sheaves

DEFINITION 1.19   Let $X$ be a topological space. A presheaf $\mathcal F$ of rings over $X$ is called a sheaf if for any open set $U\subset X$ and for any open covering $\{U_\lambda\}_{\lambda \in \Lambda}$ of $U$ , it satisfies the following conditions.

1. (``Locality'') If there is given a local section $f,g \in \mathcal F(U)$ such that
   $$\rho_{U_\lambda U}(f) =\rho_{U_\lambda U}(g)$$
   holds for all $\lambda \in \Lambda$ , then we have $f=g$
2. (``Gluing lemma''). If there is given a collection of sections $\{ f_\lambda \}_{\lambda \in \Lambda}$ such that
   $$\rho_{U_{\lambda\mu} U_\lambda }(f_\lambda) = \rho_{U_{\lambda\mu} U_\mu }(f_\mu)$$
   holds for any pair $(\lambda,\mu)\in \Lambda^2$ , then we have a section $f\in \mathcal F(U)$ such that
   $$\rho_{U_\lambda U }(f)=f_\lambda$$
   holds for all $\lambda \in \Lambda$ .

We may similarly define sheaf of sets, sheaf of modules, etc.

LEMMA 1.20   Let $X$ be a topological set with an open base $\mathfrak{U}$ . To define a sheaf $\mathcal F$ over $X$ we only need to define $\mathcal F(U)$ for every member $U$ of $\mathfrak{U}$ and check the sheaf axiom for open bases. In precise, given such data, we may always find a unique sheaf $\mathcal G$ on $X$ such that $G(U)\cong F(U)$ holds in a natural way. (That means, the isomorphism commutes with restrictions wherever they are defined.)

PROOF.. Let $\mathcal F$ be such. For any open set $U\subset X$ , we define a presheaf $\mathcal G$ by the following formula.

$$\mathcal G(U)= \left \{(s_V)\in \prod_{V\in \mathfrak{U}, V\su... ...with the property $W\subset V\subset U$.} \end{aligned}\right \}$$

Restriction map of $\mathcal G$ is defined in an obvious manner.

Then it is easy to see that $\mathcal G$ satisfies the sheaf axiom and that

$$\mathcal G(U)\cong \mathcal F(U)$$

holds for any $U\in \mathfrak{U}$ in a natural way.

$\qedsymbol$

LEMMA 1.21   Let $A$ be a ring.

1. We have a sheaf $\mathcal{O}$ of rings on $\operatorname{Spec}(A)$ which is defined uniquely by the property
   $$\mathcal{O}(O_f)=A_f \qquad(\forall f\in A)$$

2. For any $A$ -module $M$ we have a sheaf $\tilde {M}$ of modules on $\operatorname{Spec}(A)$ which is defined uniquely by the property
   $$\tilde{M}(O_f)=M_f \qquad(\forall f\in A)$$

3. For any $A$ -module $M$ , the sheaf $\tilde {M}$ is a sheaf of $\mathcal{O}$ -modules on $\operatorname{Spec}(A)$ . That means, it is a sheaf of modules over $\operatorname{Spec}(A)$ with an additional $\mathcal{O}$ -action (which is defined in an obvious way.)

PROOF.. We prove (2).

From the previous Lemma, we only need to prove locality and gluing lemma for open sets of the form $O_f$ . That means, in proving the properties (1) and (2) of Definition 1.19, we may assume that $U_\lambda=O_{f_\lambda}, U=O_f$ for some elements $f_\lambda, f\in A$ .

Furthermore, in doing so we may use the identification $O_f \approx \operatorname{Spec}A_f$ . By replacing $A$ by $A_f$ , this means that we may assume that $O_f=\operatorname{Spec}(A)$ .

To sum up, we may assume

$$U=\operatorname{Spec}(A), U_\lambda=O_{f_\lambda}.$$

To simplify the notation, in the rest of the proof, we shall denote by

$$i_\lambda: M \to M_{f_\lambda}$$

the canonical map which we have formerly written $i_{f_\lambda}$ . Furthermore, for any pair $\lambda, \mu$ of the index set, we shall denote by $i_{\lambda\mu}$ the canonical map

$$i_{\lambda\mu}: M \to M_{f_\lambda f_\mu}.$$

Locality: Compactness of $\operatorname{Spec}(A)$ (Theorem 1.12) implies that there exist finitely many open sets $\{O_{f_j}\}_{j=1}^k$ among $U_\lambda$ such that $\cup_{j=1}^k O_{f_j}=\operatorname{Spec}(A)$ . In particular there exit elements $\{c_j\}_{j=1}^k$ of $A$ such that

$$c_1 f_1+c_2 f_2+\dots +c_k f_k=1$$ (PU)

holds.

