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Let $X$ be locally Noetherian Scheme, therefore for every $a \in X$ there exist an affine open set $U =\mathrm{Spec}(R)$ such that $R$ noetherian.

Why is then every $S_V:= \Gamma(V, \mathcal{O}_X)$ for each affine open $V$ noetherian?

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I will reduce this into a statement of commutative algebra for you.

It is enough to prove: If $X=\operatorname{Spec}(A)$ is locally noetherian, then $A$ is noetherian. We may cover $X$ by open affines $\operatorname{Spec}(B)$ with $B$ noetherian. Such an affine open may be covered by principal open subsets $D(f)$, with $f\in A$, of $X$. If $\varphi\colon A\to B$ denotes the morphism of rings corresponding to the inclusion $i\colon \operatorname{Spec}(B)\subset \operatorname{Spec}(A)$, then $D(f)=i^{-1}(D(f))=D(\varphi(f))$, i.e. $B_{\varphi(f)}\cong A_f$, so that $A_f$ is noetherian. This shows that we may cover $X$ by $D(f_1),\ldots,D(f_n)$ with $A_{f_i}$ noetherian, in other words: We are given elements $f_1,\ldots,f_n\in A$ generating the unit ideal such that $A_{f_i}$ is noetherian for $i=1,\ldots,n$. Deduce from this that $A$ is noetherian.

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Let $a\in X$ and $V$ be an open affine n'hood of $a$. Since $X$ is locally Noetherian, $V=Spec(R)$ for some Noetherian ring $R$ and $\Gamma(V,\mathcal{O}_{X})\cong R$ (since $V$ is affine) hence it is Noetherian.

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