Let \(A\) be a ring. Show that the following conditions are equivalent:
\(\Spec A\) is disconnected;
there exist nonzero elements \(e_1, e_2 \in A\) such that \(e_1 e_2 = 0\), \(e_1^2 = e_1\), \(e_2^2 = e_2\), and \(e_1 + e_2 = 1\) (these are called orthogonal idempotents);
\(A\) is isomorphic to a direct product \(A_1 \times A_2\) of two nonzero rings.
Solution
1. If \(\Spec A\) is disconnected, then \(A\) has nonzero orthogonal idempotents \(e_1,e_2\) with \(e_1+e_2=1\).
Write \[ \Spec A=U\amalg V \] with \(U,V\) nonempty disjoint open subsets. Since they are complements of one another, both are also closed.
The sheaf axiom for the disjoint open cover \(\{U,V\}\) gives an isomorphism of rings \[ A=\Gamma(\Spec A,\mathcal O) \xrightarrow{\sim} \Gamma(U,\mathcal O)\times\Gamma(V,\mathcal O). \] Let \(e_1,e_2\in A\) be the preimages of \[ (1,0),\qquad(0,1). \] Then \[ e_1^2=e_1, \qquad e_2^2=e_2, \qquad e_1e_2=0, \qquad e_1+e_2=1. \] Both are nonzero: if, for example, \(e_1=0\), then its restriction to the nonempty open \(U\) would be the zero section, contradicting that it restricts to \(1\) there. Thus condition (2) holds.
2. Suppose condition (2) holds. Then\[ \boxed{A\cong Ae_1\times Ae_2.} \]
Define \[ \Phi:A\longrightarrow Ae_1\times Ae_2, \qquad a\longmapsto(ae_1,ae_2). \] The rings \(Ae_i\) have identities \(e_i\).
Define \[ \Psi:Ae_1\times Ae_2\longrightarrow A, \qquad (x,y)\longmapsto x+y. \] If \(x\in Ae_1\) and \(y\in Ae_2\), then \[ xe_1=x, \quad xe_2=0, \quad ye_1=0, \quad ye_2=y. \] Therefore \[ \Phi(\Psi(x,y))=(x,y), \] while \[ \Psi(\Phi(a))=a(e_1+e_2)=a. \] Thus \(\Phi\) is an isomorphism. Since \(e_1,e_2\) are nonzero, both factor rings are nonzero. Hence condition (3) holds.
3. Suppose\[ A\cong A_1\times A_2 \]with \(A_1,A_2\ne0\). Then \(\Spec A\) is disconnected.
Let \[ e_1=(1,0), \qquad e_2=(0,1). \] These are complementary orthogonal idempotents. For every prime ideal \(\mathfrak p\subset A\), the relation \[ e_1e_2=0 \] forces \(e_1\in\mathfrak p\) or \(e_2\in\mathfrak p\), while \(e_1+e_2=1\) prevents both from lying in \(\mathfrak p\).
Hence \[ \Spec A=D(e_1)\amalg D(e_2). \] Each \(D(e_i)\) is open, and it is also closed because \[ D(e_1)=V(e_2), \qquad D(e_2)=V(e_1). \] They are nonempty because \[ D(e_1)\cong\Spec A_1, \qquad D(e_2)\cong\Spec A_2, \] and a nonzero ring has a prime ideal. Thus \(\Spec A\) is a disjoint union of two nonempty open subsets and is disconnected.
4. Therefore the three conditions are equivalent.
Step 1 proves \((1)\Rightarrow(2)\), step 2 proves \((2)\Rightarrow(3)\), and step 3 proves \((3)\Rightarrow(1)\).
5. Equivalently,\[ \boxed{ \Spec A\text{ is connected} \iff A\text{ has no idempotents other than }0,1. } \]
A nontrivial idempotent \(e\) gives the complementary pair \(e,1-e\), and every pair in condition (2) has this form.
6. Q.E.D.
Step 4 is the required equivalence.