MaxSpec tends to be geometrically nicer than Spec in my experience (e.g $\operatorname{MaxSpec}(\mathbb{C}[z])$ looks the same as $\mathbb{C}$, no extra non-closed points), but Spec is preferred because the pre image of prime ideals is prime, which makes Spec a functor (and an equivalence of categories). What is the largest setting where MaxSpec suffices, a subcategory of commutative rings where the preimage of maximals are maximal?
The largest category where MaxSpec is “nice” enough to do algebraic geometry
algebraic-geometrycommutative-algebra
Best Answer
As mentioned in the comments, Jacobson rings are the key here.
Lemma (Stacks 00GB): Let $f:R\to S$ be a finite type morphism of rings with $R$ Jacobson. Then $S$ is Jacobson, and the preimage of a maximal ideal is maximal.
Thus if one restricts to the category of rings of finite type over a Jacobson ring, this would fit your criteria. One issue here is that as you vary your base, you'll get rings which work in one context but not the other. For instance, consider taking $\Bbb Z$ as your base versus $\Bbb C$: no nontrivial $\Bbb C$-algebra is finite type over $\Bbb Z$ for cardinality reasons, and there are examples where this lemma fails: consider the canonical injection $\Bbb Z[x]\to \Bbb C[x]$ and the preimage of $(x-\pi)$, for instance. So you have a bunch of different settings where this holds, but you can't "put them all together" (one more reason to take Spec!).
I would also like to push back against your contention that MaxSpec is geometrically nicer than Spec. Sure, there are more points that are a little strange to think about at first, but these points are geometrically very useful! To make an analogy, MaxSpec is like a Riemannian integral while Spec is like the Lebesgue integral - you can get a lot more done with it, the theory is nicer and more extensible, etc.