Let $R$ be any non-Noetherian ring and $I$ an ideal of $R$ that is not finitely generated. Let $S = R[x, y]/xyI$ viewed as a graded ring with $x$ and $y$ having degree $1$. Then $X = \text{Proj}(S)$ is a projective scheme over $R$. We claim that $H^0(X, \mathcal{O}_ X)$ is not a finitely generated $R$-module.
Decomposing the rings $S_ x$, $S_ y$, and $S_{xy}$ by bidegree, we can write
where the $(a, b)$-summands sit in degree $a + b$ and we have
The $R$-module $H^0(X, \mathcal{O}_ X)$ can be computed as the kernel of the map
defined by $(F, G) \mapsto F - G$. Here the notation $A_0$ refers to the degree $0$ component of a $\mathbf{Z}$-graded ring $A$. By our explicit description of $S_ x, S_ y, S_{xy}$ above, this kernel is isomorphic to the submodule $M \subset R^2$ consisting of those pairs $(F, G)$ with $F - G \in I$. Since the map $M \to I$ of $R$-modules given by $(F, G) \mapsto F - G$ is surjective, $M$ is not a finitely generated $R$-module.
The surjection $R[x, y] \to S$ corresponds to a closed immersion $i : X \to \mathbf{P}^1_ R$. Then $\mathcal{F} = i_*\mathcal{O}_ X$ is a finite type quasi-coherent $\mathcal{O}_{\mathbf{P}^1_ R}$-module whose $H^0$ is equal to the non-finitely generated module $M$ we computed above.
Lemma 110.66.1. Non-finite $H^0$.
There exists a ring $R$ and a projective scheme $X$ over $R$ such that $H^0(X, \mathcal{O}_ X)$ is not a finite $R$-module.
There exists a ring $R$ and a finite type quasi-coherent module $\mathcal{F}$ on $\mathbf{P}^1_ R$ such that $H^0(\mathbf{P}^1_ R, \mathcal{F})$ is not a finite $R$-module.
Proof. See discussion above. $\square$
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