Zariski closure of a singleton is irreducible

Question

Closure of singleton is irreducible in Zariski topology

Let $R$ be a commutative unitary ring. For any $E \subset R$ define $V(E) = \lbrace P \in \text{spec}(A) \mid E \subset P \rbrace $. Equip $\text{spec(R)}$ with the Zariski topology, that is, the closed subsets of $\text{spec}(R)$ are precisely the $V(E)$ sets given by any $E$. Just as a remark, it turns out that restricting $E$ to the set of radical ideals still yields the same closed subsets.
Notice that if $E$ happens to be a prime ideal, then $V(E)$ is the closure of $ \lbrace E \rbrace $.


It says in my lecture notes that the closure of a singleton $\lbrace P \rbrace$ is always an irreducible component. So basically, this means :
$V(P) \subset V(E) \cup V(F) \implies V(P) \subset V(E) \quad \text{or} \quad V(P) \subset V(F)$

I cannot prove that this implication is true. I've tried many ways but always ended up finding a mistake in my proofs.

Answer 1

If $P$ is a point and $V(P) \subset V(E) \cup V(F)$, then certainly $P\in V(E)$ or $P\in V(F)$ (because $P\in V(P)$). Since $V(E)$ and $V(F)$ are closed sets, we have $$ V(P) = \overline{\{P\}} \subset \overline{V(E)} = V(E) \quad \text{or}\quad V(P) \subset V(F). $$