Proposition

For each $f \in A$, let $X_f$ denote the complement of $V(f)$ in $X = \Spec(A)$. The sets $X_f$ are open. Show that they form a basis of open sets for the Zariski topology, and that

  1. $X_f \cap X_g = X_{fg}$;
  2. $X_f = \emptyset \iff f$ is nilpotent;
  3. $X_f = X \iff f$ is a unit;
  4. $X_f = X_g \iff r((f)) = r((g))$;
  5. $X$ is quasi-compact;
  6. More generally, each $X_f$ is quasi-compact.
  7. An open subset of $X$ is quasi-compact if and only if it is a finite union of sets $X_f$.

Solution

Basis

  • $X_1 = (V(1))^c = \emptyset^c = X$. Thus the union of all $X_f$ is $X$.

  • Let $X_{f_1}, X_{f_2}$ be given.

    \[\begin{align*} X_{f_1} \cap X_{f_2} &= V(f_1)^c \cap V(f_2)^c \\ &= (V(f_1) \cup V(f_2))^c \\ &= (V(\ev{f_1}) \cup V(\ev{f_2}))^c \\ &= (V(\ev{f_1}\ev{f_2}))^c \\ &= (V(\ev{f_1f_2}))^c \\ &= (V(f_1f_2))^c \\ &= X_{f_1f_2}. \end{align*}\]

    We showed that $V(\ev{f_1}) \cup V(\ev{f_2}) = V(\ev{f_1}\ev{f_2})$ in this post. Therefore, $\forall x \in X_{f_1} \cap X_{f_2}, x \in X_{f_1f_2} \subset X_{f_1} \cap X_{f_2}$.

Hence, $\{ X_f \mid f \in A \}$ is a basis for the Zariski topology.

1

By definition, $X_f = \{ p \in \Spec(A) \mid f \notin p \}$ for each $f$. Since every set in $\Spec(A)$ is a prime ideal, $\forall p \in \Spec(A), \forall f, g \in A$,

\[\begin{align*} fg \notin p \iff f \notin p \text{ and } g \notin p. \end{align*}\]

Therefore, $X_f \cap X_g = X_{fg}$.

2

\[\begin{align*} X_f = \emptyset &\iff \forall p \in \Spec(A), f \in p \\ &\iff f \in \cap_{p \in \Spec(A)} p \\ &\iff f \in \mathfrak{R} & \text{(By Proposition 1.8[Atiyah])} \\ &\iff \text{$f$ is nilpotent} \end{align*}\]

3

\(\begin{align*} X_f = X &\iff \forall p \in \Spec(A), f \notin p. \end{align*}\)

Suppose $X_f = X$. Every non-unit element is contained in some maximal ideal by Corollary 1.5 [Atiyah]. Since every maximal ideal is prime, $f$ must be a unit element.

On the other hand, since every prime ideal is a proper ideal, a unit is not contained in any prime ideal.

Therefore, $X_f = X$ if and only if $f$ is a unit.

4

\[\begin{align*} r((f)) = r((g)) &\iff \bigcap_{p \in \Spec(A), (f) \subset p} p = \bigcap_{p \in \Spec(A), (g) \subset p} p & \text{(Proposition 1.14)} \\ &\iff \bigcap_{p \in \Spec(A), f \in p} p = \bigcap_{p \in \Spec(A), g \in p} p. \end{align*}\]

On the other hand, $(X_f = X_g \iff V(f) = V(g)) \iff (\forall p \in \Spec(A), f \in p \iff g \in p)$.

We claim that

\[\begin{align*} (\bigcap_{p \in \Spec(A), f \in p} p = \bigcap_{p \in \Spec(A), g \in p} p) \iff \forall p \in \Spec(A), f \in p \iff g \in p. \end{align*}\]

$\impliedby$ is obvious. Suppose that there exists a prime ideal $p$ such that $f \in p$ and $g \notin p$. Then $g \notin \bigcap_{p \in \Spec(A), f \in p} p$ and $g \in \bigcap_{p \in \Spec(A), g \in p} p$. Using a similar argument for the case that $f \notin p$ and $g \in p$, $\implies$ can be shown.

