0

I'm trying to arrive at a proof that every nonzero $\alpha \in \mathbb{Z}[i\sqrt{n}]$ ($n \in \mathbb{Z}^{+}$ square-free) that is not an unit can be factored into the product of irreducible elements in $\mathbb{Z}[i\sqrt{n}]$.$\\$

I also have the following norm defined: $N(a+ib\sqrt{n})=a^2+nb^2$ and the prior knowledge that $N(\alpha)=1 \iff \alpha$ is an unit of $\mathbb{Z}[i\sqrt{n}]$

I thought of proving by induction on $N(\alpha)=n \geq2$ as $\alpha$ is not an unit.

If we have $\alpha \in \mathbb{Z}[i\sqrt{n}]$ with $N(\alpha)=2$ then:$\alpha=\beta \gamma \implies N(\alpha)=N(\beta)N(\gamma) \implies 2= N(\beta)N(\gamma) \implies N(\beta)=1 \lor N(\gamma)=1 \implies \beta \in \mathbb{Z}[i\sqrt{n}]^{\ast} \lor \gamma \in \mathbb{Z}[i\sqrt{n}]^{\ast}$

Suppose that $N(\alpha)=n>2$ and that the result is valid for $N(\beta)< n$. If $\alpha$ is reducible in $\mathbb{Z}[i\sqrt{n}]$, $\alpha=\omega\zeta$ with $\omega;\zeta \notin \mathbb{Z}[i\sqrt{n}]^{\ast} \cup \{0\} $. But now as $1<N(\omega);N(\zeta)<n$, we have by I.H. that $\omega$ and $\zeta$ are product of irreducibles in $\mathbb{Z}[i\sqrt{n}]$ and it follows that $\alpha$ is product of irreducibles in $\mathbb{Z}[i\sqrt{n}]$.

J P
  • 385
  • But these rings are not UFD for $n\ge 3$, see here. – Dietrich Burde Jan 09 '24 at 17:03
  • @DietrichBurde, is the result above a sufficient condition for $\mathbb{Z}[i\sqrt(n)]$ with $n \in \mathbb{Z}^{+}$ square-free to be an UFD? – J P Jan 09 '24 at 17:06
  • 1
    No. In fact, it is enough that $R$ is Noetherian, so that every nonunit can be written as a finite product of irreducible elements. Of course your ring is Noetherian, because it is a quotient of $\Bbb Z[x]$, which is Noetherian. So what you are trying to prove follows from this post. – Dietrich Burde Jan 09 '24 at 17:08

0 Answers0