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Background

Exercise: Let $R$ be an integral domain Prove that $R$ is a PID if and only if (i) every ideal of $R$ is finitely generated and (ii) whenever $a,b\in R$, the sum ideal $(a)+(b)$ is principal.

Solution: If $I$ is an ideal of $R$ then by (i), $I$ is generated by some $a_1,\ldots,a_n$. That is $I=(a_1)+\cdots+(a_n)$, Choose $n$ minimal here and suppose $n>1$. Then by (ii) $(a_1)+(a_2)=(c)$ for some $c,$ and $I=(c)+(a_3)+\cdots+(a_n)$ is a sum of $n-1$ principal ideals, contrary to the minimality. Therefore $n=1$ and $I$ is principal.

Questions:

At first I thought the solution to the exercise above, I would solve the exercise, I would simply use induction on the number of ideals in (ii) along with the condition of (i) to show that the ideal $I$ is principal. So I am not sure what contradiction is trying to derive in the presented solution, also why does the issue of minimality on the number of ideals matter?

Seth
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    An ideal is not a PID. A,single ideal is principal, but only the whole ring can be called a PID. – Thomas Andrews Apr 08 '25 at 16:30
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    Your proof uses induction on the natural numbers, their proof uses that the natural numbers are "well-ordered." Well-ordering and induction are equivalent, and well-ordering proofs usually involve proof by contradiction. – Thomas Andrews Apr 08 '25 at 16:34
  • @ThomasAndrews can you elaborate on how that is related to the minimallity issue. The solution is a bit confusing. Because I really thought that I would just apply induction on condition (ii) to arrive at the solution. Maybe there is something else that I am missing. – Seth Apr 08 '25 at 16:35
  • The well-ordering principle says every non-empty set of natural elements has a minimal element. So it saying, "if there are counter-examples, there is a least counter-example." – Thomas Andrews Apr 08 '25 at 16:37
  • But it is really exactly the same as a proof by induction. – Thomas Andrews Apr 08 '25 at 16:39
  • Related (but not really helpful). Seth, I agree with you that their proof is overcomplicated compared to yours. As for the contradiction in their proof: they choose $n$ minimal, and prove that $n$ cannot be $>1$, so $n=1$ (or $0$). – Anne Bauval Apr 08 '25 at 16:39
  • @ThomasAndrews If I bring in the equivalence of finitely generated ideals: Noetherian and maximal condition (for ideals), then I simply apply induction. The equivalence is asked to be shown in other exercises for an integral domain that is PID. – Seth Apr 08 '25 at 16:40
  • @ThomasAndrews I did not know the minimality condition has to do with reference to well ordering. I thought it was minimality on the number of ideals which has nothing to do with well ordering. Or at least it did not explicitly stated. I feel llke reading that proof is an exercise in textual hermeneutics. – Seth Apr 08 '25 at 16:47
  • @AnneBauval thank you for the clarification. – Seth Apr 08 '25 at 16:47
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    The most natural reading of "contrary to minimality" is that it contradicts an assumption that something is minimal that we made earlier in the proof. Well-ordering is just cultural context; you do not need to know anything about it to follow this proof, you just need to know about it to see in which way this is similar to (but, in my opinion, simpler than) a proof by induction. – Misha Lavrov Apr 08 '25 at 16:50
  • @ThomasAndrews but why does it need minimality on the number of ideals for showing an ideal is PID? If the proof can be done by induction, it seems strange. – Seth Apr 08 '25 at 16:53
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    @Seth Some people are taught to prefer well-ordering proofs rather than induction. It is really that simple. – Thomas Andrews Apr 08 '25 at 17:03
  • @ThomasAndrews ah kk. I understand what you mean. Thank you. – Seth Apr 08 '25 at 17:05
  • @MishaLavrov Thank you also for the explanation. – Seth Apr 08 '25 at 17:07
  • @Misha But to be rigorous you do need to invoke some equivalent to $\Bbb N$ being well ordered, whether it be nonempty subsets have a minimal element, or no infinite descending chains, or (transfinite) induction, or various pigeonhole principles. – Bill Dubuque Apr 08 '25 at 17:09
  • @Seth This is the ubiquitous minimal criminal form of induction. That you are studying commutative algebra w/o knowing such basics is yet another example of your attempting to run before learning to walk. I've lost count of how many of your questions are due to lacking study of necessary prerequisites. Doing so is pedagogically maximal criminal! $\ \ $ – Bill Dubuque Apr 08 '25 at 17:20

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