Prove that $\alpha^2-3\alpha+1$ is not in the ideal $(5,\alpha+2)$ in $\mathbb{Z[\alpha]}$ where $\alpha=2^{1/3}$

Question

I'm thinking about the theorem 3.24 in Marcus 'Number Fields', that is,

Let $p$ be a prime in $\mathbb{Z}$, and suppose $p$ is ramified in a number ring $R$. Then $p|disc(R)$.

I am considering $\mathbb{Z[\alpha]}$ where $\alpha=2^{1/3}$ and $p=5$ as an example, and I know that $5\mathbb{Z[\alpha]}=(5,\alpha+2)(5,\alpha^2-3\alpha+1)$. Also I proved that $||(5,\alpha^2-3\alpha+1)||=25$, so $(5,\alpha+2)$ is prime.

To verify that $5\mathbb{Z[\alpha]}$ is not prime, I should prove that $\alpha^2-3\alpha+1$ is not in the ideal $(5,\alpha+2)$ and $(5,\alpha^2-3\alpha+1)=Q^2$ where Q is prime in $\mathbb{Z}[\alpha]$ but I can't do this.

Anyone help me?

Answer 1

Suppose that $\alpha^2-3\alpha+1$ is in that ideat that I call $I$,

$\alpha(\alpha+2)$ is in $I$ implies that $\alpha^2-3\alpha+1-\alpha(\alpha+2)=-5\alpha+1\in I$, this implies that $5(\alpha+2)-5\alpha+1=9\in I$, this implies that $2(5)-9=1$ in $I$, can you show that $I$ is not $\mathbb{Z}[\alpha]$?

Answer 2

Let $\alpha=2^{\frac{1}{3}}$, and let $R=\mathbb{Z}[\alpha]$.

In the ring $R$, let $I=(5,\alpha+2)$, and let $J=(I,\alpha^2+3a+1)$.

We want to show that $\alpha^2+3\alpha+1 \notin I$.

Equivalently, we want to show $I\ne J$.

Analyzing the ideal $J$, we have \begin{align*} &\alpha^2+3\alpha+1 \in J\\[4pt] \implies\;&(\alpha+1)(\alpha+2)-1\in J\\[4pt] \implies\;&-1\in J\qquad\text{[since $\alpha+2\in J$]}\\[4pt] \implies\;&J=(1)\\[4pt] \end{align*}

Thus, we want to show $I \ne (1)$.

Suppose instead that in the ring $R$, we have $I = (1)$.

Our goal is to derive a contradiction.

Note that $R$ is isomorphic to the ring $\mathbb{Z}[x]/(x^3-2)$.

By the correspondence theorem, $I=(1)$ in the ring $R$ if and only if $I'=(1)$ in the ring $\mathbb{Z}[x]$, where $I'=(5,x+2,x^3-2)$.

Working in the ring $\mathbb{Z}[x]$, we have $$I'=(5,x+2,x^3-2)=(5,x+2)$$ since $x^3-2 = (x^2-2x+4)(x+2)-(2)(5)$.

By assumption, we have $1\in I'$, hence we can write $$f(x)(5) + g(x)(x+2)=1$$ for some $f,g\in \mathbb{Z}[x]$.

But then, substituting $x=-2$ into the above equation, we get $$f(-2)(5) = 1$$ contradiction.

This completes the proof.