----- > [!proposition] Proposition. ([[irreducible closed subspaces of Spec are precisely the vanishing of primes]]) > Let $R$ be a ([[commutative ring|commutative]]) [[ring]] and consider $\text{Spec }R$, endowed with the [[Zariski topology on a ring spectrum|Zariski topology]]. Let $Z$ be a [[closed set|closed subset]] of $\text{Spec }R$. Then $Z \text{ is irreducible } \iff Z= V(\mathfrak{p}) \text{ for some prime ideal }\mathfrak{p}.$ > > > In fact, $\mathfrak{p}$ is the *only* [[ideal]] for which $Z=V(\mathfrak{p})$, that is, we have a [[bijection]] $\{ \text{irreducible closed subsets }Z \subset \text{Spec }R \} \leftrightarrow \text{Spec }R.$ > [!proof]- Proof. ([[irreducible closed subspaces of Spec are precisely the vanishing of primes]]) > > $\impliedby$. Recall from properties in [[Zariski topology on a ring spectrum|Zariski topology]] that $\overline{\{ \mathfrak{p} \}}=V(\mathfrak{p})$ for any [[prime ideal]] $\mathfrak{p}$. Hence $V(\mathfrak{p})$ is [[irreducible topological space|irreducible]] (as it is the [[closure]] of an irreducible). > > $\implies.$ Let $Z=V(I) \subset \text{Spec }R$ be any irreducible [[closed set|closed subset]] of $\text{Spec }R$. As $V(I)=V(\sqrt{ I })$ in general, WLOG we may assume $I$ is a [[radical of an ideal|radical ideal]]. WTS $I$ is prime (will show contrapositive). If $I$ is not [[prime ideal|prime]], then choose $a,b \in R$ such that $a \notin I, b \notin I$, $ab \in I$. In this case, since $(I+\langle a \rangle) \cap (I+ \langle b \rangle)=I$ ([[radical of an ideal#^4e9b7c|remember]] $I$ is [[radical of an ideal|radical]]!), $V(I+\langle a \rangle)\cup V(I + \langle b \rangle)=V(I)$. So by irreducibility, $V(I)=V(I+ \langle a \rangle ) \text{ or } V(I)=V(I + \langle b \rangle ).$ > Neither of these is true, since $a \notin I$ and $b \notin I$. Hence $V(I)$ must not be irreducible. > > "In fact..." Suppose $Z=V(\mathfrak{q})$ and $Z=V(\mathfrak{p})$ for $\mathfrak{q,p} \in\text{Spec } R$. Recall that, in general, $V(I)=V(J) \iff \sqrt{ I }=\sqrt{ J }$. But here $\mathfrak{q}$ and $\mathfrak{p}$ are prime, hence [[radical of an ideal|radical]], and so we have $\mathfrak{q}=\mathfrak{p}$. ----- #### ---- #### 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 > ```