Here is the definition.
Definition 55.4.1. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. The Picard group of $T$ is the cokernel of the matrix $(a_{ij}/w_ i)$, more precisely where $e_ i$ denotes the $i$th standard basis vector for $\mathbf{Z}^{\oplus n}$.
\[ \mathop{\mathrm{Pic}}\nolimits (T) = \mathop{\mathrm{Coker}}\left( \mathbf{Z}^{\oplus n} \to \mathbf{Z}^{\oplus n},\quad e_ i \mapsto \sum \frac{a_{ij}}{w_ j}e_ j \right) \]
Lemma 55.4.2. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. The Picard group of $T$ is a finitely generated abelian group of rank $1$.
Proof. If $n = 1$, then $A = (a_{ij})$ is the zero matrix and the result is clear. For $n > 1$ the matrix $A$ has rank $n - 1$ by either Lemma 55.2.2 or Lemma 55.2.3. Of course the rank is not affected by scaling the rows by $1/w_ i$. This proves the lemma. $\square$
Lemma 55.4.3. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. Then $\mathop{\mathrm{Pic}}\nolimits (T) \subset \mathop{\mathrm{Coker}}(A)$ where $A = (a_{ij})$.
Proof. Since $\mathop{\mathrm{Pic}}\nolimits (T)$ is the cokernel of $(a_{ij}/w_ i)$ we see that there is a commutative diagram
with exact rows. By the snake lemma we conclude that $\mathop{\mathrm{Pic}}\nolimits (T) \subset \mathop{\mathrm{Coker}}(A)$. $\square$
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 whose cokernel is an elementary abelian $2$-group.
\[ \mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (T') \]
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
and we set
A computation (which we omit) shows there is a commutative diagram
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. $\square$
Lemma 55.4.5. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. If the genus $g$ of $T$ is $\leq 0$, then $\mathop{\mathrm{Pic}}\nolimits (T) = \mathbf{Z}$.
Proof. By induction on $n$. If $n = 1$, then the assertion is clear. If $n > 1$, then $T$ is not minimal by Lemma 55.3.13. After replacing $T$ by an equivalent type we may assume $n$ is a $(-1)$-index. By Lemma 55.4.4 we find $\mathop{\mathrm{Pic}}\nolimits (T) \subset \mathop{\mathrm{Pic}}\nolimits (T')$. By Lemma 55.3.9 we see that the genus of $T'$ is equal to the genus of $T$ and we conclude by induction. $\square$
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