I would like to show if $R$ is a comm ring, and $Y \subset\operatorname{Spec}R$ is a closed irreducible subset, then there is a prime ideal $p \subset R$ such that $Y = V(p)$, i.e. $Y = \{ q \in \operatorname{Spec}R : q \supseteq p\}$.
I'm not quite sure where to begin, other than saying
- $Y = V(I)$ for some ideal $I$ because $Y$ is closed;
- and $I$ cannot be factored into a product of ideals $J,J'$ each with vanishing set different from $V(I)$, since $Y$ is irreducible and $V(JJ') = V(J) \cup V(J')$.
Would anyone know how to approach this?
Note: This question seems related, but what I would like to show is roughly a converse to what that question is trying to show.
As you noted, if $Y$ is closed, $Y=V(I)$ for some $I$, and one can assume that $I$ is a radical ideal. Suppose $V(I)$ is irreducible, and suppose $f,g\in R$ and $fg\in I$. If $X_{fg}$ denotes the usual basic open set of primes not containing $fg$, then $$ \emptyset=X_{fg}\cap V(I)=(X_f\cap V(I))\cap (X_g\cap V(I)) $$ as $X_{fg}=X_f\cap X_g$.
Since $V(I)$ is irreducible, at least one of $X_f\cap V(I)$ or $X_g\cap V(I)$ must be empty, since two nonempty open subsets of an irreducible space cannot be disjoint. If $X_f\cap V(I)=\emptyset$, then $f$ is contained in every prime ideal containing $I$, i.e., $f$ is in the radical of $I$, i.e., $f\in I$ since $I$ is radical. Similarly, if $X_g\cap V(I)$ is empty, $g\in I$. In any case, $I$ is a prime ideal.