Quotient ring of $\Bbb Z[x]$ by an irreducible polynomial is a PID

Question

I don't know what can I do with this problem.

How can I prove that $\mathbb{Z}[x]/(x^{3}-4x+2)$ is PID?

Answer 1

Let $K=\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $f(x)=x^3-4x+2$. Then $f(x)$ has discriminant $-148$, so that $disc(1,\alpha,\alpha^2)=148=2^2\cdot 37$. Since $disc(\mathcal{O}_K)$ differs from $disc(1,\alpha,\alpha^2)$ only by a square, it follows $disc(\mathcal{O}_K)=37$, or $disc(\mathcal{O}_K)=148$.

Consider the ideal $P=(2,\alpha)$. Then $P^3=(2)(4,2\alpha,\alpha^2,\alpha-1)=(2)$, since the second ideal contains $1$ and is the whole ring. Hence $2$ ramifies, and this is equivalent with $2\mid disc(\mathcal{O}_K)$, so that $disc(\mathcal{O}_K)=disc(1,\alpha,\alpha^2)$. It follows that $\mathcal{O}_K=\mathbb{Z}[\alpha]\simeq \mathbb{Z}[x]/(x^3-4x+2)$, and this is a Dedekind ring.

Then Minkowski’s bound is given by $3!\sqrt{148}/3^3\simeq 2.703$, which shows that every class in the class group is represented by an ideal whose norm at most $2$, where it is used that $f(x)$ has three real roots. Thus every ideal class is represented either by id or by $P$. But $P$ is the unique ideal with norm $2$, and it is easy to see that it is principal (use that $N(\alpha)=-2$). It follows that the ring is a PID.