What does mean by closed points of a smooth proper curve $X$ over a field $k$?
I got this in an article on modular form and Galois representation which is as follows:
If $X$ is a smooth curve over a field $k$, the Picard group $\text{Pic}(X)$ is equal to the divisor class group $$ \text{Coker (div}: k(X)^{\times} \to \bigoplus_{x: \ \text{closed points of X}} \mathbb Z).$$
where $k(X)^{\times}$ is the set of a non-zero rational functions and the divisor $ \text{div}(f)$ is defined to be divisor of $k(X)^{\times}$.
My question:
What does mean by closed points of $X$ here?
Is it the $cocycles$ in cohomology theory i.e, the elements of $\text{Ker}(\delta)$ for coboundary map $\delta$ of cohomology groups?
Can you explain what does mean here by $\text{closed points of X}$ ?
Closed points are easier to describe using the language of schemes.
For a scheme $X$, a closed point on $X$ is simply a point $x \in X$ such that $\{x\}$ is closed under the Zariski topology on $X$.
If $X = \operatorname{Spec}A$ is an affine scheme, then the closed points correspond exactly to the maximal ideals of $A$.
Using the language of varieties, it's slightly complicated to define a closed point. Let $\overline k$ be the algebraic closure of $k$. We then have the set $X(\overline k)$ of $\overline k$-points on $X$. Two points in $X(\overline k)$ is equivalent, if there is an automorphism of $\overline k / k$ which sends one to the other. The set of closed points on $X$ is the set $X(\overline k)$ modulo this equivalence.
As an example, if $X$ is an affine variety and $k[X]$ is the coordinate ring of $X$, then there is a bijection between the set of closed points and the maximal ideals of $k[X]$:
We first define a map from $X(\overline k)$ to the maximal ideals of $k[X]$, by sending any point $x$ to the ideal $\{f \in k[X]: f(x) = 0\}$. It is then clear that equivalent points are sent to the same ideal, hence the map induces a map from the set of closed points to the maximal ideals of $k[X]$.
In the other direction, if $M$ is any maximal ideal of $k[X]$, then the quotient $k[X]/M$ is a finite extension of $k$ and hence there is an embedding $k[X]/M \hookrightarrow \overline k$. The induced map $k[X] \rightarrow \overline k$ then gives a point in $X(\overline k)$.
It is routine to check that the two maps are inverses one to the other, hence are bijections.