Exercise 3.11

Question

Let $A$ be a ring. Prove that TFAE:
   \begin{enumerate}[label=\roman*)]
     \item $A/\mathfrak{N}$ is absolutely flat ($\mathfrak{N}$ being the nilradical of $A$)
     \item Every prime ideal of $A$ is maximal. 
     \item $\operatorname{Spec} (A)$ is a $T_1$-space (i.e. every subset consisting of a single point is closed).
     \item $\operatorname{Spec}(A)$ is Hausdorff
   \end{enumerate}

\par   If these conditions are satisfied, show that $\operatorname{Spec} (A)$ is compact and totally disconnected (i.e. the only connected subsets of $\operatorname{Spec}(A)$ are those consisting of a single point).

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of equivalence]
 We will show $iv) \Rightarrow  iii) \Rightarrow ii) \Rightarrow i) \Rightarrow iv)$
 
 \par $iv) \Rightarrow iii)$: Trivial
 \par $iii) \Rightarrow ii)$: Assume that $\operatorname{Spec}(A)$ is a $T_1$-space. Let $\mathfrak{p} \in \operatorname{Spec}(A)$. Then, $V(\mathfrak{p}) = \overline{\{\mathfrak{p}\}} = \{\mathfrak{p}\}$ because $\operatorname{Spec}(A)$ is $T_1$. By definition of $V(\mathfrak{p})$, this means that $\mathfrak{p}$ is maximal. (AM Exercise 1.18). Hence, $iii)$ implies $ii)$. 
 \par $ii) \Rightarrow i)$: Assume that every prime ideal of $A$ is maximal. Choose an element $x \in A$ and any $\mathfrak{p} \in \operatorname{Spec}(A)$. Then, either $x \in \mathfrak{p}$ or $x 
otin \mathfrak{p} \Rightarrow x+ \mathfrak{p} = (1)$ because $\mathfrak{p}$ is maximal. Then, $x -1 \in \mathfrak{p}$. So for all $x \in A$ and $\mathfrak{p} \in \operatorname{Spec}(A)$, $x(x-1) \in \mathfrak{p}$. Therefore, any $\overline{x} \in A/\mathfrak{N}$ is idempotent. Using the proposition from the previous problem, we conclude that $A/\mathfrak{N}$ is absolutely flat.

\par $i) \Rightarrow iv)$. From AM Exercise 1.21 $iv)$, $\operatorname{Spec}(A)$ and $\operatorname{Spec}(A/\mathfrak{N})$ are naturally homeomorphic. Hence, we are done if we show that if a ring $A$ is absolutely flat, $\operatorname{Spec}(A)$ is Hausdorff. 
 
 If $A$ is absolutely flat, then for any $x \in A$, there exists $u \in A^{	imes}$ such that $x^2 = ux$ (again, from the proposition). Then $x(x-u) = 0 \in \mathfrak{p}$ for any prime ideal $\mathfrak{p}$. Thus either $x \in \mathfrak{p}$ or $x-u \in \mathfrak{p}$. If both $x$ and $x-u$ is in $\mathfrak{p}$, then $x-(x-u) = u \in \mathfrak{p}$ so $\mathfrak{p} = (1)$, as $u$ is a unit. Hence, $x$ and $x-u$ cannot be in a same prime ideal, so $\operatorname{Spec}(A) = V(x) \sqcup V(x-u) = (V(x-u))^c \sqcup (V(x))^c = X_{x-u} \sqcup X_x$. Hence, $\operatorname{Spec}(A)$ is Hausdorff.   
 
\end{proof}
\begin{proof}[Proof of the last claim]
  $\operatorname{Spec} A$ is always quasi-compact from AM Exercise 17 $v)$. As $\operatorname{Spec} A$ is Hausdorff by the assumption, it is compact. 
  \par Assume that $C \subset \operatorname{Spec}(A)$ is a subset of $\operatorname{Spec}(A)$ with more than one element. Let $\mathfrak{p} 
eq \mathfrak{q} \in C$. Without loss of generality assume $\mathfrak{p} 
eq \emptyset$. Choose $x \in \mathfrak{p}/\mathfrak{q}$. Then we see from the last proof that $\mathfrak{p} \in V(x) = X_{x-u}$, and $\mathfrak{q} \in X_{x}$ where $X_{x-u} \sqcup X_x = \operatorname{Spec}(A)$. Then $C = (C \cap X_x) \sqcup (C \cap X_{x-u})$, so $C$ is disconnected. Hence, the only subsets of $\operatorname{Spec}(A)$ are those consisting of a single point.

\end{proof}

Exercise 3.11

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Exercise 3.11

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