# Insight of some concepts in commutative algebra

I really enjoyed the basic algebra course and wanted to teach myself a little more. So I am trying to learn commutative algebra from Atiyah-MacDonald and Eisenbud.

The department in our university is very good for analysis based subjects. I have enjoyed courses like basic analysis, measure theory and probability theory. I have a clean insight of what a theorem is trying to say, in these subjects, and I can make a mental picture before I formulate and prove propositions rigorously. In algebra, I dont seem to have this insight. I wish I could speak to people proficient in commutative algebra to get an insight and find the right style of thinking in this field. By the way, I have read this thread.

I will make the question specific:

1) I do not understand the idea of an ideal quotient. I know it's definition and I can prove the properties listed in Atiyah-MacDonald's book. But yet I have no idea of the big picture. What is it's purpose? How can I spot an ideal quotient?

2) I came across an idea called 'exact sequences' in Eisenbud's intro to modules. In two swift examples, he constructs exact sequences. I can verify that the second example is indeed an exact sequence. But I could not figure out how he constructed such an example!!

The second example was the exact sequence:

Given a ring $R$, an ideal $I \subset R$, $a \in R$

$0 \to \dfrac{R}{(I : a)} \to \dfrac{R}{I} \to \dfrac{R}{I + (a)} \to 0$

Is there an insight to this construction that, sort of, lets me guess the 'exact sequence' relation between the objects?

P.S: I will have more specific questions as I read along. Thank you for your answers.

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To answer your first question: I never really got the feel for ideal quotients until I realised it can be used to prove the following two facts:

1. The set of zero divisors in a commutative ring $A$ is a union of prime ideals. The way I proved this at first was using Krull's Lemma, but you can also use ideal quotients as in here:

2. A ring in which every prime ideal is finitely generated is Noetherian.

If you would like general advice for commutative algebra, I am not an expert but here's what I can say. When I first started commutative algebra I had 0 intuition. One thing I realised that was helpful as I was going along was to draw a commutative diagram. For example, you should be aware that if $A \subset B$ is a finite ring extension, then given any $Q \subset A$ a prime ideal there is always a prime ideal $P \subset B$ lying over $A$. If you draw a commutative diagram, the proof of this fact is not hard. Also, since you mentioned Eisenbud, I remember proving some isomorphism in there using just a big diagram chase. So, draw a diagram if you can!

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Thank you for your comment. I proved 1) from chapter 1, exercise 14 in Atiyah-MacDonald, (using Zorn's Lemma). I don't know Krull's lemma. Lemme try using ideal quotients to prove those problems. – Isomorphism Jul 13 '12 at 20:12
@BenjaLim: Would you mind reopening your recent algebraic geometry question? I was just about to post an answer that I'm not sure if it works, and I was hoping to see your and others' feedback. – Zev Chonoles Jun 15 '13 at 9:57
@ZevChonoles Dear Zev, the reason why I deleted the question was because I solved it. Also, I don't know how to get to my deleted question. Do you have a link? – user38268 Jun 15 '13 at 11:16
@BenjaLim: Here's a link (I just went through my browser history). Also, I didn't mean to pressure you - if you don't want to reopen now that you've solved it, that's fine of course. – Zev Chonoles Jun 15 '13 at 17:32

2) contains the ideal quotient $(I:a)$, it really comes from multiplication with an element in the ring: $$\mu_a : R \rightarrow R \qquad r \mapsto ar$$ then you compose with the projection of an arbitrary ideal $I$ $$\mu_a : R \rightarrow R/I \qquad r \mapsto [ar]=ar+I$$ this has the image $(a)+I/I$ in $R/I$, and the kernel is by definition $(I:a)$. If you use the isomorphism theorems you immediately get the exact sequence, which is a short way to state a lot of information

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Actually I have worked that out. My question was: how do we guess, a priori, that there is a relation between these different quotient rings? So an exact sequence is just a short hand notation? Thank you :) – Isomorphism Jul 13 '12 at 20:07
I think it is guessing and a matter of experience to spot exact sequences, there is no general rule. – Blah Jul 14 '12 at 10:53
Actually, it is much more then a short hand notation - but this takes you into the realm of homological algebra! – Blah Jul 14 '12 at 10:54