Let \(A\) be a ring. We have corresponding topological space \(X=\text{Spec}(A)\), the collection of all prime ideals of \(A\) with Zariski Topology. We now define a sheaf on \(X\) called the structure sheaf, denoted by \(\mathcal{O}_X\). This \(X\) with this structure sheaf \(\mathcal{O}_X\) is called an affine scheme, These affine schemes are building blocks of what is called an arbitrary scheme.
To define a sheaf on \(X\) we need to associate a ring for each \(U\) open in \(X\). We do that as follows :
where the condition \(\dagger\) says that given \(p\in U\) we have \(s(p)\in A_p\) and that \(s\) is locally a fraction i.e., given \(p\in U\) there exists an open set \(U(p)\subseteq U\) and \(a\in A, f\in A\) such that \(s(q)=\frac{a}{f}\in A_q\) for all \(q\in U(p)\).
The verification that this gives a sheaf on \(X\) is same as that of the verification that sheafification of a sheaf is a sheaf. We can see the similarity between the definitions.
More details can be found here about the similarity. So, \((X,\mathcal{O}_X)\) forms a ringed space, which we call an affine scheme.
We will now see results about the global sections, stalks of structure sheaf and what does structure sheaf give on basic open subsets of \(X=\text{Spec}(A)\).
Proposition : Let \(A\) be a ring, and \((\text{Spec A},\mathcal{O})\) its spectrum.
- For any \(\mathfrak{p}\in \text{Spec A}\), the stalk \(\mathcal{O}_{\mathfrak{p}}\) of the sheaf \(\mathcal{O}_{}\) is isomorphic to the local ring \(A_{\mathfrak{p}}\) i.e., \(\mathcal{O}_{\mathfrak{p}}\cong A_{\mathfrak{p}}\).
- For any element \(f\in A\), the ring \(\mathcal{O}(D(f))\) is isomorphic to the localized ring \(A_f\) i.e., \(\mathcal{O}(D(f))\cong A_f\).
- In particular, \(\Gamma(\text{Spec A}, \mathcal{O})\cong A\).
Proof : Let \(\mathfrak{p}\in X\). We define a map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\) and show that this is a bijection.
Defining the map - Let \([(U,s)]\in \mathcal{O}_{\mathfrak{p}}\) i.e., \(U\) is an open set in \(X\) containing \(p\) and \(s\in \mathcal{O}(U)\). By definition, \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\). To get an element in \(A_{\mathfrak{p}}\) given \(s\), its only natural to consider image of \(\mathfrak{p}\) under \(s\) namely \(s(\mathfrak{p})\in A\). Defining \(s\mapsto s(\mathfrak{p})\) gives a map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\).
Showing that the map is well defined - Suppose \([(U,s)]=[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\) i.e., there is an open set \(W\subset U\cap V\) containing \(\mathfrak{p}\) such that \(s|_{W}=t|_W\). As \(\mathfrak{p}\in W\), we have in particular \(s(\mathfrak{p})=t(\mathfrak{p})\). So, there is a well defined map \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\).
Showing that the map is Injective - For \([(U,s)],[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\) with \(s(\mathfrak{p})=t(\mathfrak{p})\), we show that \([(U,s)]=[(V,t)]\in \mathcal{O}_{\mathfrak{p}}\).
As \(\mathfrak{p}\in U\), for \(s: U\rightarrow\bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) there exists open \(U(\mathfrak{p})\subset U\) containing \(\mathfrak{p}\) and \(a,f\in A\) such that \(s(\mathfrak{q})=\frac{a}{f}\) for all \(\mathfrak{q}\in U(\mathfrak{p})\). Similarly, for \(t:V\rightarrow\bigsqcup_{\mathfrak{q}\in V}A_{\mathfrak{q}}\) there exists open \(V(\mathfrak{q})\subseteq V\) and \(b,g\in A\) such that \(t(\mathfrak{q})=\frac{b}{g}\) for all \(\mathfrak{q}\in V(\mathfrak{p})\). In particular, \(\frac{a} {f}=s(\mathfrak{p})=t(\mathfrak{p})=\frac{b}{g}\).
