# Is there an elementary way to prove that the algebraic integers are a Bézout domain?

Well, the title of my question says all, but let me give some context. Right now I'm writing some lecture notes on ring theory with a little of commutative algebra. I wrote a few results about integral extensions and that led me to define the algebraic integers $\overline{\Bbb Z}$ as the integral closure of the extension $\Bbb Z\subseteq \Bbb C$. Since I included a topic on factorization in integral domains where I defined what is a Bézout domain, it seems very natural to me to show that $\overline{\Bbb Z}$ is an example of a Bézout domain.

However, and here comes my problem, I haven't found an elementary proof of the above fact. The only reference that I have is a theorem in Kaplansky's book "Commutative rings". I checked the proof given in that book, but I can't understand it very well, moreover Kaplansky uses terminology from Algebraic Number Theory as "class group" and certainly I can't use any of that in my notes because they are aimed to cover a basic/intermediate course of ring theory.

In summary, is there a ring-theoretic way to prove that the ring of algebraic integers is a Bézout domain?

P.S. I'm aware of this post and the only answer uses Algebraic Number Theory, which as I said, I can't use.

• Iirc Dekekind highlighted this Theorem as one that is trivial to prove using his ideal theory, but much more difficult without such. Towards such, you may find helpful related work that I cite in this Mar 31, 2005 sci.math post in the thread "JSH: Critique means slow, and thorough", which apparently sparked some comments HD1207 by M. Zafrullah. See also this answer. – Bill Dubuque Aug 18 '17 at 18:09
• Mr. @Bill Dubuque thanks for the links. – Xam Aug 18 '17 at 18:13