Localization of ring of integers at discriminant

Question

Let $K$ be a number field with ring of integers $\mathcal{O}_K$. Let $\alpha \in \mathcal{O}_K$ that has degree $n$ over $\mathbb{Q}$, and let $d = D_{K/\mathbb{Q}}(1, \alpha, ..., \alpha^{n-1})$. Then show that $$(\mathcal{O}_K)_d = \mathbb{Z}[\alpha]_d $$ where $(\mathcal{O}_K)_d, \mathbb{Z}[\alpha]_d$ are the localization of these rings at $d$.

Answer 1

Lemma: Let $A$ be an integrally closed integral domain with field of fractions $F$ and let $L$ be a finite separable extension of $F.$ Let $B$ be the integral closure of $A$ in $L.$ Let $\alpha_1,\dots,\alpha_n$ be a basis of $L/F$ which is contained in $B,$ with discriminant $d = d(\alpha_1,\dots,\alpha_n).$ Then $$ dB\subseteq A\alpha_1+A\alpha_2+\cdots+A\alpha_n. $$

Proof: Let $\beta = \sum_{i = 1}^n a_i\alpha_i\in B$ be arbitrary, with the $a_i\in K.$ It follows that $$ Tr_{L/K}(\alpha_i\beta) = \sum_{j = 1}^n Tr_{L/K}(\alpha_j\alpha_i)a_j. $$ Now, $Tr_{L/K}(\alpha_i\beta)\in A,$ so solving the system of equations above for the $a_i$ implies that the $a_i\in\frac{1}{\det(Tr_{L/K}(\alpha_i\beta))}A = \frac{1}{d}A.$ Thus, $da_i\in A,$ and hence $d\beta\in A.$ $\square$

Set $A = \Bbb{Z},$ $F = \Bbb{Q},$ $L = K,$ $B = \mathcal{O}_K,$ and $\alpha_i = \alpha^{i-1}.$ The lemma tells us that $$ d\mathcal{O}_K\subseteq \Bbb{Z}+\Bbb{Z}\alpha + \dots + \Bbb{Z}\alpha^{n-1} = \Bbb{Z}[\alpha], $$ and as $\Bbb{Z}[\alpha]\subseteq\mathcal{O}_K,$ we also have $d\Bbb{Z}[\alpha]\subseteq d\mathcal{O}_K\subseteq\Bbb{Z}[\alpha].$ Thus, we have inclusions of rings $$ \Bbb{Z}[\alpha]\hookrightarrow\mathcal{O}_K\hookrightarrow\Bbb{Z}\left[\alpha,\frac{1}{d}\right]. $$ Localizing the above away from $d,$ we obtain $$ \Bbb{Z}[\alpha]_d = \Bbb{Z}\left[\alpha,\frac{1}{d}\right]\hookrightarrow(\mathcal{O}_K)_d\hookrightarrow\Bbb{Z}\left[\alpha,\frac{1}{d}\right]_d = \Bbb{Z}\left[\alpha,\frac{1}{d}\right]. $$ Since the maps above are inclusions of sets (viewing everything as subsets of $K$), we find that $(\mathcal{O}_K)_d\subseteq\Bbb{Z}[\alpha]_d = \Bbb{Z}\left[\alpha,\frac{1}{d}\right]$ and conversely that $\Bbb{Z}[\alpha]_d = \Bbb{Z}\left[\alpha,\frac{1}{d}\right]\subseteq(\mathcal{O}_K)_d,$ which implies the desired equality.