Exercise on Fundamentals of Divisibility: Factorization Domains Set $R=\mathbb{Z}[\sqrt{10}]$. Show that in $R$ every element $\alpha\not=0$ is a product of irreducible elements, but $R$ is not a unique factorization domain.
I have shown that $R$ is not a unique factorization domain, but I don't know how to prove that $R$ is a factorization domain. Any clues?
And here's my proof that $R$ is not a unique factorization domain:
Claim: 3 is irreducible in $\mathbb{Z}[\sqrt{10}]$.
Proof:
Suppose 3 reducible in $\mathbb{Z}[\sqrt{10}]$. Then there exists $a,b\in \mathbb{Z}[\sqrt{10}]$ such that $3=ab$, and $a,b\not\in\mathbb{Z}[\sqrt{10}]^{\times}$. Let $N:\mathbb{Z}[\sqrt{10}]\rightarrow \mathbb{Z}$ and $N(\alpha)=|\alpha\bar{\alpha}|^2$. Then $N$ is multiplicative and if $\alpha$ is a unit of $\mathbb{Z}[\sqrt{10}]^{\times}$, $N(\alpha)=1$, since $1=N(\alpha\alpha^{-1})=N(\alpha)N(\alpha^{-1})$, and the only divisors of $1$ are $\pm1$, and $N(\beta)$ is either postive or $0$ in $\mathbb{Z}$ . But then $9=N(3)=N(ab)=N(a)N(b)$, implying $N(a)=\pm3$. And since $a=x+\sqrt {10} y$ where $x,y\in\mathbb{Z}$, $x^2 + 2xy\sqrt {10} +y^2 = 3$, implying that $2xy=0$ and $x^2+y^2=3$. Thus either $x$ or $y$ are zero since $\mathbb{Z}$ is an integral domain, implying in the case that $x=0$, $y^2=3$, and when $y=0$, $x^2=3$, contradiction. Thus 3 is irreducible in $\mathbb{Z}[\sqrt{10}]$. 
$\\$
Claim: 3 is not prime in $\mathbb{Z}[\sqrt{10}]$.
Proof:
Since $3(-3)=(1+\sqrt{10})(1-\sqrt{10})$, it follows that if 3 is prime then $3|(1+\sqrt{10})$ or $3|(1-\sqrt{10})$. This implies $(1+\sqrt{10})$ or $(1-\sqrt{10})$ is reducible in $\mathbb{Z}[\sqrt{10}]$ or related to 3. If $(1+\sqrt{10})$ is reducible then for some $x,y\in\mathbb{Z}[\sqrt{10}]$, $(1+\sqrt{10})=xy$ where $x,y\not\in\mathbb{Z}[\sqrt{10}]^{\times}$. then $9=N(1+\sqrt{10})=N(x)N(b)$, implying $N(x)=3$, which is a contradiction as shown before. $(1-\sqrt{10})$ follows similarly. So then $3$ and $(1+\sqrt{10})$ are related or $3$ and $(1-\sqrt{10})$. In the case where $(1+\sqrt{10})|3$, this implies there exists $a\in\mathbb{Z}[\sqrt{10}]^{\times}$ such that $(1+\sqrt{10})a=3$. So $a=x+\sqrt{10}y$, where $x,y\in\mathbb{Z}$. Then $x + 2(x+y)\sqrt{10} +10y=0$, implying $x+y=0$ and $x+10y=0$, implying $a=0$. But $N(a)=0$, so $a$ not a unit, contradiction, and similarly shown for $(1-\sqrt{10})$. 
 A: Since $10\neq 1 \pmod 4,$ $\mathbb{Z}[\sqrt{10}]$ is the ring of integers of $\mathbb{Q}(\sqrt{10}).$ The ring of integers of a number field is always a Noetherian domain.

We show that every Noetherian domain $R$ is a factorization domain like this:
Suppose $a_0\in R$ is not zero or a unit, but not expressible as a product of irreducibles. Then in particular, it can not be irreducible itself, so we can factorize $a_0 = bc$ where neither $b$ or $c$ are units. Hence we have strict containments $Ra_0 \subset Rb$ and $Ra_0 \subset Rb.$ Now if both $b$ and $c$ could be written as a product of irreducibles, $a_0$ could be, so let $a_1=b$ if $b$ is not a product of irreducibles, otherwise set $a_1=c.$ Thus $Ra_0 \subset Ra_1$ and $a_1$ is not zero or a unit and not expressible as a product of irreducibles. We can repeat this process indefinitely to get the strictly ascending chain $$ Ra_0 \subset Ra_1 \subset Ra_2 \subset Ra_3 \cdots $$
contradicting the fact that $R$ is Noetherian. 

Also, about how you did the rest of the problem, have you checked the norm you used is indeed a multiplicative norm? Usually the norm on that ring is $N(a+\sqrt{10}b) = a^2 - 10b^2.$ To show $3$ is irreducible with this norm we have to show $a^2-10b^2=3$ has no solutions, which follows from the fact that $3$ is not a quadratic residue mod 5. 
For the second part proving $3$ is not prime, once you see $3|1\pm\sqrt{10}$ it is very easy to finish off: Your ring lies inside $\mathbb{C}$ and $3$ can only divide those numbers if $\dfrac{1\pm \sqrt{10}}{3} \in \mathbb{Z}[\sqrt{10}].$
A: The existence of factorizations into irreducibles follows easily using the norm function $N(a+\sqrt{10}b) = a^2 - 10b^2$, in the same way for $\mathbb Z$ with the absolute value. The argument is along the lines below:


*

*Prove that $\delta \mid \alpha \implies |N(\delta)| \le |N(\alpha)|$.  This follows from the multiplicativity of the norm.

*Prove that $\delta \mid \alpha, \ \delta \mbox{ not a unit} \implies |N(\delta)| < |N(\alpha)|$. This follows from $|N(\delta)|=1$ if $\delta$ is a unit.

*Note that every $\alpha$ is either irreducible or can be written $\alpha = \beta \gamma$, with $\beta$ and $\gamma$ not units. Conclude by induction on $|N(\alpha)|$ that $\beta$ and $\gamma$ can be factored as a product of irreducibles and so can $\alpha$.
A: The ring $\Bbb{Z}[x]$ is Noetherian by the Hilbert Basis Theorem. It follows that 
$\Bbb{Z}[\sqrt{10}] = \Bbb{Z}[x]/(x^2 + 10)$ is also Noetherian. It now suffices to understand why a Noetherian domain is also a factorisation domain.
A: Hint $\ $ If not,  a nonunit $\alpha\ne 0$ has factorizations with an unbounded number of nonunit factors (keep splitting non-atoms) hence unbounded norm, since each nonunit factor has norm $\ge 2$.
