Since the lemmas and theorems of algebraic geometry are beyond my knowledge and mathematical maturity at this point, I have struggled to somehow intuitively/visually grasp its basic concepts from the well-written book An Invitation to Algebraic Geometry by Karen E. Smith, et al and my best friends, Wikipedia and the MSE community.

Generally speaking, when something ends to the suffix -ization, it suggests the performance of an action. So, my question is:

Is the localization of a commutative ring at a prime ideal (or other multiplicative sets) in some sense a way of looking at the things locally?

I already have some thoughts on this that I would like to share them with you. Also, a discussion with @rschwieb on this question provoked my curiosity.

So, Hilbert's Nullstellensatz gives an algebro-geometric dictionary to think about varieties. In particular, when we work over an algebraically closed field, solutions of polynomial equations define geometric varieties that correspond to radical ideals in the polynomial ring over the field. This dictionary reverses the order of subset inclusion, translating bigger varieties to smaller radical ideals and vice-versa. A maximal ideal which is in some sense the biggest radical ideal becomes the smallest variety, a point. This dictionary also translates unions of varieties to intersections of their radical ideals. Therefore, a prime ideal is a radical ideal that its variety is in some sense indecomposable. Krull's dimension becomes a number that's supposedly the maximum dimension of these indecomposable sub-varieties as expected.

Now, when we localize a commutative ring by a multiplicative set, we get a ring that has only one maximal ideal. In particular, when we do this with the complement of a prime ideal $P$ as a multiplicative set, it feels like we are crushing every point of the variety outside of the indecomposable component corresponding to $P$ to a single point, which is the maximal ideal of the localized polynomial ring. It seems like the big picture works.

So, I am looking for more evidence about this. Maybe a few elementary explanations and well-versed related proofs. I am looking for more intuition. What results in algebraic geometry, in particular about localization, can be thought of as examples along this line of thought? Are the things I have said correct? Why is the localization of $\mathbb{Z}$ equal to $\mathbb{Q}$? Any pictorial explanation? Also, any more information regarding other concepts explained with this algebro-geometric language is welcome.

At the end, if moderators found my question suitable for Math Overflow, please do move my question to Math Overflow if you think I can get better explanations and answers there.

  • $\begingroup$ This question would not be appropriate for Math Overflow $\endgroup$ – Andres Mejia Dec 12 '17 at 23:43
  • $\begingroup$ @AndresMejia: And I haven't asked it there. :P $\endgroup$ – stressed out Dec 13 '17 at 0:07

I'm not an expert in this field, but here is my take:

Let's restrict our attention to a variety $X=\{x :P(x)=0\}$ for some polynomial $P \in k[x_1, \dots x_n]$. When we think about polynomials on this, they should belong to the ring $A:=k[x_1, \dots,x_n]/(P)$ (the quotient), which simply says that two polynomials $f(x) \sim g(x)$ if they agree at all points in $X$ (for all maximal ideals that contain $P$, which is to say that they are the same in the quotient.

However, we may want to know something about polynomials near some $\mathbf{a} \in X$. Well, the regular functions around the point are invertible iff they do not vanish at the point. Likewise, there are all the functions that happen to vanish at $a$. This is precisely what $A_{\mathfrak{m}}$, the localization at $\mathfrak{m}=(x_1-a_1, \dots x_n-a_n)$ describes! What these functions look like around that point is that they are invertible if they are not in $\mathfrak{m}$, and not if they are in $\mathfrak{m}$.

In this sense, if you believe that the general philosophy is: the algebraic content in the functions on the space $X$ tell us something about the geometry of $X$, then studying the local ring of functions, tells us what certain points look like.

For example, if given a curve $X$, a point $p \in X$ is nonsingular (local information) if and only if the local co-ordinate ring is a DVR (local ring that is also a principal ideal domain.) The process of localization does this.

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  • $\begingroup$ I do not understand the notation used in the first paragraph of your answer. :| I understand the second paragraph and the third paragraph, but again I have no idea about what a DVR is. $\endgroup$ – stressed out Dec 11 '17 at 9:05
  • $\begingroup$ @stressed-out for our purposes, a local ring that is also a principal ideal domain (the unique maximal ideal is principal.) $\endgroup$ – Andres Mejia Dec 11 '17 at 14:06
  • $\begingroup$ @stressed-out I also tried to finagle the first paragraph. I'm not clear what notation you were unfamiliar with, but I just changed some of it around. $\endgroup$ – Andres Mejia Dec 11 '17 at 14:06
  • $\begingroup$ The notation used in $Q(x) \mid_X=Q^{\prime} \mid_X$ was unfamiliar and a bit ambiguous at first. :P Plus, there are still a few other inconsistencies in notation. For example, you use $K[x_1,\cdots,k_n]$ and then $k[x_1,\cdots,x_n]$. The first one has a $k_n$ appearing in the last coordinate and in the last one $k$ has not been capitalized. + $X$ has not been defined in your first paragraph. I guess $X=\{x: P(x)=0\}$. Is DVR the short for a Discrete Valuation Ring? $\endgroup$ – stressed out Dec 11 '17 at 22:00
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    $\begingroup$ yes, it is short for a discrete valuation ring. $\endgroup$ – Andres Mejia Dec 12 '17 at 23:43

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