the ring of integers: prove that (2) is a prime ideal and that it is a pid

Question

Consider a real root $\alpha$ of $f(X)=X^3-3X+1$. Consider the ring of integers $A_K$ for $K=\mathbf{Q}[\alpha]$. I showed that the ideal $(1+\alpha)$ is prime in $A_K$ and that $A_K=\mathbf{Z}[\alpha]+(\alpha +1)A_K$. It is clear that the discriminant of $\mathbf{Z}[\alpha]$ is $3^4$. How can I prove that $A_K=\mathbf{Z}[\alpha]$ and that the ideal $(2)$ is prime in $A_K$ and $A_K$ is a PID? Thanks everyone for the help.

Answer 1

Let $f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_0$ be Eisenstein at $p$ (all $a_j$ are integers divisible by $p$ and $p^2 \nmid a_0$) and let $\alpha$ be a root of $f$. Then $p$ does not divide the index of $\alpha$, that is, if $$ \xi = \frac{r_1 + r_2 \alpha + \ldots +r_{n-1}\alpha^{n-1}}p $$ is an algebraic integer, then $p$ divides all the $r_j$. The proof is simple: $\xi \alpha$ is an algebraic integer, but $\alpha^n$ is divisible by $p$, hence $$ \xi_1 = \frac{r_1 \alpha + \ldots +r_{n-2}\alpha^{n-2}}p $$ is an algebraic integer. Repeating this step the claim follows.

In your case, $f(X-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$, and since the discriminant of this polynomial is a power of $3$, we find that $\{1, \alpha, \alpha^2\}$ must be an integral basis.