A topological space is quasi-compact if every open cover has a finite subcover.
Show that a topological space is noetherian if and only if every open subset is quasi-compact.
If \(X\) is an affine scheme, show that \(\operatorname{sp}(X)\) is quasi-compact, but not in general noetherian. We say a scheme \(X\) is quasi-compact if \(\operatorname{sp}(X)\) is.
If \(A\) is a noetherian ring, show that \(\operatorname{sp}(\Spec A)\) is a noetherian topological space.
Give an example to show that \(\operatorname{sp}(\Spec A)\) can be noetherian even when \(A\) is not.
Solution
1. If a topological space \(X\) is noetherian, then every open subset of \(X\) is quasi-compact.
Let \(U\subseteq X\) be open and suppose \[ U=\bigcup_{\alpha\in A}U_\alpha \] is an open cover, with each \(U_\alpha\) open in the subspace \(U\). Since \(U\) itself is open in \(X\), every \(U_\alpha\) is open in \(X\).
Suppose there is no finite subcover. Choose \(\alpha_1\). Since \(U_{\alpha_1}\ne U\), choose \(\alpha_2\) such that \[ U_{\alpha_2}\not\subseteq U_{\alpha_1}. \] Inductively, since no finite union covers \(U\), choose \(\alpha_{n+1}\) with \[ U_{\alpha_{n+1}} \not\subseteq U_{\alpha_1}\cup\cdots\cup U_{\alpha_n}. \] Then \[ U_{\alpha_1} \subsetneq U_{\alpha_1}\cup U_{\alpha_2} \subsetneq U_{\alpha_1}\cup U_{\alpha_2}\cup U_{\alpha_3} \subsetneq\cdots \] is a strictly ascending chain of open subsets of \(X\), contradicting noetherianity. Hence a finite subcover exists.
2. Conversely, if every open subset of \(X\) is quasi-compact, then \(X\) is noetherian.
Let \[ U_1\subseteq U_2\subseteq U_3\subseteq\cdots \] be an ascending chain of open subsets and put \[ U=\bigcup_{n\ge1}U_n. \] The set \(U\) is open. By hypothesis it is quasi-compact, and the family \(\{U_n\}\) is an open cover of \(U\). Hence finitely many of the \(U_n\) cover \(U\).
Because the chain is nested, if \(N\) is the largest index among that finite subcover, then \[ U=U_N. \] Thus \[ U_n=U_N \] for every \(n\ge N\), so every ascending chain of opens stabilizes. This is the noetherian condition.
3. Therefore\[ \boxed{ X\text{ is noetherian} \iff \text{every open subset of }X\text{ is quasi-compact}. } \]
Combine 1 and 2.
4. For every ring \(A\), the affine spectrum\[ X=\operatorname{Spec}A \]is quasi-compact.
Let \[ X=\bigcup_{\alpha\in A_0}U_\alpha \] be an arbitrary open cover. Distinguished opens form a basis, so refine this cover by distinguished opens \[ X=\bigcup_{\lambda\in\Lambda}D(f_\lambda), \] with each \(D(f_\lambda)\) contained in one of the original \(U_\alpha\).
The equality of the union with all of \(X\) means that no prime ideal contains every \(f_\lambda\). Equivalently, \[ V((f_\lambda)_{\lambda\in\Lambda})=\varnothing. \] Hence \[ \sqrt{(f_\lambda)_{\lambda\in\Lambda}}=A, \] so \[ 1\in(f_\lambda)_{\lambda\in\Lambda}. \] Membership in an ideal is a finite linear combination, so there exist finitely many indices \[ \lambda_1,\ldots,\lambda_r \] and coefficients \(a_i\in A\) such that \[ 1=\sum_{i=1}^ra_if_{\lambda_i}. \] No prime can contain all these finitely many \(f_{\lambda_i}\), hence \[ X=D(f_{\lambda_1})\cup\cdots\cup D(f_{\lambda_r}). \] Choosing the corresponding finitely many original cover members gives a finite subcover.
5. An affine spectrum need not be noetherian. For example,\[ A=k[x_1,x_2,x_3,\ldots] \]has non-noetherian spectrum.
Consider the ideals \[ I_n=(x_1,\ldots,x_n). \] Then \[ I_1\subsetneq I_2\subsetneq I_3\subsetneq\cdots. \] The corresponding closed subsets form a descending chain \[ V(I_1) \supseteq V(I_2) \supseteq V(I_3) \supseteq\cdots. \] This chain is strict. Indeed, the prime ideal \[ \mathfrak p_n=(x_1,\ldots,x_n) \] belongs to \(V(I_n)\) but not to \(V(I_{n+1})\), since \[ x_{n+1}\notin\mathfrak p_n. \] Thus the descending chain of closed subsets does not stabilize, so \(\operatorname{Spec}A\) is not noetherian.
6. If \(A\) is a noetherian ring, then\[ \operatorname{Spec}A \]is a noetherian topological space.
Let \[ V(I_1) \supseteq V(I_2) \supseteq V(I_3) \supseteq\cdots \] be a descending chain of closed subsets. Since \[ V(I)=V(\sqrt I), \] this corresponds to an ascending chain of radical ideals \[ \sqrt{I_1} \subseteq \sqrt{I_2} \subseteq \sqrt{I_3} \subseteq\cdots. \] Because \(A\) is noetherian, every ascending chain of ideals stabilizes. Hence the radical ideals stabilize and therefore the original closed subsets stabilize. Thus the spectrum is noetherian.
7. The converse to 6 fails. Let\[ A = k[x_1,x_2,x_3,\ldots] /(\,x_ix_j: i,j\ge1\,). \]Then \(A\) is not noetherian, but\[ \operatorname{Spec}A \]has one point and is therefore noetherian.
Let \[ \mathfrak m=(\bar x_1,\bar x_2,\ldots)\subseteq A. \] By construction, \[ \mathfrak m^2=0, \] and \[ A/\mathfrak m\cong k. \] Thus \(\mathfrak m\) is maximal.
Every element of \(\mathfrak m\) is nilpotent, so every prime ideal of \(A\) contains \(\mathfrak m\). Since \(\mathfrak m\) is maximal, it is the unique prime ideal. Hence \[ \operatorname{Spec}A=\{\mathfrak m\}, \] a one-point topological space, which is noetherian.
On the other hand, \(\mathfrak m\) is not finitely generated as an ideal. Because \[ \mathfrak m^2=0, \] multiplication of a generator in \(\mathfrak m\) by an arbitrary element of \(A\) only scales it by the image of that element in \[ A/\mathfrak m=k. \] Thus an ideal generated by finitely many elements of \(\mathfrak m\) is merely their finite-dimensional \(k\)-linear span. The vectors \[ \bar x_1,\bar x_2,\ldots \] are linearly independent, so they cannot be generated by finitely many of them. Hence \(A\) is not noetherian.
8. Q.E.D.
Steps <1>1–<1>3 prove part (a), <1>4–<1>5 prove part (b), 6 proves part (c), and 7 proves part (d).