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 $(1-\sqrt{10})$.