Classical affine algebraic sets correspond to radical ideals, and irreducible sets to prime ideals

The ring is a finite-variable polynomial ring over algebraically closed $k$, so the strong Nullstellensatz applies with the assumed AC and gives $I(V(J))=\sqrt{J}$. The other composite is the identity on closed sets by F2. For radical $J$ the first composite is also the identity, proving the stated inverse bijections and their inclusion reversal.

If $X$ is nonempty irreducible and $fg\in I(X)$, then $X=(X\cap V(f))\cup (X\cap V(g))$. Irreducibility forces one of these closed sets to be $X$, hence $f\in I(X)$ or $g\in I(X)$. Also $1\mathrm{\notin }I(X)$ because $X$ has a point. Thus $I(X)$ is prime.

Conversely suppose $P=I(X)$ is prime. Then $X$ is nonempty, since $I(\mathrm{\varnothing })=R$. If $X=C\cup D$ with $C,D$ proper closed subsets, choose $f\in I(C)\setminus I(X)$ and $g\in I(D)\setminus I(X)$; these exist by the injective reversing correspondence in step 1.1. The product vanishes on $C\cup D=X$, contradicting primality. Thus $X$ is irreducible. For any prime $P$, the elementary implication $f^r\in P\Rightarrow f\in P$ makes $P$ radical, so step 1.1 realizes it by such an $X$.

For a point $a$, evaluation onto $k$ has kernel $I(\{a\})$. A proper ideal strictly containing this kernel would contain an $f$ with $f(a)\mathrm{\ne }0$; subtracting $f-f(a)$ in the kernel puts a nonzero constant in that ideal, hence 1. Thus the kernel is maximal. Conversely F5 writes each maximal ideal as $(x_i-a_i{)}_i$, whose locus is exactly $\{a\}$. The unit ideal has empty locus, and the empty set has vanishing ideal $R$.

Let $\pi :R\to A$ be the quotient and $H⊴A$. Its inverse image contains $I(X)$, so its zero locus lies in $X$ and equals $V_X(H)$. Polynomial vanishing upstairs gives $I(V_X(H))=\sqrt{{\pi }^{-1}H}$. Passing to the quotient using F6 and F7 yields $I_X(V_X(H))=\sqrt{H}$. This also proves the relative closed-set correspondence.