Exercise 1.22

Question

Let $A = \prod_{i=1}^n A_i$ be the direct product of rings $A_i$. Show that $\operatorname{Spec}(A)$ is the disjoint union of open (and closed) subspaces $X_i$, where $X_i$ is canonically homeomorphic with $\operatorname{Spec}(A_i)$.
  \par Conversely, let $A$ be any ring. Show that the following statements are equivalent:
  \begin{enumerate}[label=oman*)]
    \item $X = \operatorname{Spec}(A)$ is disconnected.
    \item $A \cong A_1 	imes A_2$ where neither of the rings $A_1,A_2$ is the zero ring.
    \item $A$ contains an idempotent $
e 0,1$.
  \end{enumerate}
  \par In particular, the spectrum of a local ring is always connected (Exercise 12).

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of first statement]
  Consider the projection map $\pi_i\colon A 	o A_i$. $\pi_i$ is surjective, hence induces a homeomorphism $\pi_i^*\colon\operatorname{Spec} A_i \overset{\sim}{	o} V(\operatorname{Ker}(\pi_i))$ by Exercise $1.21iv)$. Letting $X_i \coloneqq V(\operatorname{Ker}(\pi_i))$, we have $\operatorname{Spec}(A) \supseteq \bigcup_{i=1}^n X_i$. By Exercise 1.15, we have
  \begin{gather*}
    \bigcup_{i=1}^n X_i = V\left(\bigcap_{i=1}^n\operatorname{Ker}(\pi_i)ight) = V(0) = \operatorname{Spec}(A),\
    X_i \cap X_j = V( \operatorname{Ker}(\pi_i) + \operatorname{Ker}(\pi_j)) = V(1) = \emptyset,
  \end{gather*}
  and so $\operatorname{Spec}(A) = \coprod_{i=1}^n X_i$, where each $X_i$ is closed in $\operatorname{Spec}(A)$ hence open since $\operatorname{Spec}(A) \setminus X_i$ is a finite union of closed subsets of $\operatorname{Spec}(A)$.
\end{proof}
\begin{proof}[Proof of second statement]
  $(ii) \Rightarrow (i)$. This is the first statement.
  \par $(i) \Rightarrow (iii)$. Suppose $X = X_1 \amalg X_2$, where $X_1 = V(\mathfrak{a}_1)$, $X_2 = V(\mathfrak{a}_2)$ for ideals $\mathfrak{a}_i \subseteq A$. By Exercise $ef{exc:1.15}$, $V(1) = \emptyset = V(\mathfrak{a}_1 + \mathfrak{a}_2)$ implies $\sqrt{\mathfrak{a}_1 + \mathfrak{a}_2} = A$, so by Exercise $1.13iv)$ in the text, $\mathfrak{a}_1 + \mathfrak{a}_2 = 1$, i.e., $x_1 + x_2 = 1$ for some $x_1 \in \mathfrak{a}_1,x_2\in\mathfrak{a}_2$. Now $X = V(0) = V(\mathfrak{a}_1\cdot\mathfrak{a}_2)$ implies $\mathfrak{a}_1\mathfrak{a}_2 \subseteq \sqrt{0}$, and so $(x_1x_2)^m = 0$ for some $m$. Thus, define
  \begin{equation*}
    1 = (x_1 + x_2)^{2m} = \underbrace{x_1^{2m} + \cdots + \binom{2m}{m+1} x_1^{m+1}x_2^{m-1}}_{e_1} + \underbrace{\binom{2m}{m}x_1^mx_2^m + \cdots + x_2^{2m}}_{e_2}
  \end{equation*}
  We have $e_1e_2 = 0$ since each term in the product contains $(x_1x_2)^m$. Then, $e_1 = e_1(e_1 + e_2) = e_1^2$ and so $A$ contains an idempotent $
e 0,1$.
  \par $(iii) \Rightarrow (ii)$. Let $A_1 = A/(e)$ and $A_2 = A/(1-e)$, where $e 
e0,1$ is idempotent. $A_1 
e 0$ since if $e \in A^	imes$, $1 = ee^{-1} = e^2e^{-1} = e$, a contradiction; similarly, $A_2 
e 0$ since if $1-e \in A^	imes$, $1 = (1-e)(1-e)^{-1} = (1-e)^2(1-e)^{-1} = 1-e$, implying $e=0$. Thus, $A 	o A_1 	imes A_2$ is an isomorphism by Prop.~1.10 since $(e) + (1-e) = (1)$ and $(e) \cap (1-e) = (e(1-e)) = (e-e^2) = (0)$.
\end{proof}

Exercise 1.22

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Exercise 1.22

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