The Stacks project

Proof. Observe that $n > 1$ for example by Lemma 55.3.5 and hence $n' \geq 1$. We check conditions (1) – (5) of Definition 55.3.1 for $n', m'_ i, a'_{ij}, w'_ i, g'_ i$.

Condition (1) is immediate.

Condition (2). Symmetry of $A' = (a'_{ij})$ is immediate and since $a_{nn} < 0$ by Lemma 55.3.6 we see that $a'_{ij} \geq a_{ij} \geq 0$ if $i \not= j$.

Condition (3). Suppose that $I \subset \{ 1, \ldots , n - 1\}$ such that $a'_{ii'} = 0$ for $i \in I$ and $i' \in \{ 1, \ldots , n - 1\} \setminus I$. Then we see that for each $i \in I$ and $i' \in I'$ we have $a_{in}a_{i'n} = 0$. Thus either $a_{in} = 0$ for all $i \in I$ and $I \subset \{ 1, \ldots , n\}$ is a contradiction for property (3) for $T$, or $a_{i'n} = 0$ for all $i' \in \{ 1, \ldots , n - 1\} \setminus I$ and $I \cup \{ n\} \subset \{ 1, \ldots , n\}$ is a contradiction for property (3) of $T$. Hence (3) holds for $T'$.

Condition (4). We compute

as desired.

Condition (5). We have to show that $w'_ i$ divides $a_{in}a_{jn}/a_{nn}$. This is clear because $a_{nn} = -w_ n$ and $w_ n | a_{jn}$ and $w_ i | a_{in}$.

To show that $g = g'$ we first write

Comparing with the expression for $g'$ we see that it suffices if

for $i \leq n - 1$. In other words, we have

It is elementary to check that this is an integer $\geq 0$ if we choose $w'_ i$ as in (4). $\square$