Is $\mathbb{Z}[\sqrt[3]{2}]$ a UFD?

Question

Is the quotient ring $\mathbb{Z}[X]/(X^3-2)$ a UFD? I don’t have any idea to find whether some ring is a UFD or not.

Answer 1

Let $K=\mathbb{Q}(\sqrt[3]{2})$ and $O=\mathbb{Z}[\sqrt[3]{2}]$. Assume first that $O=O_K$, then $O$ is a UFD iff it is a PID.

The absolute discriminant can now be easily computed: it’s $108$. The Minkowski bound is $\sqrt{108}\frac{4}{\pi}\frac{6}{27} = \sqrt{3}\frac{16}{3\pi} < 4$.

So it is enough to show that any ideal of norm less than $4$ is principal.

Well, $(2)=(\sqrt[3]{2})^3$, $(3)=(1+\sqrt[3]{2})^3$ so by standard arguments on number fields, $O$ is a PID and we’re done.

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Now, we only need to show that $O$ is normal. Indeed, let $z=a+bx+cx^2 \in O_K$ (with $x=\sqrt[3]{2}$) with $a,b,c \in \mathbb{Q}$, we want to show that $a,b,c$ are integers.

Multiplying by $1,x$ or $x^2$ and taking the trace, we find that $3a,6b,6c$ are integers. If $3b$ or $3c$ isn’t an integer, then $3z$ equals, up to an element of $O$, $\frac{x}{2}$, $\frac{x^2}{2}$, or $\frac{x+x^2}{2}$, none of which is in $O_K$ (by norm). So $3z \in O$.

As I’ve written above, $3 \in O^{\times}(1+x)^3$, so if $\mathfrak{p}$ is a prime above $3$, it contains $1+x$ and $3z \in \mathfrak{p}$ so $3a-3b+3c=3z - 3b(x+1)-3c(x-1)(x+1) \in \mathfrak{p} \cap \mathbb{Z}=(3)$ so $a-b+c$ is an integer.

It leaves nine possibilities for $3z$ mod $3O$: $0,1+x,1-x^2,x+x^2,1-x+x^2,2+2x,2-2x^2,2x+2x^2,2-2x+2x^2$. If $y$ is a member of this list and $3z \in 3O+y$ then $y/3 \in O_K$ and thus $27$ divides the norm of $y$. But it’s easy to check that it’s only the case when $y=0$, and thus $z \in O$, QED.