In his great answer to this MO question James Borger gives a geometric characterization of non-unital $k$-algebras (i.e. not necessarily unital) : They correspond to affine schemes $X$ over $k$ together with a morphism $\text{Spec}(k) \to X$ over $k$. More generally, what about defining a non-unital scheme over a base scheme $S$ as a usual scheme $X$ over $S$ together with a morphism $S \to X$ over $S$? The idea is that this section "cuts out the unital part" of $X$.

Question: In the same spirit as algebraic geometry answers questions about commutative algebra and vice versa, is there some non-unital algebraic geometry (perhaps with the above definition of non-unital schemes) which interacts with non-unital commutative algebra? What are explicit applications?

See this MO question for some important examples of non-unital rings. So my question is basically if we can study them via some kind of spectral theory. Note that the naive approach, just imitating the unital case by considering the set of prime ideals, does not work at all since there are not enough prime ideals to establish some basic facts about the structure sheaf, even if we assume something like $A*A = A$.

For non-unital rings in which every element is idempotent, i.e. boolean rings, the spectral theory should yield the non-unital Stone duality out of the unital Stone duality. Likewise with some modifications (restriction to $\mathbb{C}$-points) we should get the non-unital Gelfand duality out of the unital Gelfand duality.

  • $\begingroup$ Talking from a distant memory, isn't it that the category of all modules over a nonunital ring is equivalent to the category of unital modules over the unital extension. So if one is to describe spectra from categories of modules, it does not buy that much. On the other hand, for the relative situations, the modules over (graded) non-unital monads are sometimes useful in description of quasicoherent modules over projective schemes. $\endgroup$ – Zoran Skoda Mar 23 '11 at 20:04
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    $\begingroup$ @Zoran: I was about to object, and then realized that I was confusing bundles and sheaves. As you may know, the intuition for this free unital extension is that it's the "one-point compactification" of the "(possibly) noncompact space" encoded by the (possibly) nonunital algebra. (If the space is compact, the one-point compactification is just disjoint unioning a point; if it's not compact, then the new point is near any end of the space.) The one-point compactification has the property that it has the same collection of open sets as the original space had; hence they have the same sheaves. $\endgroup$ – Theo Johnson-Freyd Mar 23 '11 at 22:12
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    $\begingroup$ Theo, while your reasoning is correct I think that this is the intuition from the case of (commutative) Banach algebras. The unital and nonunital case there are radically different; the Zariski topology would come from exact localizations and the nonunitality does not bring any news there. I was talking about bare rings, not topological rings, algebraic geometry, not operator algebras, so I am a bit cautious about the "compactification" here. I also have in mind the case of noncommutative rings. $\endgroup$ – Zoran Skoda Mar 24 '11 at 0:20

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