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Let $K = \mathbb Q(\alpha)$ is a number field. Suppose $\alpha \in O_K$ and let $f \in \mathbb Z[X]$ be its minimal polynomial. Show that if the discriminant of $f$ is a square-free integer, then $O_K = \mathbb Z[\alpha]$.

I don't really know how to approach this problem. I have observed that:

i) $ \mathbb Z[\alpha] \subseteq O_K$ trivially

ii) The problem is equivalent to showing that $\{1,\alpha,\ldots ,\alpha^{n-1} \}$ is an integral basis for $K$

I can't see how the discriminant is at all related, and I suppose this is where my understanding of this topic lacks. Any push in the right direction would be appreciated.

Thanks

Matt
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1 Answers1

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Let $\beta_1,\ldots,\beta_n \in O_K$ be an integral basis for $O_K$. Also, let $\alpha_1,\ldots,\alpha_n$ be the roots of $f$, where $\alpha = \alpha_1$. Then there exists $A \in \mathcal{M}_{n\times n}(\mathbb{Z})$ st $$\alpha_{j} = \sum^{n}_{i=1} A_{ij}\beta_{i}$$ for $i\leq j\leq n$. Then we have that $$\text{disc}(\alpha_{1},\ldots,\alpha_{n}) = \text{det}(A)^{2}\text{disc}(\beta_1,\ldots,\beta_n)$$Note that if $\text{disc}(\alpha_{1},\ldots,\alpha_{n})$ is squarefree, then we have that $\text{det}(A)^{2} = \pm 1$. So therefore $\alpha_{1},\ldots,\alpha_{n}$ generate $O_K$, and thus form an integral basis.

Finally, we note that $1,\alpha,\ldots,\alpha^{n-1} \in O_K$ are a basis for $K$, and we have that $$\text{disc}(1,\alpha,\ldots,\alpha^{n-1})=\text{disc}(f)=\text{disc}(\alpha_{1},\ldots,\alpha_{n})$$So therefore $1,\alpha,\ldots,\alpha^{n-1}$ are an integral basis for $O_K$, and $O_K=\mathbb{Z}[\alpha]$.

James Sibbit
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  • One possible definition of $Disc(O_K)=Disc(\alpha_{1},\ldots,\alpha_{n})$ is the $\det$ of the matrix $M_{ij}=Tr(\alpha_i\alpha_j)$, it needs more work to relate it to the ideal of $\Bbb{Z}$ generated by all the $Disc(\gamma)=\prod_{i\ne j} (\sigma_i(\gamma)-\sigma_j(\gamma)),\gamma \in O_K$ and $\sigma_j$ the complex embeddings – reuns Sep 07 '19 at 14:39