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Those are the definitions given to us during our lecture:

Zariski Topology

We call a Zariski Topology in $\mathbb{K}^n$ a family of complements of $V\left(I\right)=\left\{a\in \mathbb{K}^n\::\:\forall _{f\in I}\:\:f\left(a\right)=0\right\}$, where $I\subset \mathbb{K}\left[X_1,\:...,\:X_n\right]$ is an arbitrary ideal. In other words, the algebraic sets $V\left(I\right)$ are a family of closed sets

Prime Spectrum

Let R be an arbitrary ring. For any ideal $I\subset R$ we define the set $V\left(I\right)=\left\{p\in Spec\:R\::\:I\:\subset p\right\}$ as the prime spectrum

What I'd like to know is what the motivation behind those concepts is, why were they created in the first place? What's their use, and how can I interpret them (e.g. geometrically or visually)?

user
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  • See answers and comments here: https://mathoverflow.net/questions/21502/what-is-the-zariski-topology-good-bad-for/21529 and https://math.stackexchange.com/questions/1798667/why-was-sheaf-cohomology-invented – KCd Mar 28 '21 at 13:18
  • Not quite for what I am looking for. I am just looking for the simplest explanation and interpretation of Zariski Topology, with a simple example so I can build an intuition to this. The link you've send me is too complicated, we haven't had that much material yet (that's the reason I avoid MathOverflow because apparently it's too advanced). The thing I get from Zariski Topology is that its sets (open and closed) are polynomials, but because we have a topology here, what I don't quite see is how the properties should hold here (like metrics), and maybe even how I can draw it/show it visually – user Mar 28 '21 at 13:24
  • I mean it's clear that the set of algebraic sets are just some polynomials, so it's easy to visualize it. But here, although the closed sets are also similarly defined, we have the additional condition that our polynomials should also belong to an ideal I, and this is what I don't quite see visually. – user Mar 28 '21 at 13:29
  • The intuition comes from looking at zero sets of polynomials in affine or projective spaces. Zero sets have properties if the closed sets for a topology, and classically zero sets of a “function” should be closed. Wouldn’t you want the graph of solutions to a polynomial equation to be closed? Since zero sets (except that of the zero polynomial are so small compared to the whole space, the open sets are really huge (complements of curves in the plane are an example), so it makes sense that the topology is not Hausdorff. Exercise: show a ball in $\mathbf C^n$ is dense in the Zariski topology. – KCd Mar 28 '21 at 13:34
  • The intuition comes from classical algebraic geometry over algebraically closed fields: subsets of points in affine and projective space, not the abstraction to sets of polynomials in a polynomial ring. An ideal in a polynomial ring “corresponds” to the set of common zeros of the polynomials in the ideal. That is, an ideal wants to be the set of polynomials vanishing on a common set in affine or projective space. The Zariski closure of a subset of affine or projective space is related to the full zero set of the ideal. – KCd Mar 28 '21 at 13:38
  • Ok, and what about the Prime Spectrum? What is the motivation to this, because at least as we've defined it it's just the set of all prime ideals in which I is a subset of them. But why do we even bother in the first place? – user Mar 28 '21 at 13:39
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    The motivation for that is not purely geometric, but in part algebraic: in classical algebraic geometry (have you studied that before: Nullstellensatz and so forth?) maximal ideals correspond to points and prime ideals correspond to irreducible varieties, but for a ring homomorphism $f:A\to B$, the inverse image of a maximal ideal need not be maximal while the inverse image of a prime ideal is always a prime ideal. Prime ideals are more robust than maximal ideals. – KCd Mar 28 '21 at 13:49
  • "(have you studied that before: Nullstellensatz and so forth?)" no, I'm just in my sixth semester, and the only things we had was a basic introductory course into Topology, and now having Commutative Algebra our Prof has introduced those concepts because he wants us to do some exercises (so in the end nothing was really explained and I just don't find it satisfying studying and learning something I don't even know the motivation of). "the inverse image of a maximal ideal need not be maximal" that's strange because we just proved in our course that it's always the case – user Mar 28 '21 at 13:52
  • A subring of a field need not be a field, but a subring of an integral domain is always an integral domain. That is why the inverse image of a maximal ideal need not be a maximal ideal. If they are for the particular rings you use, that just reflects something special about those rings. It is definitely not true all the time (try inclusion map of $\mathbf Z$ into $\mathbf Q$). – KCd Mar 28 '21 at 13:57
  • I think we just proved that for f being an epimorphism, so you might be right – user Mar 28 '21 at 13:58
  • I’m sorry to say that without taking some time to learn classical algebraic geometry (at least the Nullstellensatz and its immediate consequences for varieties in affine or projective space) then the motivation and intuition for the Zariski topology will be completely missing for you. It would be like trying to learn topology while having no experience with metric spaces. – KCd Mar 28 '21 at 14:01
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    See https://math.stackexchange.com/questions/161884/why-zariski-topology Maybe that page has better explanations for you than the earlier ones I pointed out: – KCd Mar 28 '21 at 14:06
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    That's also kind of the feeling I have right now, reading the links you've sent me. But then it's strange from our Prof to introduce us to that just for the sake of being able to do some exercises (proving some properties in rings and ideals). But on the other hand, I've seen this multiple times where Profs just started using Functionals, Manifolds etc without knowing that we don't know this yet – user Mar 28 '21 at 14:06
  • That link was useful, thanks, now I know how to imagine it – user Mar 28 '21 at 14:16

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