You have proper descent for IndCoh (which in your setting is just the ind-completion of the bounded derived category of coherent sheaves.) Does that suffice for your purposes?

It sounds like you're asking for compactifications of the space of conics. One compactification that is often useful is the space of complete conics. This can be described in various other ways (e.g., as the blowup of $\mathbb{P}^5$ along the Veronese, as a Kontsevich space, etc...)

A comment: One reason why base change fails is that the classical fiber product $X\times_S S'$ may forget some information remembered by the derived fiber product. If instead you work with derived schemes you can get more general statements; see the derived algebraic geometry section of en.wikipedia.org/wiki/Base_change_theorems.

@DmitriPavlov That being said, the unfortunate reality at the moment is that the published literature lags behind "folklore."

@DmitriPavlov See for instance Section 1.2 of arxiv.org/abs/1310.5127v2. The original philosophical source is arxiv.org/abs/math/0508382, but that was written at a time when the foundations were insufficient to make such a statement rigorous.

...(reality is not quite as clean as this outline, but hopefully it shows how useful birational geometry is.) I think all these considerations are geometric in nature. At the very least, it's very hard for me to see any of the above reasoning as algebraic (no equations show up anywhere). Maybe it doesn't fit "easily visualized" (because algebraic surfaces have 4 real dimensions) but in that case I'm not sure anything in algebraic geometry past curves is geometric by your definition.

...very important for understanding the behavior of such curves. E.g., if I want to classify rational curves on some given surface $X$, the first step to do so is to know where in the classification of surfaces $X$ fits into. Say for instance that $X$ is rational.... then the classification tells me that it's an iterated blowup of $\mathbb{P}^2$ or a Hirzebruch surface, so if I can understand 1. curves on those surfaces and 2. how the behavior changes upon blowing-up then I can say a lot about rational curves on any rational surface...

Re: "given a picture of a surface without its equation, I doubt that you'll be able to tell whether it is rational, Enriques or K3." Sure I can: if the surface contains families of rational curves, it's rational. (OK, this is a bit facetious, since maybe you give me a picture where I can't tell if it contains rational curves.) It's still unclear to me what you consider geometric, but since you mention enumerative geometry, maybe you'd include the study of rational curves on varieties? In which case it is worth mentioning that birational geometry is...

Can you give examples of questions you consider "geometric"? (I suspect the answer to your question will overwhelmingly be that birational geometers think of the field geometrically, but your question seems to imply that you don't consider something like "are these two varieties birationally equivalent?" a geometric question.)

The problem with this plan is that the Diophantine equations generated from discrete computational problems are usually a mess, i.e., very complicated with no clear arithmetic structure. Modern number theory doesn't really have anything to say about such equations.

It is indeed possible that such a curve class cannot be perturbed away from $\Sigma.$ But what does happen is that such a curve class will be the difference of curve classes that can be perturbed away. In the setting that Will mentions (surfaces), you can do this by adding a sufficiently large multiple of an ample divisor.

@Mare I suspect that in general, imposing a relation like $xy+2yx$ instead of $xy+yx$ will make it much easier to find finite dimensional algebras. (The reason being that the quotient has trivial center, unlike the case of your post.)

I suspect what Hidalgo has in mind is the following representation-theoretic argument (which can also be rephrased in elementary terms): The automorphisms $a_1,a_2,a_3$ are part of a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ action. The pullback formula tells you that each $\theta_{r,\alpha}$ generates a character (1-dim representation) of the group, and the characters corresponding to different $(r,\alpha)$ are different. Therefore there can be no nontrivial relations between them. I don't really see why the divisor formula is necessary though.