Let \(\mathfrak{q}\in U(\mathfrak{p})\cap V(\mathfrak{p})\). Then, \(s(q)=\frac{a}{f}=\frac{b}{g}=t(q)\). Thus, we have \(s|_{U(\mathfrak{p})\cap V(\mathfrak{q})}=t|_{U(\mathfrak{p})\cap V(\mathfrak{q})}\). Thus, \([(U,s)]=[(V,t)]\). So, \(s\mapsto s(\mathfrak{p})\) is injective.
Showing that the map is surjective - Let \(\frac{a}{f}\in A_{\mathfrak{p}}\), we want to choose an open set \(U\) containing \(\mathfrak{p}\) and \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) such that \(s(\mathfrak{p})=\frac{a}{f}\). One choice for \(s\) is sending \(q\) to image of \(\frac{a}{f}\) in \(A_{\mathfrak{q}}\). For this, we need \(f\notin\mathfrak{q}\) i.e., \(\mathfrak{q}\in D(f)\). Let \(U=D(f)\) and consider \(s:U\rightarrow \bigsqcup_{\mathfrak{q}\in U}A_{\mathfrak{q}}\) sending \(\mathfrak{q}\) to image of \(\frac{a}{f}\) in \(A_{\mathfrak{q}}\). We then have \(s(\mathfrak{p})=\frac{a}{f}\in A_{\mathfrak{p}}\). Thus, the map is surjective.
So, we have isomorphism \(\mathcal{O}_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}\) given by \(s\mapsto s(\mathfrak{p})\).
Let \(f\in A\). We define a map \(A_f\rightarrow \mathcal{O}(D(f))\) and show that this is a bijection.
Defining the map - Given \(\frac{a}{f}\in A_f\) we assign \(s\in \mathcal{O}(D(f))\) where \(s:D(f)\rightarrow \bigsqcup_{q\in D(f)}A_q\). Let \(q\in D(f)\) then, \(f\notin q\). So, \(\frac{a}{f}\) is defined in \(A_q\). So, define \(s(q)\) to be the image of \(\frac{a}{f}\) in \(A_q\) for each \(q\in D(f)\). It is clearly a well defined function. Similarly we define for \(\frac{a}{f^n}\in A_f\) a map \(s:D(f^n)=D(f)\rightarrow \bigsqcup_{q\in D(f)}A_q\) as \(q\mapsto \frac{a}{f^n}\in A_q\).
Showing that the map is injective - Suppose \(\frac{a}{f^n},\frac{b}{f^m}\in A_f\) is such that the corresponding maps \(s,t\) are equal i.e., \(\frac{a}{f^n}=\frac{b}{f^m}\in A_q~\forall q\in D(f)\) i.e., given \(q\in D(f)\) there exists \(t_q\notin q\) such that \(t_q(af^m-bf^n)=0\).
Consider the case when \(D(f)=\{q\}\). As \(t\notin q\), we have \(q\in D(t)\) i.e., \(D(f)\subseteq D(t)\) i.e., \(V(t)\subseteq V(f)\) i.e., \(\sqrt{(f)}\subseteq \sqrt{(t)}\). As \(f\in \sqrt{(f)}\) we have \(f^l=td\) for some \(d\in A\). We have \(t(af^m-bf^n)=0\) which implies \(td(af^m-bf^n)=0\) i.e., \(f^l(af^m-bf^n)=0\) i.e., \(\frac{a}{f^n}=\frac{b}{f^m}\in A_f\) and we are done.
Suppose \(D(f)=\{q_i\}_{i\in \Lambda}\). As \(t_i\notin q_i\) we have \(q_i\in D(t_i)\) i.e., \(D(f)\subseteq \bigcup_{i\in \Lambda} D(t_i)\). As in previous observation, this means \(f^l\) is in the ideal generated by \(\{t_i\}\) for some \(l\in \mathbb{N}\). So, we have (after rearranging indices in \(\Lambda\)) \(f^l=a_1t_1+\cdots+a_nt_n\) for some \(a_i\in A\). As \(t_i(af^m-bf^n)=0\), we have \(a_it_i(af^m-bf^n)=0\) for all \(i\). So, \(\sum_{i=1}^na_it_i(af^m-bf^n)=0\) i.e., \(f^l(af^m-bf^n)=0\). Thus, \(\frac{a}{f^n}=\frac{b}{f^m}\in A_f\). Thus, the map \(A_f\rightarrow \mathcal{O}(D(f))\) is injective.