Let $\Delta$ denote the discriminant and $\Delta(R/\mathbb{Z})$ denote the discriminant ideal of $R$ over $\mathbb{Z}$. I'm having trouble proving the following:
Suppose $[\mathbb{Q}(\theta) : \mathbb{Q}] = n$ and $\theta \in R$ which is the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\theta)$. Then $\Delta(\theta)\mathbb{Z} = \Delta(R/\mathbb{Z}) \implies R = \mathbb{Z}[\theta]$.
Attempt: Let $X=\{x_0, \dots, x_{n-1}\}$ be a $\mathbb{Z}$-basis for $R$. I know from a lemma that $\Delta(\theta) = \det A^2 \Delta(X)$, where $A$ is the change of basis matrix between $X$ and $\{1, \theta, \dots, \theta^{n-1}\}$. So $\Delta(X) \mid \Delta(\theta)$, but also, $\Delta(\theta) \mid \Delta(X)$ since $\Delta(R/\mathbb{Z}) = \Delta(\theta)\mathbb{Z}$. This implies that $\det A = \pm1$.
Now, consider a typical element $y \in R$. Then \begin{align*} y &= \sum_{k=0}^{n-1} a_k x_k \\ & = \sum_{k=0}^{n-1} \sum_{j=0}^{n-1}a_kr_{jk}\theta^j \end{align*} where $(r_{jk}) = A^{-1}$. Since $A^{-1} = \frac{1}{\det A} C^T = \pm C^T$, where $C$ is the cofactor matrix of $A$. Since $C$ has only entries from $\mathbb{Z}$ we have that the $r_{jk} \in \mathbb{Z}$. Thus, $y \in \mathbb{Z}[\theta]$.
Question: Is this right? Also, someone told me that in general $|\det A| = [R : \mathbb{Z}[\theta]]$. I can't see why this would be true. Is there a way to understand this without going into Smith normal form(see here) ?
