Hilbert's Nullstellensatz says that if $A $ is every ideal in $k [x_1,\dots x_n] $, $k $ an algebraically closed field, then $I (V (A)) = \sqrt A;$ I'm using the notation of Mumford, that should be well known.
Now take the ring $\mathcal D$ defined in this way: its elements are the ideals of $k [x_1,\dots x_n] $, the product $A\cdot B $ is the ideal $ (A \cup B )$, and the sum $A+B $ is the ideal $A\cap B $. With these definitions, $\mathcal D$ is actually a ring. Then take the ring $\mathcal S $ defined in this way: its elements are the algebraically closed sets of $\mathbb A^n $, the product $\Sigma_1\cdot \Sigma_2 $ is $ \Sigma_1 \cap \Sigma_2 $, and the sum $\Sigma_1+\Sigma_2 $ is $\Sigma_1\cup \Sigma_2 $. Clearly $\mathcal S$ is a ring too. Take the homomorphism of rings $V$ and $I $ as above: consider a sequence $$\dots \to \mathcal D \xrightarrow {V } \mathcal S \xrightarrow {I } \mathcal D \xrightarrow {V } \mathcal S \to \dots$$ One cannot say that it is exact. However if we restrict to the set $\mathcal A \subset \mathcal D $ of ideals in a radical ideal $A'$ (so $A'=\sqrt {A'}$), we have the sequence $$\dots \to \mathcal A \xrightarrow {V } 0 \xrightarrow {I } \mathcal A \xrightarrow {V } 0 \to \dots$$ Notice that this sequence is well defined because every element in $\mathcal A $ is sent from $V $ to the same set $\Sigma $ (so we use the ring with only one element, namely $0$) and $I (0)=0$. In fact, $I (\Sigma)=A' $, and considering the way we defined products and sums in $\mathcal D $, if we restrict to $\mathcal A$ is clear that $A'$ is the $0$ of $\mathcal A$. So it's trivial that the sequence is exact.
I have a very poor knowledge of algebraic geometry and categories; does what I wrote makes sense? I haven't found anything about it. Thanks in advance for any help.