Blow up of the arithmetic surface $xy = t^e$ Let $R$ be a DVR with the uniformizer $t$. Consider the arithmetic surface:
$$C_e : xy = t^e.$$
I want to repeat the computation of Liu, Algebraic Geometry, Example 3.53, Chapter 8 and compute the special fiber of the minimal regular model of $C_e$. Unfortunately, I have some problems with understanding why the components intersect in the given way.
Suppose that $e \ge 3$. If we blow up $\mathcal C_e$ on $(x, y, t)$, we obtain three charts. One of the charts (say, third) has the equation $C_{e - 2}$, so we can proceed inductively. The special fiber of the blow-up of $C_e$ has four components $F_1^{(e)}, F_2^{(e)}, F_3^{(e)}, F_4^{(e)}$ with the following intersection pattern: 
$$ F_1^{(e)} - F_2^{(e)} - F_3^{(e)} - F_4^{(e)} $$
(components are connected in the above line iff they intersect) where $$F_1^{(e)}, F_4^{(e)}$$ do not pass through the third chart. One easily checks that the strict transform of $F_2^{(e)}$ in the next blow-up is $F_2^{(e-2)}$ and of $F_3^{(e)}$ is $F_3^{(e-2)}$. Also, since $$F_1^{(e)}, F_4^{(e)}$$ do not pass through the third chart, they do not intersect with $F_1^{(i)}$ and $F_4^{(i)}$ for $i < e$. Thus it seems to me that the intersection graph should be of the form:

This is wrong, since (according to Liu) the intersection graph is a path on $(e - 1)$ vertices. Where is my mistake?
 A: I've finally found my mistake. My "naive" way to compute the strict transform for the arithmetic setting was wrong. The problem comes down to the following:

Let $C : xy = t^e$ and let $\pi : \widetilde C \to C$ be the blow-up along $\mathfrak m = (x, y, t)$. Compute the strict transform of $F := F_2^{(e+2)} : x = 0$ (one of the two components of the special fiber).

Recall that $\widetilde C$ has three charts:


*

*$y_1 = t_1^e x^{e - 2}, t = x t_1$ (the components of the special fiber are $G_1: x = y_1 = 0$ and $G_2 : t_1 = y_1 = 0$),

*$x_2 = t_2^e y^{e - 2}, t = y t_2$ (the components of the special fiber are $G_3: y = x_2 = 0$ and $G_4 : t_2 = x_2 = 0$),

*$x_3 y_3 = t^{e - 2}$ (the components of the special fiber are $G_5: x_3 = 0$ and $G_6 : y_3 = 0$).


By using the formulas:
$$y_1 = 1/x_2 = y_3/x_3, \quad t_1 = t_2/x_2 = 1/x_3, \quad x = y x_2 = t x_3,$$
etc. one checks that $G_1, \ldots, G_6$ glue to four components:


*

*$F_1 = F_1^{(e)}$ (made of $G_4$)

*$F_2 = F_2^{(e)}$ (made of $G_5$ and $G_3$)

*$F_3 = F_3^{(e)}$ (made of $G_1$ and $G_6$)

*$F_4 = F_4^{(e)}$ (made of $G_2$)


My (wrong) reasoning: 
on the third chart we have: $F : x_3 t = 0$. To obtain the strict transform, we divide by $t$ to obtain: $\widetilde F : x_3 = 0$. This is $F_2^{(e)}$ in the notation of the question.
The correct reasoning:
let $R[\overline x, \overline y] = R[x, y]/(xy - t^e)$, $R[\overline x, \overline y_1, \overline t_1] = R[y_1, t_1]/(y_1 - t_1^e x^{e - 2}, xt_1 - t)$, etc.


*

*on the first chart the blow-up is given by:
$$\varphi^* : Spec(R[\overline x, \overline y_1, \overline t_1] ) \to Spec(R[\overline x, \overline y] ),$$
where $$\varphi : R[\overline x, \overline y] \to R[\overline x, \overline y_1, \overline t_1], \quad \overline x \mapsto \overline x, \quad \overline y \mapsto xy_1, \quad t \mapsto t_1 x $$
We compute from definition the strict transform. It is the Zariski closure of:
$$ (\varphi^*)^{-1}(V(x, t) \setminus \{\mathfrak m\}) = (\varphi^*)^{-1}(V(x, t)) \setminus (\varphi^*)^{-1}((x, y, t)) = V((x, t_1 x)) \setminus V((x, x y_1, x t_1)) = V((x)) \setminus V((x)) = \varnothing$$
i.e. $\widetilde F$ doesn't pass through the first chart.

*on the second chart, the blow up is induced by:
$$ \varphi : R[\overline x, \overline y] \to R[\overline x_2, \overline y, \overline t_2], \quad \overline x \mapsto \overline x_2 \overline y, \, \overline y \mapsto \overline y, \, t \mapsto \overline t_2 \overline x $$
and $\widetilde F$ is the Zariski closure of:
$$ (\varphi^*)^{-1}(V(x, t) \setminus \{\mathfrak m\}) = (\varphi^*)^{-1}(V(x, t)) \setminus (\varphi^*)^{-1}((x, y, t)) = V((x_2 y, t_2 y)) \setminus \{(x_2 y, y, t_2 y )\} = V((y) (x_2, t_2)) \setminus V((y)) = V((x_2, t_2)) \setminus V((y))$$
i.e. $V((x_2, t_2))$, or in other words $\widetilde F$ on the second chart is given by: $x_2 = t_2 = 0$.

*we check (analogously as in the first chart) that $\widetilde F$ does not pass through the third chart.


Finally, $\widetilde F$ is (in the notation of the problem) $F_1^{(e)}$. 