Let $m,n\in M$ be elements such that

$$i_{j}(m) =i_{j} (n) \qquad($$ in $$M_{f_j}.)$$

With the help of the ``module version'' of Lemma 1.8, we see that for each $j$ , there exist positive integers $N_j$ such that

$$f_j^{N_j}(m-n)=0$$

holds for all $j\in \{1,2,3,\dots,k\}$ . Let us take the maximum $N$ of $\{N_j\}$ . It is easy to see that

$$f_j^{N}(m-n)=0$$

holds for any $j$ . On the other hand, taking $(kN)$ -th power of the equation (PU) above, we may find elements $\{a_j\} \subset A$ such that

$$a_1 f_1^N+a_2 f_2^N+\dots +a_k f_k^N=1$$

holds. Then we compute

$$m-n= (a_1 f_1^N+a_2 f_2^N+\dots +a_k f_k^N)(m-n)=0$$

to conclude that $m=n$ .

Gluing lemma:

Let $\{m_\lambda \in M_{f_\lambda}\}$ be given such that they satisfy

$$i_{\lambda \mu}(m_{\lambda})= i_{\lambda \mu}(m_{\mu})$$

for any $\lambda, \mu$ . We fist choose a finite subcovering $\{O_{f_j}=U_{\lambda_j}\}_{j=1}^k$ of $\{U_\lambda\}$ . Then we may choose a positive integer $N_1$ such that

$$m_{\lambda_j} = x_j /f_j^{N_1} \qquad (\exists x_j \in M)$$

holds for all $j\in \{1,2,3,\dots,k\}$ .

$$i_{j l}(x_j f_l^{N_1})=i_{j l}(x_l f_j^N)$$

Then by the same argument which appears in the ``locality" part, there exists a positive integer $N_2$ such that

$$(f_i f_j)^{N_2}(x_j f_l^{N_1}-x_l f_j^{N_1}) =0$$

holds for all $j,l\in \{1,2,3,\dots,k\}$ . We rewrite the above equation as follows.

$$(f_j^{N_2} x_j) f_l^{{N_2}+{N_1}}- (f_l^{N_2} x_l) f_j^{{N_2}+{N_1}} =0.$$

On the other hand, by taking $k({N_1}+{N_2})$ -th power of the equation (PU), we may see that there exist elements $\{b_j\} \in A$ such that

$$\sum_{j=1}^k b_j f_j^{{N_1}+{N_2}}=1$$

holds.

Now we put

$$n=\sum_j b_j (f_j^{N_2} x_j).$$

Then since for any $l$

$$(f_j^{N_2} x_j)= (f_l^{N_2} x_l)f_j^{{N_2}+{N_1}}/f_l^{{N_2}+{N_1}} =f_j^{{N_2}+{N_1}} m_{\lambda_l}$$

holds on $O_l$ , we have $i_{l}(n)=m_{\lambda_l}$ .

Now, take any other open set $O_{f_\mu}=U_\mu$ from the covering $\{U_\lambda\}$ . $\{O_{f_j}\}_{j=1}^k \cup \{ O_{f_\mu}\}$ is again a finite open covering of $\operatorname{Spec}(A)$ . We thus know from the argument above that there exists an element $n_1$ of $M$ such that

$$i_{j}(n_1)= m_{f_j},\quad i_{\mu}(n_1)=m_{\mu}.$$

From the locality, $n_1$ coincides with $n$ . In particular, $i_{\mu}(n)=m_{\mu}$ holds. This means $n$ satisfies the requirement for the ``glued object''.

$\qedsymbol$

COROLLARY 1.22   Let $A$ be a commutative ring. Let $B$ be a non-commutative ring which contains $A$ as a central subalgebra (that means, $Z(B)\supset A$ ). Then there exists a sheaf $\tilde B$ of $\mathcal{O}$ -algebras over $\operatorname{Spec}(A)$