----- > [!proposition] Proposition. ([[disconnected spectrum is that of a product ring]]) > Let $A$ be a [[ring]]. The following are equivalent: > > 1. [[prime ideal|Spec]] $A$ is [[connected|disconnected]]; > 2. There exists 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 $e_{1}+e_{2}=1$; > 3. $A$ is [[ring isomorphism|isomorphic]] to $A_{1} \times A_{2}$ for some [[ring|rings]] $A_{1},A_{2}$. > [!proof]- Proof. ([[disconnected spectrum is that of a product ring]]) > $(1) \implies (2)$. Suppose $\text{Spec }A$ is [[connected|disconnected]], i.e., there exist [[ideal|ideals]] $I,J$ of $A$ such that $\text{Spec }A=V(I) \cup V(I)$ with $V(I) \cap V(J)=\emptyset$. > > Now $V(I) \cap V(J)=V(I+J)$ is empty, meaning that no [[prime ideal]] contains $I+J$. [[existence of maximal ideals in commutative rings|Since any proper ideal has a maximal ideal containing it]], and [[maximal ideal|maximal ideals]] are [[prime ideal|prime]], $I+J$ must not be proper: $I+J=(1)$. > > Moreover, $V(I) \cup V(J)=V(I \cap J)$, meaning that $\text{Spec }A=V(I \cap J)$: every [[prime ideal]] contains $I\cap J$. [[nilradical equals intersection of all prime ideals|Thus]], $I \cap J$ is contained in the [[nilradical of a ring|nilradical]] $\sqrt{ 0 }$ of $A$. > > Therefore, we know that there exist $a \in I$, $b \in J$ with $a+b=1$, forcing $V(\langle a \rangle) \cap V(\langle b \rangle)=\emptyset$. And $(ab)^{n}=0$ for some $n \geq 0$. Since $V(\langle a \rangle)=V(\langle a^{n} \rangle)$ and $V(\langle b \rangle)=V(\langle b^{n} \rangle)$, we have $V(\langle a^{n} \rangle) \cap V(\langle b^{n} \rangle)=\emptyset=V(\langle a^{n} \rangle + \langle b^{n} \rangle)$. So $\langle a^{n} \rangle+\langle b^{n} \rangle=(1)$. So there exist $e_{1} \in \langle a^{n} \rangle$, $e_{2} \in \langle b^{n} \rangle$, such that $e_{1}+e_{2}=1$, $e_{1}e_{2}=ra^{n}sb^{n}=(rs)(a^{n}b^{n})=(rs)(ab)^{n}=0$. It follows that > $e_{1}^{2}=e_{1}(e_{1}+e_{2})=e_{1}$ and $e_{2}^{2}=e_{2}(e_{1}+e_{2})=e_{2}$. > > > $(2) \implies (3)$. As $\langle e_{1} \rangle$ and $\langle e_{2} \rangle$ are [[relatively prime integers|coprime]], by [[Chinese remainder theorem|CRT]] we have $\frac{A}{\langle e_{1} \rangle \langle e_{2} \rangle } \cong \frac{A}{\langle e_{1} \rangle } \times \frac{A}{\langle e_{2} \rangle }$ > but $\langle e_{1} \rangle \langle e_{2} \rangle=\langle e_{1}e_{2} \rangle=0$, so in fact $A$ is isomorphic to $\frac{A}{\langle e_{1} \rangle} \times \frac{A}{\langle e_{2} \rangle}$. > > > $(3) \implies(1)$. [[ideals of product ring are products of ideals|The prime ideals]] of $A_{1} \times A_{2}$ take the form $\mathfrak{p} \times (1)$ for $\mathfrak{p} \in \text{Spec }A_{1}$ and $(1) \times \mathfrak{q}$ for $\mathfrak{q} \in \text{Spec }A_{2}$. Every prime ideal of $A_{1}$ contains $\text{Nil}_{A_{1}}$ and every prime ideal of $A_{2}$ contains $\text{Nil}_{A_{2}}$; put $I_{1}=\langle (1, 0) \rangle$ and $I_{2}=\langle (0,1) \rangle$, then every prime ideal of $A_{1} \times A_{2}$ contains exactly one of $I_{1}$ or $I_{2}$, witnessing that $\text{Spec}(A_{1} \times A_{2})=V(I_{1}) \sqcup V(I_{2})$ is disconnected. > > ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```