$\newcommand\Spec{\operatorname{Spec}}
\newcommand\p{\mathfrak{p}}
\newcommand{\longmapsfrom}{\longleftarrow\!\shortmid}
\newcommand{\a}{\mathfrak{a}}
\newcommand{\p}{\mathfrak{p}}$Hello,
I am trying to find references for this result regarding properties of the closed sets of the spectrum of a ring:
Proposition. Let $A$ be a ring (commutative and unital). There is a one-to-one correspondence
\begin{align*}
\begin{Bmatrix}
\text{Closed sub-}\\\text{sets of }\Spec A
\end{Bmatrix}
&\longleftrightarrow
\begin{Bmatrix}
\text{Radical}\\\text{ideals of }A
\end{Bmatrix}\\
X&\longmapsto I(X)=\bigcap_{\p\in X}\p\\
V(\a)&\longmapsfrom\a
\end{align*}
Moreover, the maps $I(-)$ and $V(-)$ restrict to a one-to-one correspondence between the irreducible closed subsets of $\Spec A$ and the prime ideals of $A$, and we have $V(\p)=\overline{\{\p\}}$ for $\p\subset A$ prime. The latter implies that $\Spec A$ is a sober topological space.
(The first claim is "the Nullstellensatz for the spectrum of a ring.")
This result appears in Vakil's book, Theorem 3.7.1 and Exercise 3.7.F, and the proof is left as an exercise. However, I wanted to know if this proposition shows up in other standard AlgGeom or CommAlg books or references, if possible with proof (I already know the proof, I just want references). In particular, I haven't found it in Hartshorne's book.
Even though I am interested in finding a mention of the result in the literature, for completeness, here is a proof:
By logical reasons, for any ideal $\a\subset A$ one can verify that $V(I(V(\a)))=V(\a)$ (so the map $I(-)$ in the statement of the proposition is injective), and that $I(V(\a))\supset\sqrt{\a}$. If we use the axiom of choice, we can deduce the other contention and therefore that $I(V(\a))=\sqrt{\a}$ (Atiyah-MacDonald, Proposition 1.14). (This part of the proof was suggested by LT1918.)
Suppose now that $\p\subset A$ is prime. Then $V(\p)\cong\Spec(A/\p)$ as topological spaces (SP, tag 00E5), and the latter is irreducible since the spectrum of an integral domain is irreducible. Conversely, suppose $\a\subset A$ is a radical but non-prime ideal, and let $f,g\in A$ be such that $fg\in\a$ but $f,g\not\in\a$. Then $D(f)\cap V(\a)$ and $D(g)\cap V(\a)$ are non-empty (use that $\a=\bigcap_{\substack{\p\in\Spec A\\\p\supset\a}}\p$) open sets of $V(\a)$ whose intersection is $D(f)\cap D(g)\cap V(\a)=D(fg)\cap V(\a)=\varnothing$, so $V(\a)$ is reducible.
Finally, for the last claim, let $\p\in\Spec A$ and suppose $\p\in V(\a)$. Then $\p\supset\a$, so $V(\p)\subset V(\a)$. This means that $V(\p)$ is the smallest closed set containing $\p$, i.e., $V(\p)=\overline{\{\p\}}$.
Best Answer
This is Proposition 12.10 of these lecture notes from A. Gathmann.