5

$X_1 = X$ by Part 3. By Part 6, $X$ is quasi-compact.

6

Let $f \in X$ be given. Let \(\{ U_{\alpha} \}_{\alpha \in I}\) be an open cover of $X_f$. Let $V_{\alpha} = U_{\alpha} \cap X_f$ for each $\alpha \in I$. Then $\{ V_{\alpha} \}$ is an open cover of $X_f$ such that $X_f = \bigcup_{\alpha \in I} V_{\alpha}$. We will show that $\{ V_{\alpha} \}$ contains a finite cover.

Then $X \setminus V_{\alpha}$ is closed for each $\alpha \in I$. By definition, each closed set in the Zariski topology is of the form of $V(a)$ for some ideal $a$. Thus there exists an ideal $a_{\alpha}$ such that $X \setminus V_{\alpha} = V(a_{\alpha})$ for each $\alpha \in I$.

\[\begin{align*} X_f &= \bigcup_{\alpha \in I} (X \setminus V(a_{\alpha})) \\ &= X \setminus (\bigcap_{\alpha \in I} V(a_{\alpha})) \\ &= X \setminus V(\bigcup_{\alpha \in I} a_{\alpha}) \\ &= X \setminus V(\sum_{\alpha \in I} a_{\alpha}). \end{align*}\]

This implies that $V(f) = V(\sum a_{\alpha})$, thus

\[\begin{align*} \bigcap_{p \in \Spec(A), f \in p} p = \bigcap_{p \in \Spec(A), \sum_{\alpha \in I} a_{\alpha} \subset p} p. \end{align*}\]

By Proposition 1.8, $r((f)) = r(\sum a_{\alpha})$. Then $f \in r((f)) = r(\sum a_{\alpha})$, so $f^m \in \sum a_{\alpha}$. This implies the existence of $x_{\alpha_1} \in a_{\alpha_1}, \cdots, x_{\alpha_n} \in a_{\alpha_n}$ for some $\alpha_1, \cdots, \alpha_n \in I$ such that $f^m = x_{\alpha_1} + \cdots + x_{\alpha_n}$.

\[\begin{align*} (f^m) \subset (x_{\alpha_1}) + \cdots + (x_{\alpha_n}) \subset a_{\alpha_1} + \cdots + a_{\alpha_n} &\implies V((f^m)) \supset V(a_{\alpha_1} + \cdots + a_{\alpha_n}) \\ &\implies V((f)) \supset V(a_{\alpha_1} + \cdots + a_{\alpha_n}) \\ &\implies (X \setminus V((f))) \subset (X \setminus V(a_{\alpha_1} + \cdots + a_{\alpha_n})) \\ &\implies X_f \subset (X \setminus V(a_{\alpha_1} \cup \cdots \cup a_{\alpha_n})) \\ &\implies X_f \subset (X \setminus \cap V(a_{\alpha_i})) \\ &\implies X_f \subset \bigcup_{i=1}^n V_{\alpha_i} \\ &\implies X_f \subset \bigcup_{i=1}^n U_{\alpha_i}. \end{align*}\]

7

A finite union of compact sets is compact, so a finite union of sets $X_f$ is compact.

Let $U$ be a compact, open subset of $X$. We will show that $U$ is a finite union of sets $X_f$. Let $C = \{ X_f \mid f \in A, X_f \subset U \}$. $C$ is an open cover because $\{ X_f \mid f \in A \}$ is a basis, so every open set is the union of basis elements. Since $U$ is compact, $U \subset X_{f_1} \cup \cdots \cup X_{f_n}$ for some $X_{f_1}, \cdots, X_{f_n} \in C$. On the other hand, $X_{f_1} \cup \cdots \cup X_{f_n} \subset U$. Therefore, $U = X_{f_1} \cup \cdots \cup X_{f_n}$.