Decomposition of a prime in a tower of extension Let $ \mathbb{Q} \subseteq \mathbb{E} \subseteq \mathbb{F}$ be tower of extensions such that $[\mathbb{F}: \mathbb{E}]=[\mathbb{E}:\mathbb{Q}]=2$. Let $p$ be a prime such that $p \nmid \text{disc}_{\mathbb{E}/\mathbb{Q}}$ but $p | \text{disc}_{\mathbb{F}/\mathbb{Q}}$. Hence $p$ is unramified in $\mathbb{E}$ i.e. either $p\mathcal{O}_\mathbb{E}$ is a prime ideal or $p\mathcal{O}_\mathbb{E}=\mathfrak{q}_1\mathfrak{q}_2$ for some prime ideals $\mathfrak{q}_1,\mathfrak{q}_2 \subseteq \mathcal{O}_\mathbb{E}$ but $p$ ramifies in $\mathbb{F}$.
I would like to know how would the decomposition of $p{\mathcal{O}_\mathbb{F}}$ look like as a product of prime ideals of $\mathcal{O}_\mathbb{F}$.
Since $p$ ramifies in ${\mathcal{O}_\mathbb{F}}$, $p{\mathcal{O}_\mathbb{F}}= \prod \mathfrak{p}_i^{e_i}$ with at least one $e_i>1$. Can we be more specific about the decomposition of $p\mathcal{O}_\mathbb{F}$ considering the decomposition of $p\mathcal{O}_\mathbb{E}$? For example, can we say that $p\mathcal{O}_\mathbb{F}$ is either $\mathfrak{p}_1^2$ or $\mathfrak{p}_1^4$ or $\mathfrak{p}_1^2\mathfrak{p}_2^2$ for some prime ideals $\mathfrak{p}_1,\mathfrak{p}_2 \subseteq \mathcal{O}_\mathbb{F}$?
Any suggestions will be appreciated.
Thanks in advance.
 A: If $\newcommand{\OO}{\mathcal{O}}\newcommand{\EE}{\mathbb{E}}\newcommand{\FF}{\mathbb{F}}\newcommand{\QQ}{\mathbb{Q}}\newcommand{\pri}{\mathfrak{p}}\newcommand{\qri}{\mathfrak{q}}p\OO_\EE$ is prime, then $p\OO_\FF = \pri^2$, where $\pri$ is a prime ideal of $\OO_\FF$. This is because this is the only way for a prime to ramify in the quadratic extension $\FF/\EE$.
If $p\OO_\EE = \qri_1 \qri_2$, where $\qri_i$ are distinct prime ideals of $\OO_\EE$, then by assumption, one of these two prime ideals has to ramify in $\FF/\EE$, but the other one could be split, inert, or also ramified. So the factorization of $p \OO_\FF$ has one of the following three forms (where $\pri_1, \pri_2, \pri_3$ are prime ideals of $\OO_\FF$): $\pri_1^2 \pri_2 \pri_3$, $\pri_1^2 \pri_2$, or $\pri_1^2 \pri_2^2$. If $\FF/\QQ$ is a Galois extension, then only the last possibility (both $\qri_1$ and $\qri_2$ are ramified) can occur, since there's a $\QQ$-automorphism of $\FF$ interchanging $\qri_1$ and $\qri_2$.
Here's an example of how one of the other two possibilities can occur. Let $\EE = \QQ(i)$, where $i^2 = -1$. In $\mathbb{Z}[i]$, we have $5 = (2 + i)(2 - i)$. Let $\FF = \EE(\sqrt{2+i})$. (This is a $D_4$ extension of $\QQ$.) A quick computation in Sage (or you can do it by hand) shows that
$$(2 - i) \OO_\FF = (i\sqrt{2+i} - i + 1)(i \sqrt{2+i} + i - 1),$$
and these are distinct prime ideals of $\OO_\FF$. So
$$5 \OO_\FF = (\sqrt{2+i})^2 (i\sqrt{2+i} - i + 1)(i \sqrt{2+i} + i - 1).$$
I'll leave it to you to find an example where one prime ramifies and the other is inert, rather than split.
