# Structure Sheaf of the Spectrum of a Ring

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.

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The second question has a negative answer because if $X = \mathrm{Spec} A$ then $\mathcal{O}_X(X) = A$. –  Andrea Feb 9 '12 at 20:04
Well, let's at least get started: define a map $A_f \to \mathscr O(X_f)$ by sending $a/f^n$ to the function on $X_f$ sending $\mathfrak p$ to the image of $a/f^n$ in $A_\mathfrak p$. Injectivity shouldn't be so bad. Surjectivity is more finicky. –  Dylan Moreland Feb 9 '12 at 20:10
It is prooved (questin 2) in chapter 2 in Hartshorne's Algebraic geometry. –  rafaelm Feb 9 '12 at 21:01
Maybe one more comment: We have an (anti-)equivalence between the category of commutative rings (with unity) and the category of affine schemes (as is proven in any scheme-theoretic treatise on algebraic geometry), and so a positive answer to Question 2 would contradict this equivalence. –  M Turgeon Feb 10 '12 at 18:57
In some texts the structure sheaf is defined to be the the localized ring $A_f$ on $X_f$. This is done in Görtz& Wedhorn for example. –  Carsten Feb 10 '12 at 19:05

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 $\mathcal O_{X_{red}}(X)=k[Y]$.