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While wandering through math stack exchange I found an interesting question, namely this one:

Why are the algebraic integers a Bezout Domain?

(Found here: Is there an elementary way to prove that the algebraic integers are a Bézout domain?)

Now, this follows easily from both the principal ideal theorem or by the fact that the class group of any ring of algebraic integers of a finite extension has torsion class group, as shown here: https://math.stackexchange.com/a/1303278/917010. This is a obvious consequence of the finiteness of the class number, but it is a much weaker fact, which I would like to prove indipendently.

I've now tried for a few days to prove this without any success. Of course we must at some point use the fact that we are not dealing with any Dedekind ring but with one which is the integral closure of the integers in a finite extension of $\Bbb{Q}.$ (Which is actually what makes of this a number theory question.) But I have found no plausible way to do this.

My motivation for this is that I am trying to understand Algebraic number theory and Class field theory, and such a proof would give a maybe clearer idea of the Bezout property.

Of course the argument with the Minkowski bound is clear to me, but I must admit that it does not give me a great understanding of what is going on.

In conclusion, and for the sake of clarity, I will re-state the question:

Why is the class group torsion? (Without using the finiteness of the class group)

I must thank you in advance for any effort. Worst case scenario, this will be a good a lesson on how strong is the theorem on the finiteness of the class group.

Jabberwocky
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    It's not that much weaker. Do you know any natural examples of infinite abelian groups which have finite exponent (which aren't just, say, $\mathbb{F}_p$-vector spaces)? It's not a condition that comes up much. – Qiaochu Yuan Dec 31 '22 at 02:41
  • Okay maybe I can grant you that "much" is maybe an overstatement, yet there should be quite natural examples (Besides the large and easy family you rule out). For instance the group of all roots of unity or $\mathbb{Q}/\mathbb{Z}$ should work. – Jabberwocky Dec 31 '22 at 11:27
  • I don't understand why you want to avoid showing that the class group is finite, the proof is easy (there is $C$ such that for all ideal $I$ there is $a\in I,(a)=IJ, N(J)\le C$ and $O_K/(C!)$ is a finite ring so it has finitely many ideals). – reuns Dec 31 '22 at 11:57
  • Because the existence of that C is derived through calculations which doesn't really clarifies a lot about what is algebraically going on the Bezout matter I stated as main reason for this question. This could be stated just as: Given two algebraic integers, they are two multiples of another algebraic number. It would be nice to have a clear way to explain why this is true. – Jabberwocky Dec 31 '22 at 17:16
  • Furthermore, two different proofs are always better than one, right? – Jabberwocky Dec 31 '22 at 17:16
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    @Jabberwocky: $\mathbb{Q}/\mathbb{Z}$ does not have finite exponent. I really do not believe there is a natural example here. – Qiaochu Yuan Dec 31 '22 at 19:05
  • @QiaochuYuan From the beginning I thought that by finite exponent we mean that any element of the group has finite order, which Is what I wanted to prove. Now I will edit the question accordingly. Thanks for making me notice this issue. – Jabberwocky Dec 31 '22 at 19:40
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    That condition is called "torsion." (Finite exponent means there is a fixed $n$ such that every element of the group is $n$-torsion.) Plausibly there might be a proof of this for the class group that doesn't prove finiteness but I'm not aware of one. – Qiaochu Yuan Dec 31 '22 at 20:50
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    I'd be curious to hear an example of (a reasonably natural) Dedekind domain with non-finite, but torsion, class group... At first I thought something like an appropriate infinite ascending union of rings of algebraic integers, but then I realized that "there're some technical issues..." :) – paul garrett Dec 31 '22 at 22:07
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    @paulgarrett Let $p$ be an odd prime, let $F$ be the algebraic closure of $\mathbb{F}p$ and let $A = F[x,y]/\langle y^2 - x^3-x \rangle$. The class group is isomorphic to the group of infinitely many $F$ points on the elliptic curve $y^2 = x^3+x$. However, any particular $F$ point is in the finite field $\mathbb{F}{p^k}$ for some $k$ and therefore lies in the subgroup of $\mathbb{F}_{p^k}$ points, which is finite. (This is basically your idea of an infinite ascending union, but noticing that the function field case is easier than the number field case.) – David E Speyer Jan 02 '23 at 16:31
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    @DavidESpeyer, ah, good! Thanks! :) – paul garrett Jan 02 '23 at 17:09

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