Another example: If $X$ is a smooth projective variety over a field $k$, then the relative Picard functor is representable by a smooth group scheme $Pic_{X/k}$. This is the moduli space of line bundles on $X$, in particular $Pic_{X/k}(k)$ (Edit: If $X(k)\neq \emptyset$) is the usual Picard group. The geometry of the Picard scheme tells us a lot about the line bundles; for example one can look at the connected component $Pic_{X/k}^0$ of the identity of $Pic_{X/k}$, and the points $L\in Pic_{X/k}^0(k)$ are precisely the line bundles belonging to divisors algebraically equivalent to $0$.

$\DeclareMathOperator\Pic{Pic}$Another example: If $X$ is a smooth projective variety over a field $k$, then the relative Picard functor is representable by a smooth group scheme $\Pic_{X/k}$. This is the moduli space of line bundles on $X$, in particular $\Pic_{X/k}(k)$ (Edit: If $X(k)\neq \emptyset$) is the usual Picard group. The geometry of the Picard scheme tells us a lot about the line bundles; for example one can look at the connected component $\Pic_{X/k}^0$ of the identity of $\Pic_{X/k}$, and the points $L\in \Pic_{X/k}^0(k)$ are precisely the line bundles belonging to divisors algebraically equivalent to $0$.

Another example: If $X$ is a smooth projective variety over a field $k$, then the relative Picard functor is representable by a smooth group scheme $Pic_{X/k}$. This is the moduli space of line bundles on $X$, in particular $Pic_{X/k}(k)$ is the usual Picard group. The geometry of the Picard scheme tells us a lot about the line bundles; for example one can look at the connected component $Pic_{X/k}^0$ of the identity of $Pic_{X/k}$, and the points $L\in Pic_{X/k}^0(k)$ are precisely the line bundles belonging to divisors algebraically equivalent to $0$

Another example: If $X$ is a smooth projective variety over a field $k$, then the relative Picard functor is representable by a smooth group scheme $Pic_{X/k}$. This is the moduli space of line bundles on $X$, in particular $Pic_{X/k}(k)$ (Edit: If $X(k)\neq \emptyset$) is the usual Picard group. The geometry of the Picard scheme tells us a lot about the line bundles; for example one can look at the connected component $Pic_{X/k}^0$ of the identity of $Pic_{X/k}$, and the points $L\in Pic_{X/k}^0(k)$ are precisely the line bundles belonging to divisors algebraically equivalent to $0$.