TITLE: Structure Sheaf of the Spectrum of a Ring
QUESTION [6 upvotes]: Let $A$ be a ring and $X$ be the spectrum of $A$ with the Zariski topology. For an element $f\in A$ let $X_f:=\{p\subset A\text{ prime ideal }\,|\,f\notin p\}$; the $X_f$ form a basis of the topology on $X$. Finally let $\mathcal{O}$ be the structure sheaf of $X$ (a sheaf or rings).
I have managed to show that the stalk of $\mathcal{O}$ at $\mathfrak{p}\in X$ is isomorphic to to the local ring $A_{\mathfrak{p}}$. What I'm struggling to understand though is the following:

Question 1: For $f\in A$ the ring $\mathcal{O}(X_f)$ is isomorphic to the localized ring $A_f$. Why?

Let $\mathfrak{N}$ be the nilradical of $A$. I know that the spectra of $A$ and $A/\mathfrak{N}$ are homeomorphic, but I have trouble answering this:

Question 2: Are the structure sheaves of the spectra of $A$ and $A/\mathfrak{N}$ the same?

I am very thankful for any hints, references or full out answers.

REPLY [3 votes]: Question 1 is done in Hartshorne, proposition 2.2 on pages 71-72, as rafaelm mentioned.  Dylan is exactly right in the first step of the proof.  Surjectivity is more difficult.
Question 2 is not true as Andrea noted.  For simplicity, take any ring with non-zero nilradical, i.e. a non-reduced ring, for example $A=k[X,Y]/(X^2)$.  Then $A/nil(A)=A/(\bar{X})=k[Y] \neq A$.  Then $\mathcal O_X \neq \mathcal O_{X_{red}}$, where $X=Spec ~A$ and $X_{red}=Spec ~(A/nil(A))$, because $\mathcal O_X(X) = A$ and $\mathcal O_{X_{red}}(X)=k[Y]$.