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Let $K=\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $f(x)=x^3+x+1$. If $p$ is a rational prime. What can you say about factorization of $p\mathcal{O}_K$ in $\mathcal{O}_K$?

I have this: The discriminant is -31 and the minkowski bound, $M_K \leq 1.57$. Then $N(\mathcal{A})=1$, $\mathcal{A}$ a ideal class. I can say anything about the question?

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The Discriminant of the $\mathbb{Q}$-basis $\{1, \alpha, \alpha^2\}$ is $-31$ which is square free, and contained in $\mathcal{O}_K$. From this can you conclude that $\{1, \alpha, \alpha^2\}$ is a basis for $\mathcal{O}_K$? You should be able to (this is a standard result).

If you convince yourself that the ring of integers is indeed $\mathbb{Z}[\alpha]$, then try to use the following (also well-known ?) result:

Let $p$ be a rational prime. Now $f = x^3 + x + 1$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Reduce $f$ mod $p$ to obtain $\overline{f}$, and let $\overline{f} = \prod_i h^{e_i}_i$, be an irreducible factorization of $\overline{f}$ in $\mathbb{F}_p[x]$. Let $f_i \in \mathbb{Z}[x]$ such that $f_i$ mod p $= h_i$. Then the ideal $P_i = (p, f_i(\alpha))$ is prime in $\mathcal{O}_k$, and

\begin{equation}

(p) = \prod_i P^{e_i}_i

\end{equation}

From your observation that $M_K \leq 1.57$, you further know that $\mathcal{O}_K$ is a PID, so all you $P_i$ are principal.

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