compactness

THEOREM 1.12   For any commutative ring $A$ , the spectrum $\operatorname{Spec}(A)$ of $A$ (equipped with the Zariski topology) is a compact set.

PROOF.. Let $\mathfrak{U}=\{U_\lambda\}$ be an open covering of $\operatorname{Spec}(A)$ . We want to find a finite subcovering of $\mathfrak{U}$ .

For any $x \in \operatorname{Spec}(A)$ , we have a index $\lambda_x$ and an open subset $O_{f_x}$ of $U_{\lambda_x}$ such that

$$x\in O_{f_x} \subset U_{\lambda_x}$$

holds. Replacing $\mathfrak{U}$ by $\{U_{\lambda_x}\}_{x \in \operatorname{Spec}(A)}$ if necessary, we may assume each $U_\lambda$ is of the form $O_{f_\lambda}$ for some $f_\lambda\in A$ .

Now,

$$\cup O_{f_\lambda}=\operatorname{Spec}(A)$$

implies that

$$\forall x \in \operatorname{Spec}(A) \exists \lambda$$    such that $$\rho_x(f_\lambda)\neq 0 \quad($$that means, $$f_\lambda \notin x.)$$

Now we would like to show from this fact that the ideal $I$ defined by

$$I=\{f_\lambda\}_{\lambda \in \Lambda}$$

is equal to $A$ . Assume the contrary. Using Zorn's lemma we may always obtain an maximal ideal $\mathfrak{m}$ of $A$ which contains $I$ . This is a contradiction to the fact mentioned above.

Thus we have proved that $I=A$ . In particular, we may find a relation

$$1=\sum_{j=0}^N a_j f_{\lambda_j}$$

for some positive integer $N$ , index sets $\{\lambda_j\}_{j=0}^N$ , and elements $a_j\in A$ . This clearly means that

$$\bigcup_{j=0}^N O_{f_{\lambda_j}}=\operatorname{Spec}(A)$$

as required. $\qedsymbol$