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I wrote a proof and would like to see how sturdy it is. I'm very new to this subject, and am curious. I wrote: We have $I \vartriangleleft Z$. Then for $x,y\in I$ and $z\in \mathbb{Z}$, $x - y \in I$, $xz,yz\in I$. Then the set $I$ can be represented as $\{ax+by\mid a,b \in \mathbb{Z}, x,y \in I\}$. Without loss of generality, consider only positive linear combinations. Then, by the well ordering axiom, $I\ge0$ has a least element. Call it $d$. Then by Bezout's theorem, all elements in $I$ are a multiple of $d$, and is the ideal generated by $d$. Thus every ideal in the ring of integers is a principal ideal.

I would greatly appreciate any feedback!

ultralegend5385
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rlaivsezlt
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  • This is false. Rings of integers are not in general a PID. Maybe you are thinking about ideals in $\mathbb{Z}$. – ShyamalSayak Feb 25 '24 at 22:16
  • Everything between "$I\triangleleft \mathbb{Z}$" up to just before "by the well ordering axiom" is unnecessary and never used anyway. – Arturo Magidin Feb 25 '24 at 22:25
  • Please search for answers before posting questions. For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Feb 25 '24 at 22:42
  • Apply the Theorem in the linked dupe, using $\Bbb Z$ is Euclidean using the magnitude $|n|$ as the Euclidean size of $n,,$ i.e. remainders have smaller magnitude. – Bill Dubuque Feb 25 '24 at 22:44

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