Lemma 54.6.4. Let $S, s_ i, S_ i$ be as in (54.6.0.1) and assume $f : X \to S$ corresponds to $g_ i : Y_ i \to S_ i$ under $F$. Then there exists a factorization
\[ X = Z_ m \to Z_{m - 1} \to \ldots \to Z_1 \to Z_0 = S \]
of $f$ where $Z_{j + 1} \to Z_ j$ is the blowing up of $Z_ j$ at a closed point $z_ j$ lying over $\{ s_1, \ldots , s_ n\} $ if and only if for each $i$ there exists a factorization
\[ Y_ i = Z_{i, m_ i} \to Z_{i, m_ i - 1} \to \ldots \to Z_{i, 1} \to Z_{i, 0} = S_ i \]
of $g_ i$ where $Z_{i, j + 1} \to Z_{i, j}$ is the blowing up of $Z_{i, j}$ at a closed point $z_{i, j}$ lying over $s_ i$.
Proof.
Let's start with a sequence of blowups $Z_ m \to Z_{m - 1} \to \ldots \to Z_1 \to Z_0 = S$. The first morphism $Z_1 \to S$ is given by blowing up one of the $s_ i$, say $s_1$. Applying $F$ to $Z_1 \to S$ we find a blowup $Z_{1, 1} \to S_1$ at $s_1$ is the blowing up at $s_1$ and otherwise $Z_{i, 0} = S_ i$ for $i > 1$. In the next step, we either blow up one of the $s_ i$, $i \geq 2$ on $Z_1$ or we pick a closed point $z_1$ of the fibre of $Z_1 \to S$ over $s_1$. In the first case it is clear what to do and in the second case we use that $(Z_1)_{s_1} \cong (Z_{1, 1})_{s_1}$ (Lemma 54.6.3) to get a closed point $z_{1, 1} \in Z_{1, 1}$ corresponding to $z_1$. Then we set $Z_{1, 2} \to Z_{1, 1}$ equal to the blowing up in $z_{1, 1}$. Continuing in this manner we construct the factorizations of each $g_ i$.
Conversely, given sequences of blowups $Z_{i, m_ i} \to Z_{i, m_ i - 1} \to \ldots \to Z_{i, 1} \to Z_{i, 0} = S_ i$ we construct the sequence of blowing ups of $S$ in exactly the same manner.
$\square$
Comments (0)