Structure sheaf of $\operatorname{Spec}A$

Question

I was trying to do the construction of the structure sheaf of $\operatorname{Spec}A$ by myself when I encountered a difficulty.

To do this I plan to construct the sheaf on a basis of $SpecA$ and then make the canonical extension. The problem comes with the restriction maps. Let $\mathcal{O}_A$ denote the structure sheaf. Then $\mathcal{O}_A(D(f))=A_f$.

If $D(f) \subset D(g)$ I want to find an homomorphism $A \xrightarrow{\varphi} A_f$ sending $g$ to an unity and then use the universal property of the localization to get a morphism between localizations.

My problem is that I just know that $rad(g) \subset rad(f)$ and I think is not enough to find the desired $\varphi$. What can I do?

Answer 1

Just use the obvious map $\varphi(a) = a/1$. Indeed, if $\varphi(g)$ is not a unit in $A_f$, then there is a prime ideal $p$ in $A_f$ containing $g/1$. Its preimage $q = \varphi^{-1}[p]$ is a prime ideal of $A$ which contains $g$ but not $f$ (if $f\in q$, then $f/1\in p$, but $f/1$ is a unit in $A_f$, so $p$ is not prime). This contradicts $D(f)\subseteq D(g)$.

Answer 2

No, I am wrong in my previous comment!

I fix it here: \begin{equation} D(f)\subset D(g)\iff\exists n\in\mathbb{N},h\in A\setminus\{0\}\mid f^n=gh; \end{equation} therefore the restriction morphisms are \begin{equation} \varphi_{gf}:\frac{a}{g^k}\in A_g=\mathcal{O}_{\operatorname{Spec}A}(D(g))\to \frac{ah^k}{f^{kn}}\in A_f=\mathcal{O}_{\operatorname{Spec}A}(D(f)); \end{equation} in particular, for $g=1$ one finds again $\varphi_{1f}=\varphi:A\to A_f$.

If $f\in rad(A)$ then $A_f=\{0\}$ and $\varphi$ is the zero morphism.