Let $R = \mathbb{C}[[t]]$ and let $\mathcal{X} \hookrightarrow \mathbb{P}_R^2$ be the arithmetic surface defined by the equation $$(X^2 - 2Y^2 + Z^2)(X^2 - Z^2) + tY^3Z = 0.$$ The generic fiber of $\mathcal{X}$ is a smooth quartic plane curve $C$ - so $\mathcal{X}$ is a model for $C$, but it is not regular. The special fiber of this surface is $$\mathcal{X}_{\mathbb{C}} : (X^2 - 2Y^2 + Z^2)(X^2 - Z^2) = 0.$$ I am trying to blow the surface up at the point $[0,1,0]$ in the special fiber to get a regular model for $C$.

Blowing this point up in the special fiber can be achieved by looking in the affine patch $Y\neq0$ and then blowing up the origin $(X,Z) = (0,0)$. First we can blow up the ambient space at this point to get $$\{(X,Z,[a_0,a_1])\mid a_0Z = a_1X\}\hookrightarrow \mathbb{A}_R^2\times\mathbb{P}_R^1$$ and then look at the patch of $\mathbb{P}^1$ where $a_0\neq0$ which will give $Z = sX$ where we are now using coordinates $[1,s]$ for the affine patch. Restricting to $\mathcal{X}$, we end up with the equation $$X\left[(X^2-2+s^2X^2)(X-s^2X) + ts\right] = 0$$ If we look in the other affine path where $a_1\neq 0$ with coordinates $[r,1]$ we get a similar equation $$Z\left[(r^2Z^2 - 2 + Z^2)(r^2Z - Z) + t\right] = 0$$ These two affine equations should glue to give an arithmetic surface having generic fiber $C$ and special fiber a conic meeting two lines, each at two points. But the special fiber when $a_0\neq 0$ has equation $$X^2(X^2-2+s^2X^2)(1 - s^2) = 0,$$ and when $a_1\neq 0$, the equation is $$Z^2(Z^2 - 2 + r^2Z^2)(r^2 - 1) = 0.$$

I guess the conic is given affine locally by the equation $Z^2-2 + r^2Z^2$ and the two lines are given by $1-s$ and $1+s$? Something still seems wrong to me. Any suggestions are welcome.


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