{P}^3_{\mathbb{A}^1}$?

Whenever $g = (d-1)(d-2)/2$ I would think you can get an example in $\mathbb{P}^2_{\mathbb{A}^1}$, is there anyway to control the number of nodes you get in the special fiber?

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The fibre over $t=0$ of your curve given by $xy=t$ is the curve $xy=0$. Is this really a stable curve in the usual sense? It seems to consist of two rational lines. I thought stable curves had the property that each rational curve intersects at least three other components. Maybe I'm wrong. –  Harry Jul 21 '12 at 14:44
    
You are correct but this is fixed just by adding marked points. I'm not too concerned with the marked points so I didn't mention it but your point is absolutely right. –  solbap Jul 21 '12 at 16:44
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1 Answer

For curves of genus 2 (or more generally hyperelliptic curves), it is easier to describe them as normalization of $\mathbb P^1$ in a quadratic extension than writting explictely equations in $\mathbb{P}^3$. Let $g\ge 2$. If $X$ is the normalization of $\mathbb P^{1}_{k[t]}$ (with $\mathrm{char}(k)\ne 2$) in the extension $$L:=k(t)(x)[y]/(y^2-(x^2-t)\prod_{1\le i\le 2g}(x-\lambda_i)) $$ where the $\lambda_i\in k^*$ are pairwise distinct, then $X$ is stable, with only one singular fiber (above $t=0$). It can be obtained by glueing two affine open subsets $$ \mathrm{Spec} (k[t][x,y]), \quad \mathrm{Spec} (k[t][1/x, y/x^{g+1}]). $$ In the first chart, $x,y$ are related by the above equation. In the second one, $1/x, y/x^{g+1}$ satisfy the relation $$ (y/x^{g+1})^2=(1-t(1/x)^2)\prod_{1\le i\le 2g}(1-\lambda_i (1/x)).$$ They are respectively the normalization of $\mathrm{Spec}(k[t][x])$ and $\mathrm{Spec}(k[t][1/x])$ in $L$.

For plane curves of degree $d\ge 3$ (in $\mathrm{char}(k)=0$ or $>d$), take the equation $$ (y^2-x^2)(x^{d-2}+y^{d-2}+z^{d-2})+t(x^{d}+y^{d}+z^{d}) $$ which defines a relative curve in $\mathbb P^2_{k[t]}$ over $k[t]$. The generic fiber is smooth by the Jacobian criterion. The fiber above $t=0$ is stable with three irreducible components: two straight lines and a Fermat curve of degree $d-2$. There might be other singular non-stable fibers, I did'nt check.

Edit I forgot the second part of the question (not appearing in the title). The answer in the general case is :

If $X$ is a stable curve of genus $g\ge 2$, then the number of singular points is bounded by $3g-3$. This bound is reached for all $g$.

I don't think being plane curve improves the bound.

To prove this bound, there is either a direct elementary way or a theoretical way. The elementary one is by induction on the number of singular points. If $X$ is irreducible, then the number of singular points in bounded by $g$. Now suppose $X$ is reducible and fix an intersection point $s\in X$. Let $Y$ be the normalization of $X$ at $s$. It consists in separate the two irreducible components $\Gamma_1, \Gamma_2$ meeting at $s$. Then $Y$ has one or two connected components. Each of these connected components might fail to be stable. This happens if $\Gamma_1$ or $\Gamma_2$ are projective lines and meets the other irreducible components at exactly two points (plus $s$). Then we contract such a component, compute the new genus/genera, and the new number of singular points, and we check that $3g-3$ is an acceptable bound.

The bound $3g-3$ is actually reached for any $g\ge 2$. Construct a stable curve with $2g-2$ rational smooth irreducible components $\Gamma_1,\dots, \Gamma_{g-1}$, $\Gamma'_1,\dots, \Gamma'_{g-1}$, and make $\Gamma_i$ intersect $\Gamma'_i$, $\Gamma_{i-1}$ and $\Gamma_{i+1}$ with the convent $\Gamma_{i+g-1}=\Gamma_{i}$. Similarly $\Gamma'_i$ intersects (only) $\Gamma_i$, $\Gamma'_{i\pm 1}$. This construction can cause problem for $g=2$. Then take two projective lines meeting at three points.

Now some ideas for the theoretical proof. Note that $3g-3$ is the dimension of the moduli space of stable curves of genus $g$. Then use the computation of the dimension of deformation spaces as in Deligne-Mumford (The irreducibility of the space of curves of given genus), §1.

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