$B_{\mathfrak p}$ not always a simple extension of $A_{\mathfrak p}$? Let $B$ be the integral closure of some ring of integers $A$ in an extension of number fields, and let $\mathfrak p$ be a prime of $A$.  I've seen an example where $B$ is not a simple extension of $A$.  And I know that given $\alpha \in B_{\mathfrak p}$, we always have $B_{\mathfrak p} = A_{\mathfrak p}[\alpha]$ provided the product of discriminants $D(\alpha)D(B/A)^{-1}$ is a unit at $\mathfrak p$.  
However, is there an example of a prime $\mathfrak p$ where $B_{\mathfrak p}$ cannot be generated as a simple extension of $A_{\mathfrak p}$ by any element of $B_{\mathfrak p}$ whatsoever?  For example, what if $D(B/A)$ had value $2$ at $\mathfrak p$, and the discriminant of any element in $B_{\mathfrak p}$ could never be made to have equal value?
Serge Lang writes in ANT: "The hypothesis that $B = A[\alpha]$ is not always satisfied, but if we are interested in the decomposition of a single prime $\mathfrak p$, then it suffices to look at the localization $B_{\mathfrak p}$ over $A_{\mathfrak p}$, and in that case $B_{\mathfrak p}$ can be generated by a single element except for a finite number of exceptions."
So, does anyone have an idea of such an exception?  Obviously this would have to happen when $\mathfrak p$ is ramified.  EDIT: Oops, the exception I found is when $\mathfrak p$ is not ramified.
 A: Let $K = \mathbb{Q}$ and $E = \mathbb{Q}(\sqrt{7}, \sqrt{10})$.  We will show that $S := \mathcal O_{E,(3)}$ is not a simple extension of $R := \mathbb{Z}_{(3)}$.  We define $$\alpha_1 = (1 + \sqrt{7})(1+ \sqrt{10})$$ $$\alpha_2 = (1 + \sqrt{7})(1 - \sqrt{10})$$ $$\alpha_3 = (1 - \sqrt{7})(1 + \sqrt{10})$$  $$\alpha_4 = (1 - \sqrt{7})(1 - \sqrt{10})$$ These are all elements of $S$, all conjugate over $\mathbb{Q}$, for which the product of any two $\alpha_i \alpha_j$ ($i \neq j$) is divisible in $S$ by $3$.  For example, $\alpha_1 \alpha_2 = (1 + \sqrt{7})^2 \cdot (-9)$.  It follows that modulo $3S$, we have for any $n \geq 1$ and any $i$ that $$Tr_{K/\mathbb{Q}}(\alpha_i^n) = Tr_{K/\mathbb{Q}}(\alpha_1^n) = \alpha_1^n + \alpha_2^n + \alpha_3^n + \alpha_4^n \equiv (\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4)^n$$ where this last congruence holds because, as we said, any product $\alpha_i \alpha_j$ for $i \neq j$ is $0$ modulo $3S$.  We compute $\sum\limits_{i=1}^4 \alpha_i$ and see that this is exactly $4$.  Thus $Tr_{K/\mathbb{Q}}(\alpha_i) \equiv 4^n \equiv 1$ modulo $3S$, hence modulo $3S \cap R = 3 \mathbb{Z}_{(3)}$, which implies that $\frac{\alpha_i^n}{3}$ does not lie in $S$ (if $\alpha_i^n = 3x$ for $x \in S$, we would have $Tr_{K/\mathbb{Q}}(\alpha_i^n) = 3 Tr_{K/\mathbb{Q}}(x) \in 3 \mathbb{Z}_{(3)}$).  \
Now, we suppose by contradiction that $S = R[\alpha]$ for some $\alpha \in S$.  Let $f \in R[X]$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$.  For $g \in R[X]$, let $\overline{g}$ be the image in $R/3R[X]$, which we can identify with $\mathbb{F}_3[X]$.  It is clear that $g(\alpha)$ is divisible by $3$ in $S = R[\alpha]$ if and only if $\overline{g}$ is divisible by $\overline{f}$ in $\mathbb{F}_3[X]$.  For $1 \leq i \leq 4$, we can write $\alpha_i$ as $f_i(\alpha)$ for some $f_i \in R[X]$.  \
Since $\alpha_i \alpha_j = f_i(\alpha) f_j(\alpha)$ (for $i \neq j$) is divisible by $3$, and $\alpha_i^n = f_i(\alpha)^n$ is not, we have that $\overline{f}$ divides each product $\overline{f_i} \cdot \overline{f_j}$, but does not divide any power $\overline{f_i}^n$.  The extension $\mathbb{Q}(\sqrt{7}, \sqrt{10})/\mathbb{Q}$ is Galois, so $\overline{f}$ should factor into irreducibles as either $$\overline{f} = \overline{p_1}^2 \overline{p_2}^2$$ $$\overline{f} = \overline{p_1} \cdot \overline{p_2} \cdot \overline{p_3} \cdot \overline{p_4}$$ $$\overline{f} = \overline{f}$$ With the situation we described, the only possibility is that $\overline{f}$ splits into four linear factors.  But this is impossible, since there are only three linear factors in $\mathbb{F}_3[X]$.  \
This also shows that $\mathcal O_E$ is not a simple extension of $\mathbb{Z}$, and that the discriminant $D(1, \alpha, \alpha^2, \alpha^3)$ of $E/ \mathbb{Q}$ for any $\alpha \in S$ with $E = \mathbb{Q}(\alpha)$ is divisible by $3$ (use Lang, Chapter III, Proposition 16), even though the discriminant of $E/ \mathbb{Q}$ (which will divide each $D(1, \alpha, \alpha^2, \alpha^3)$ when $\alpha \in \mathcal O_E$) is not divisible by $3$.  This is because $E$ is a compositum of two unramified extensions which are both unramified at $3$, hence $3$ is also unramified in $E$.
