Reading around about the Nullstellensatz theorem, I saw that we can interpret the maps $V\to I(V)$ (where $V$ is an algebraic set) and $S\to Z(S)$ (where $S$ is an ideal), as functors which are adjoint, in some sense. I didn't find, however, more then side remarks about this. So I'm having a hard time trying to understand the full picture: what are the categories here, exactly? I can see how we can define the category of algebraic affine sets, but how do we think about the ideals as a category? Do we think of it as a subcategory of the modules over $k[\bar x]$? Looking at this question gives a way to do it, but why is it a good way- don't we want to save some of the algebraic structure?

Also, the radical ideals have some special property - the image of $I$ consists only of radical ideals. I wonder if there is a universal property for radical ideals, and whether it says something of value in the category of algebraic sets? Can it explain this special property?

I don't know a lot of category theory- just some basic definitions and constructions, so I am sorry if this make no sense at all, or if I am missing something trivial. Thanks in advance to anyone who can help in clarifying any of the points above, or maybe provide some references.

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    $\begingroup$ The categories are partially ordered sets and this special type of pair of adjoint functors is called Galois connection. $\endgroup$ – Ennar Oct 30 '18 at 20:50

As Ennar points out in their comment, the categories at hand here are the partially ordered sets of algebraic subsets and of ideals. Actually, one of them has to be the dual of said poset in order for $I$ and $Z$ to be order-preserving, i.e. functors.

Indeed, for any two posets $P,Q$, a functor between them seen as categories is the same as a (weakly) monotone map $P\to Q$. Here if we let $Alg$ be the poset of algebraic subsets, $Id$ the poset of ideals, we have $I: Alg^{op} \leftrightarrow Id : Z$ that are monotone (if $K\subset J$ are ideals, $Z(J)\subset Z(K)$ thus in $Alg^{op}, Z(K)\leq Z(J)$).

Still as Ennar points out, these two functors form an adjunction, better known in order theory as a Galois connection. This means that for any ideal $J$ and algebraic set $V$, $J\subset I(V) \iff V\subset Z(J)$, or in terms of orders (using the order of $Alg^{op}$ !) $J\leq I(V)\iff Z(J)\leq V$

The radical ideal here (I'm assuming we're working over an algebraically closed field) is nothing but the composition $I\circ Z: Id \to Id$. Standard category theory (or order theory) yields that for all $J\in Id$, $J\leq I(Z(J))=\sqrt{J}$ (which can be interpreted as the unit of the adjunction) .

What's very interesting is that, as we are faced with a Galois connection, the monad $I\circ Z$ on the category $Id$ is actually an idempotent monad, aka (for orders) a closure operator: $(I\circ Z)^2 = I\circ Z$, which is just the basic fact that $\sqrt{\sqrt{J}} = \sqrt{J}$.

How to see this ? Well $J\leq IZ(J)$ so $Z(J)\leq ZIZ(J)$, but also by the counit ($ZI(V)\leq V$) of the adjunction $ZIZ(J)\leq Z(J)$, so $Z(J) = ZIZ(J)$, and so the rest follows.

So what are radical ideals ? Radical ideals are fixed point of $IZ$; formulated differently they're the objects such that the unit ($J\leq IZ(J)$ ) is actually an isomorphism; formulated yet again in a different way they are the algebras for the monad $IZ: Id\to Id$ . This gives different points of view that lead to different generalizations.

Note that I could actually have replaced $Alg^{op}$ by $\mathcal{P}(\mathbb{A}^n)^{op}$ and similarly $Id$ by $\mathcal{P}(k[X_1,...,X_n])$ to have the "full" Galois-connection. Then restricting to the fixed points of the monad on one side, and the comonad on the other, we get the known bijection between algebraic subsets and radical ideals.


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