Proposition

Show that

the set $\{ x \}$ is closed in $\Spec(A)$ if and only if $p_x$ is maximal.
$\overline{\{ x \}} = V(p_x)$.
$y \in \overline{\{ x \}} \iff p_x \subset p_y$.
$X$ is a $T_0$-space.

Solution

1

Suppose $x$ is maximal. Clearly, $x \in V(x)$. Let $y \in V(x)$. Then $y \in \Spec(A)$ and $x \subset y$. Since $x$ is maximal, $x = y$. Therefore, $V(x) = \{ x \}$, so $\{ x \}$ is closed in $\Spec(A)$.

On the other hand, suppose $x$ is not maximal. Then there exists $y \in \Spec(A)$ such that $x \subsetneq y$. For any $E \subset A$, if $E \subset x$, then $E \subset y$. Therefore, $V(E) \ne \{ x \}$ for any $E \subset A$.

Therefore, the set $\{ x \}$ is closed in $\Spec(A)$ if and only if $x$ is maximal.

2

Let $y = p_y \in \Spec(A)$.

\[\begin{align*} y \in \overline{ \{ x \} } &\iff \forall E \subset A, (x \in V(E) \implies y \in V(E)) \\ &\iff \forall E \subset A, (E \subset p_x \implies E \subset p_y) \\ &\iff p_x \subset p_y \\ &\iff p_y \in V(p_x). \end{align*}\]

Therefore, $\overline{ { x } } = V(p_x)$.

3

Let $y = p_y \in \Spec(A)$.

Therefore, $\overline{ { x } } = V(p_x)$.

4

Let $p_x \ne p_y \in \Spec(A)$ be given. Let $E = p_x \setminus p_y$. We will assume that $E$ is nonempty without loss of generality.

Then $x \in V(E)$ and $y \notin V(E)$. Thus $x \notin (V(E))^c$ and $y \in (V(E))^c$, so $V(E)^c$ is a neighborhood of $y$ that does not contain $x$.