Lemma 55.4.4. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. Assume $n$ is a $(-1)$-index. Let $T'$ be the numerical type constructed in Lemma 55.3.9. There exists an injective map

$\mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (T')$

whose cokernel is an elementary abelian $2$-group.

Proof. Recall that $n' = n - 1$. Let $e_ i$, resp., $e'_ i$ be the $i$th basis vector of $\mathbf{Z}^{\oplus n}$, resp. $\mathbf{Z}^{\oplus n - 1}$. First we denote

$q : \mathbf{Z}^{\oplus n} \to \mathbf{Z}^{\oplus n - 1}, \quad e_ n \mapsto 0\text{ and }e_ i \mapsto e'_ i\text{ for }i \leq n - 1$

and we set

$p : \mathbf{Z}^{\oplus n} \to \mathbf{Z}^{\oplus n - 1},\quad e_ n \mapsto \sum \nolimits _{j = 1}^{n - 1} \frac{a_{nj}}{w'_ j} e'_ j \text{ and } e_ i \mapsto \frac{w_ i}{w'_ i} e'_ i\text{ for }i \leq n - 1$

A computation (which we omit) shows there is a commutative diagram

$\xymatrix{ \mathbf{Z}^{\oplus n} \ar[rr]_{(a_{ij}/w_ i)} \ar[d]_ q & & \mathbf{Z}^{\oplus n} \ar[d]^ p \\ \mathbf{Z}^{\oplus n'} \ar[rr]^{(a'_{ij}/w'_ i)} & & \mathbf{Z}^{\oplus n'} }$

Since the cokernel of the top arrow is $\mathop{\mathrm{Pic}}\nolimits (T)$ and the cokernel of the bottom arrow is $\mathop{\mathrm{Pic}}\nolimits (T')$, we obtain the desired homomorphism of Picard groups. Since $\frac{w_ i}{w'_ i} \in \{ 1, 2\}$ we see that the cokernel of $\mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (T')$ is annihilated by $2$ (because $2e'_ i$ is in the image of $p$ for all $i \leq n - 1$). Finally, we show $\mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (T')$ is injective. Let $L = (l_1, \ldots , l_ n)$ be a representative of an element of $\mathop{\mathrm{Pic}}\nolimits (T)$ mapping to zero in $\mathop{\mathrm{Pic}}\nolimits (T')$. Since $q$ is surjective, a diagram chase shows that we can assume $L$ is in the kernel of $p$. This means that $l_ na_{ni}/w'_ i + l_ iw_ i/w'_ i = 0$, i.e., $l_ i = - a_{ni}/w_ i l_ n$. Thus $L$ is the image of $-l_ ne_ n$ under the map $(a_{ij}/w_ j)$ and the lemma is proved. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).