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Commutative Algebra is difficult.

Many things are agreeable and I usually can understand the proof (as syntactic manipulation), but could not create one as I don't understand any motivation at all. So I would like your help filling the missing pieces. To me, understanding the definition without understanding why it is defined in certain ways is difficult.

Specifically, (correct me if I'm wrong), I understand that we have curves in some affine space that we could "model" as affine domain, i.e. $R := k[x_1, \dots, x_n]/\mathfrak p$ for some prime ideal. The localization of the ring $R$ at some maximal ideal $\mathfrak m$ is the neighborhood of the point (in the affine domain) corresponding to $\mathfrak m$.

What is a localization at some prime $\mathfrak p$ in this picture? Are we intersecting the curve of $R$ to the curve of $\mathfrak p$? If so, is quotienting with $\mathfrak p$ similar to union?

Thanks for reading and potentially helping out! Appreciate any comments!

my99n
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    I think these are good questions, but this is too broad for this site. Even answering one of these questions properly requires a full answer. Commutative algebra and algebraic geometry are vast and difficult subjects. I think the best way for you to find answers is by reading a book, e.g. Commutative Algebra with a View Toward Algebraic Geometry or Undergraduate Commutative Algebra. – Joe May 19 '25 at 17:24
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    This question is quite broad - any one of those bullet points would make a good nucleus for a question on its own. Please consider either editing this question down substantially or posting a new question which focuses on only one of these issues. – KReiser May 19 '25 at 17:26
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    Please ask one question per post. Also, I suggest you take the tour with a focus on the heading *Get answers to practical, detailed questions*. – Lee Mosher May 19 '25 at 17:27
  • (Both of the texts that I have linked to explain commutative algebra from a geometric standpoint.) Having learned commutative algebra about a year ago, I also found (Mostly) Commutative Algebra and the classic Atiyah-Macdonald very helpful. – Joe May 19 '25 at 17:29
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    I also think this question is too broad, and should be split into multiple individual questions. Some of these individual questions are likely to be duplicates -- for instance here and here for your bullet point 1. In broad strokes, though, in addition to the excellent textbook Joe recommended, I'll recommend Richard Borcherds's very approachable lecture series on the subject. – Chris Grossack May 19 '25 at 17:30
  • Thanks. I will check out those materials first and ask again. – my99n May 19 '25 at 17:33

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I can answer the "graded ring" question partly. In topology, we have groups $H^i(X)$ associated to a topological space $X$ and called cohomology groups. We also have things called "bundles" that are like cartesian products, but possibly with a 'twist' of some sort. Just as a cylinder looks like the cartesian product of a circle, $S^1$, with the real line $\mathbb R$, a Mobius strip looks almost like that, but with a twist. We call each of these a line bundle over a circle.

If we have two bundles $\eta$ and $\phi$ over the same space $X$, we can take a "direct sum" of them (the details don't matter) to get a new bundle over $X$. Topologically, this makes a ton of sense.

To any bundle $\eta$ over a space $X$, there is, for each $i$, a cohomology class $w_i(\eta) \in H^i(X)$, called the $i$th Steiffel-Whitney class. These things are enormously important, for they provide an algebraic tool for telling whether two spaces (or bundles over those spaces) are the same or not.

There's a formula for $w_i(\eta \oplus \phi)$ that looks something like this:

$$ w_i(\eta \oplus \phi) = w_0(\eta) w_i(\phi) +w_1(\eta)w_{i-1}(\phi) + \ldots w_{i}(\eta)w_0(\phi), $$ where here we're using a "product" that takes something in $H^k$ and something in $H^p$ and gives a result in $H^{k+p}$. This formula is nice, but when we allow ourselves to think of the whole graded ring consisting of all the cohomology groups, we can define $$ w(\eta) = w_0(\eta) + w_1(\eta) + ... $$ where each term lives in a different cohomology group in the graded ring. When we do this, the 'Whitney sum formula" becomes $$ w(\eta \oplus \phi) = w(\eta)w(\phi). $$ That alone is enough to make the idea of a graded group seem natural to me.

John Hughes
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  • Unfortunately, the question was altered to no longer ask about graded rings about 2 minutes before your answer was submitted. – KReiser May 19 '25 at 17:36
  • Well, this is very helpful to me but I edited the question to the simpler one first while trying to formulate another post for graded ring. What should do I then? Edit this post again? – my99n May 19 '25 at 17:40
  • Nah...just leave it (and your comment). Leaving your comment will make me look like less of an idiot who seems to be answering an un-asked question. :) – John Hughes May 20 '25 at 00:59