Dualizing object in the duality between commutative rings and affine schemes

Many dualities between geometry and algebra arise via a dualizing object. Roughly, if $\mathcal C$ is a category of spaces and $\mathcal D$ a category of "algebras", one often finds a dualizing object $R$ which lives in both categories $\mathcal C$ and $\mathcal D$ such that the constructions $C\mapsto \hom_\mathcal C(C,R)$ and $D\mapsto \hom_\mathcal D(D,R)$ constitute an equivalence or at least adjunction between $\mathcal C$ and $\mathcal D$.

For instance, Pontrjagin duality ($R=\mathbb R/\mathbb Z$), Stone duality ($R=\mathbb Z/(2)$), Gelfand duality ($R = \mathbb C$), and the fundamental theorem of Galois theory ($R=\bar k$, for $k$ a field) arise in this way.

Question: What is the dualizing object in the duality between affine schemes and commutative rings?

(Second question: What is the dualizing object in the fundamental theorem of covering spaces, which roughly states that the category of covering spaces over a space $X$ is equivalent to the category of all sets equipped with an action of the fundamental groupoid on that set? On the one hand, it is not a duality, so maybe it doesn't have a dualizing object in the strict sense, but since this statement is very similar to the fundamental theorem of Galois theory, maybe there's something similar.)

Solution 1:

Your question presumes that a dualising object exists. My view is that it does not exist. On the affine scheme side, the affine line $\mathbb{A}^1$ has a ring structure and represents the $\textbf{Aff}^\textrm{op} \to \textbf{CRing}$ half of the equivalence – this is unproblematic. But in the philosophy of the dualising object conception of duality, there would have to be a ring $R$ that "is" in some sense also $\mathbb{A}^1$ (in a "covariant" way!) that "represents" the $\textbf{CRing}^\textrm{op} \to \textbf{Aff}$ half of the equivalence and I do not see any reasonable way of interpreting this precisely.

The situation is a lot better if we instead look at affine varieties over an algebraically closed field $k$ and integral domains finitely generated over $k$. The dualising object is then $\mathbb{A}^1_k$ as a variety and $k$ as an algebra. This is "obviously correct": as before, $\mathbb{A}^1_k$ has a ring structure, and moreover the set of points of $\mathbb{A}^1_k$ is canonically identified with the set of elements of $k$, so there is a good sense in which we can think of $\mathbb{A}^1_k$ and $k$ as being "the same". Furthermore, $k$ does in fact represent the functor that sends an integral domain finitely generated over $k$ to the set of points of the affine variety it corresponds to.

Given the above, it would seem that the ring incarnation of $\mathbb{A}^1$ should be $\mathbb{Z}$ – but, frankly, this is unconvincing. While it is true that we can canonically identify the elements of $\mathbb{Z}$ with certain points of $\mathbb{A}^1$, there are many more points besides. It is also difficult to say that the functor represented by $\mathbb{Z}$ is a functor that sends a ring to the set of points of the scheme it corresponds to – at least, it certainly does not represent the functor sending a ring $A$ to the set of prime ideals of $A$. For these (and other) reasons, I think it is not reasonable to say that there is a dualising object in this story.

Solution 2:

From affine schemes to ring the functor is represented by the affine line $\mathbb A^1=\mathrm{Spec}\mathbb Z[x]$ equipped with a ring object structure. In the reverse direction, it is unclear what you would mean since there is no obvious functor from affine schemes to sets, and hence it's unclear which functor the dualizing object would have to represent.