= Solution
An <elliptic curve> works over every <algebraically closed field>, including one algebraic over a finite field. Give it explicitly in $\mathbb P^2$ by
$$
E:\quad
\begin{cases}
y^2z=x^3-xz^2,&\operatorname{char}k\ne2,\\
y^2z+yz^2=x^3,&\operatorname{char}k=2.
\end{cases}
$$
In the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a <singular point of an algebraic variety> would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in <characteristic> different from $2$: at $0$ it is $-1$, and at $\pm1$ it is $2$. In the second case the three homogeneous <partial derivatives> are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are <smooth varieties>. A reducible plane cubic would have intersecting positive-degree components by <Bézout theorem>, and be singular at an intersection; therefore these smooth cubics are <irreducible varieties>. The <genus-degree formula> gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are <elliptic curves>.
For a <smooth algebraic curve>, its <divisor class group> is its <Picard group>. Its degree-zero subgroup is
$$
\operatorname{Pic}^0(E)=\{[D]:\deg D=0\}.
$$
The map $P\mapsto[P-O]$ gives an <isomorphism> of <abelian groups>
$$
E(k)\cong\operatorname{Pic}^0(E).
$$
To recall why it is a bijection, <Riemann-Roch theorem> in <geometric genus> one says every degree-one <divisor class> has exactly one independent section: a <canonical divisor> has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective <divisor on an algebraic curve> of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the <chord-and-tangent group law> on the <elliptic curve>.
Choose a prime integer $\ell$ different from $\operatorname{char}k$. The <multiplication-by-n morphism> $[\ell]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant <morphism of algebraic varieties> between <smooth projective curves> is surjective: its image is closed by <proper morphism> and cannot have <algebraic dimension> zero. Since $k$ is an <algebraically closed field>, this makes $[\ell]$ surjective on $k$-points. Therefore
$$
E(k)=\ell E(k).
$$
Yet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the <algebraically closed field>; since $k$ is infinite, this gives infinitely many points.
By the <Fundamental theorem of finitely generated abelian groups>, a <finitely generated abelian group> $A$ satisfying $A=\ell A$ has no free part and is a finite <torsion group> of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus $\operatorname{Pic}^0(E)$ is not finitely generated. A subgroup of a <finitely generated abelian group> is finitely generated, so
$$
\boxed{\operatorname{Cl}(E)\text{ is not a finitely generated abelian group}.}
$$
This proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable <algebraically closed fields>.
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