# Different stable degenerations of the hyperelliptic curve $y^2 = x^2(x^2-1)^2 - t^2$

I'm reading a paper, where they consider the hyperelliptic curve $$C : y^2 = x^2(x^2-1)^2 - t^2$$ over the ring $k[[t]]$, where $k$ is a field. Let $K = k((t))$, then they say:

"The stable reduction of $C_K$ is the union of two copies of $\mathbb{P}^1$ intersecting at 3 points"

This seems reasonable to me, since at $t = 0$, you get $y^2 = x^2(x^2-1)^2$, so an affine patch is given by $$\text{Spec }k[x,y]/(y-x(x^2-1)(y+x(x^2-1))$$ which is visibly two genus 0 curves intersecting at the points $(1,0),(0,0),(-1,0)$. However, the usual projectivization of the equation for $C$ yields a curve with a unique point at $\infty$ which is singular, but I assume that when they speak of $C_K$, they must be referring to the smooth model of $C_K$.

Though, I'm not sure how to carry out the computation to see that the two components of the smooth model of $C_K$ do not intersect "at infinity".

Next, they say:

"Another stable degeneration of $C_K$ is the union of two curves of genus 1 intersecting transversally at a point"

This part I don't understand. Of course if we adjoin $\sqrt{y}$, then the same computation as above will yield two nodal cubics (in $x,\sqrt{y}$) of the form $(\sqrt{y})^2 = \pm x(x^2-1)$, but I don't see how we can view this as a degeneration of $C_K$. What am I missing?

• You can compute by hand that (the projective closures of) $y = x(x^2-1)$ and $y=-x(x^2-1)$ don't intersect "at infinity". You can introduce projective coordinates, but you can also eyeball it: as $x \rightarrow \infty$, $y \rightarrow \infty$ in one case but $-\infty$ in the other, so the slopes go to $1$ and $-1$. – user84144 Feb 22 '17 at 2:12
• WHICH PAPER ARE YOU READING?, I need understand!! – Gonzalo Manzano Mar 11 '19 at 14:26

For hyperelliptic curves, it's usually easier to glue together two coordinate charts, rather than take the projective closure and desingularize. The chart with the points at $\infty$ on $y^2=\text{(6th power poly)}$ is obtained by substituting $x\to1/x$ and $y\to y/x^3$, yielding in your case $y^2=(1-x^2)^2-t^2x^6$. The points at $\infty$ are $(0,\pm1)$. The degeneration is $y^2=(1-x^2)^2$, yielding the two conics $y-1+x^2=0$ and $y+1-x^2=0$. One of the conics contains $(0,1)$, the other $(0,-1)$, so they do not intersect at the points at $\infty$.