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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 $\mathbb{Z}[\sqrt{10}]$.


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 $\mathbb{Z}[\sqrt{10}]$.


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

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up vote 2 down vote accepted

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 $\alpha$.

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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 $\dfrac{1\pm \sqrt{10}}{3} \in \mathbb{Z}[\sqrt{10}].$

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The first fact you quoted above is not trivial to prove (I think you need some algebraic number theory to prove it). – user38268 Oct 9 '12 at 10:24
@BenjaLim Certainly easier to prove than Hilbert's Basis theorem though. You only need about as much Algebraic Number theory to prove it as you do to know what the ring of integers of a number field is. In showing that the algebraic integers form a subring of the algebraic numbers, you usually show that if $\alpha \in \mathbb{A}$ then $\mathbb{Z}[\alpha]$ is finitely generated as a $\mathbb{Z}$ module, which is equivalent to being Noetherian. – Ragib Zaman Oct 9 '12 at 10:30
I agree, IIRC one just needs to exhibit a $\Bbb{Z}$ - basis for $\Bbb{Z}[\alpha]$, which can be done by considering the integral dependence relation that $\alpha$ satisfies. – user38268 Oct 9 '12 at 10:41

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.

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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 $\ge 2$.

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