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I'm trying to comprehend a proof from my Elementary Number Theory course. Here, a quadratic integer $\theta$ is a solution of an equation of the form $x^2 + bx + c = 0$ with $b$ and $c$ integers.

Let $R = \mathbb{Z}[\theta]$ be a quadratic integer ring, and suppose that for every $a, b$ in $R$ not both zero, there exists a linear combination of $a$ and $b$ that divides both $a$ and $b$. Show that every irreducible element of $R$ is prime, so that unique factorization holds in $R$.

Official solution: Let $r$ be an irreducible element of $R$, and suppose that $r$ divides a product $s t$ of elements of $R$. We must show that $r$ divides $s$ or $r$ divides $t$. Suppose that $r$ does not divide $s$. By hypothesis, there is a linear combination $n r+m s$ (for some $n, m \in R)$ that divides both $r$ and $s ;$ since no associate of $r$ can divide $s, n r+m s$ must be a unit, which without loss of generality we can assume equal to 1. Then $1=n r+m s$, so $t=n r t+m s t$, and since $r$ divides nrt and $m s t$ we must have $r$ divides $t$.

  1. I don't understand the line: ...is a linear combination $n r+m s$ (for some $n, m \in R)$ that divides both $r$ and $s ;$ since no associate of $r$ can divide $s, n r+m s$ must be a unit, which without loss of generality we can assume equal to 1. Why does "no associate of $r$ can divide $s$" imply that $n r+m s$ must be a unit? Also, is it correct that all units in quadratic integer rings are of form $\pm 1$?

  2. Does showing that every irreducible element is prime imply that the ring we are working in is a UFD? How can I come to that conclusion? (I factor into irreducibles, and using the prime property I continuously "delete" the same irreducibles from both sides of the equation, thus we have unique factorization. I hope this is correct?)

Again, partial answers (answering only one of the questions) are very welcome! I will upvote them too. If there are any other solutions to the above problem, not necessarily following the official solution, then those are also very welcome! Thanks in advance!

Albert
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    $r$ is irreducible and $nr+ms$ divides $r$ so either $nr+ms$ is a unit or it is $r$ times a unit. The latter case is impossible since we assume $nr+ms$ divides $s$ and $r$ doesn't divide $s$. – reuns Apr 10 '21 at 11:35
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    The unit group is either $\pm 1$ or $\pm 1,\pm i$ or $\pm 1,\pm e^{2i\pi /3}, \pm e^{4i\pi /3}$, but you don't need it here. – reuns Apr 10 '21 at 11:36
  • @reuns In the line, "without loss of generality we can assume equal to 1", I thought it was being used that $\pm 1$ are the only units in the ring $R$. Was I mistaken in this? Perhaps we don't need to use this, and there is some other way of understanding why assuming that the unit is $1$ loses no generality? – Albert Apr 10 '21 at 11:53
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    $Nr+Ms=u$ then $nr+ms = u^{-1}(Nr+Ms)$ – reuns Apr 10 '21 at 11:53

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It is a special case of a standard proof that irreducibles are prime in (Bezout $\Rightarrow$) gcd domains. The hypothesis implies that $\,\gcd(a,b)\,$ exists for all $\,a,b\,$ not both $\,0,\,$ since linear common divisors are always greatest, i.e. by hypothesis there are $\,r,s\in R$ with $\,d := ar+bs\mid a,b\,$ a common divisor, necessarily (divisibly) greatest, by $\,c\mid a,b\Rightarrow c\mid ra+sb = d.\,$

Below is the standard proof of Euclid's Lemma in gcd form and Bezout form

Lemma $\ \ (p,a)=1,\ p\mid ab\,\Rightarrow\, p\mid b,\ $ with proof as below

$\color{#c00}{\rm GCD}\text{ form:}\ \ \ \ \ \color{#c00}{(a,\ p)\ =\ 1},\ p\mid ab,pb\,\Rightarrow\, p\mid (ab,\ pb)\, =\ \ \color{#c00}{(a,\ p)}b\ =\ b$
$\color{#0a0}{\rm Bezout}\text{ form:}\ \ \ \color{#0a0}{ra\!+\!sp=1},\ p\mid ab,pb\,\Rightarrow\, p\mid rab\!+\!spb = (\color{#0a0}{ra\!+\!sp})b = b$

Bill Dubuque
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