Structure sheaf consists of noetherian rings Let $X\subseteq \mathbb{A}^n$ be an affine variety. The ring $k[x_1,\ldots,x_n]$ is noetherian because of Hilbert's basis theorem. 
The coordinate ring $k[X]=k[x_1,\ldots,x_n]/I(X)$ is noetherian because ideals of $k[X]$ are of the form $J/I(X)$, where $J\supseteq I(X)$ is an ideal of $k[x_1,\ldots,x_n]$.
The local ring of $X$ at $p\in X$, given by $\mathcal{O}_{X,p}=\{f \in k(X) : f \text{ regular  at } p\}$ is noetherian because it is a localization of $k[X]$, and the ideals of a ring of fractions $S^{-1}A$ are of the form $S^{-1}J$, where $J$ is an ideal of $A$.
If $U\subseteq X$ is open, let $\mathcal{O}_X(U)=\bigcap_{p\in U}\mathcal{O}_{X,p}$. Is this ring noetherian as well?
 A: There is a counterexample in section 19.11.13 of Ravi Vakil's Foundations of Algebraic Geometry https://math216.wordpress.com/
A: Since any open subscheme of a noetherian scheme is noetherian (Corollary 3.22 in Görtz/Wedhorn), we can reduce to case of global sections.
If $X = \operatorname{Spec} R$ is affine, then $X$ is noetherian if and only if $R$ is noetherian (Prop. 3.19 in Görtz/Wedhorn), hence the section ring of any affine open will be noetherian.
But in general, the answer is negative, see this answer:
Is the global section ring of a Noetherian Scheme Noetherian as well?
A: We can do the following in the local case, but the details of it are left to the reader.
Let $X$ be an irreducible affine variety over the complex numbers. The answer to your question is affirmative because by taking the intersection in the field of fractions $\mathbb{C}(X)$, we have that
$$\mathbb{C}[X]=\bigcap_{x\in X}\mathcal{O}_{x,X}.$$
This says that the local rings of $X$ determine the coordinate ring $\mathbb{C}[X]$.
Now, to see why this is true, one inclusion is clear. The other one goes as follows. Say $f\in \mathcal{O}_{x,X}$ for all $x\in X$. Let us analyze the ideal which lifts $f$ to the coordinate ring, that is $$Z=\{g\in \mathbb{C}[x_0,\ldots , x_n]| \mbox{ if } \overline{g}\in \mathbb{C}[X],\overline{g}.f\in\mathbb{C}[X]\}.$$ Observe that $f$ regular at $x$ means that $f=h_x/g_x$, where $g_x\ne 0$, which tells us that $g_x\in Z$, and therefore $x\notin V(Z)$ for all $x$ (this is an open condition). Note the ideal defining $X$, $I(X)$, is contained in $Z$, and openness tells us $V(Z)=\emptyset$. Now, Nullstellensatz says that $1\in I(V(Z))$, which implies that $f\in \mathbb{C}[X]$.
