Let $K$ be a field, $K\_s$ its separable closure, $K$ $\subseteq$ $F$ $\subseteq$ $K\_s$ an extension with $[F:K]$ $=$ $n$, $R$ $\subseteq$ $K$ a Dedekind domain with quotient field $K$, $S$ the integral closure of $R$ in $F$, and $\mathfrak{p}$ a maximal ideal of $R$.



> Can we find a $d>n$ with $(d,n)=1$ and an extension $K$ $\subseteq$ $\widetilde{K}$ $\subseteq$ $K\_s$ with $[\widetilde{K}:K]$ $=$ $d$ such that $\widetilde{K}$ and $F$ are linearly disjoint over $K$ (in $K\_s$) and, if $\widetilde{R}$ denotes the integral closure of $R$ in $\widetilde{K}$, the integral closure of $R$ in $\widetilde{K}F$ is equal to $\widetilde{R}S$, and $\mathfrak{p}$ is inert in $\widetilde{K}$ (i.e. $\mathfrak{p}\widetilde{R}$ is a prime ideal of $\widetilde{R}$) ?
> 


It is okay to assume that $K$ and $F$ are algebraic number fields, or even that $R$ $=$ $\mathbb{Z}$.


In the case where $K$ and $F$ are algebraic number fields, if $d$ $>$ $n$ is a prime number, any extension $K$ $\subseteq$ $\widetilde{K}$ of degree $d$ is linearly disjoint to $F$. And when the discriminants of $S$ and $\widetilde{R}$ over $R$ are relatively prime in $R$, the integral closure of $R$ in $\widetilde{K}F$ equals $\widetilde{R}S$. (Cf. Fröhlich and Taylor, *Algebraic Number Theory*, Ch. III, 2.13). If $\mathfrak{p}\cap\mathbb{Z}$ $=$ $p\mathbb{Z}$, by [Zsigmondy's theorem](https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem) we can find a prime number $q$ such that $p$ has order $d$ in $\mathbb{F}\_q^{\times}$. Then $p$ splits as a product of $(q-1)/d$ primes of degree $d$ in $\mathbb{Q}(\zeta\_q)$. So if $d$ and $(q-1)/d$ are relatively prime, $p$ is inert in the unique subfield $D$ of $\mathbb{Q}(\zeta\_q)$ of degree $d$ over $\mathbb{Q}$. Starting out with a prime $d$ that is larger than the residue class degree of $\mathfrak{p}$ over $p$ as well, $\mathfrak{p}$ will be inert in the compositum $\widetilde{K}$ $=$ $KD$. And taking $d$ also larger than the absolute value of the discriminant of $F$ over $\mathbb{Q}$, $q$ is relatively prime to the discriminant of $S$ over $R$, since $q$ $>$ $d$. So this field $\widetilde{K}$ will do the trick.


However, there is no guarantee that $(d,(q-1)/d)$ $=$ $1$, and I would be much obliged to learn how such a construction could be made to work (without appealing to extended Riemann hypotheses).


The existence of such a $\widetilde{K}$ in the general (Dedekind domain) setup is used in [this Journal of Algebra paper](https://www.sciencedirect.com/science/article/pii/S0021869398980887) by Ilaria Del Corso and Roberto Dvornicich. In the paragraph following their Lemma 4, the authors optimistically "let $\widetilde{K}$ be an unramified extension of $K$ of degree $d$ such that $\mathfrak{p}$ is inert in $\widetilde{K}$" (with any $d$ that is large enough and relatively prime to $n$). They then show (Lemma 5) that $\widetilde{K}$ and $F$ are linearly disjoint and that the integral closure of $R$ in $\widetilde{K}F$ is $\widetilde{R}S$, using the fact that $\widetilde{K}$ is unramified over $K$. But, aside from $\mathbb{Q}$, various other fields $K$ exist that do not admit any non-trivial unramified extensions whatsoever (see [this answer](https://mathoverflow.net/questions/26491/is-there-a-ring-of-integers-except-for-z-such-that-every-extension-of-it-is-ram/26526#26526)).



Finding field extensions in which a given prime is inert
Thanks to Arno Fehm's observation above, the answer in the general case is **NO**.


In the number field case, $R$ $=$ $\mathfrak{O}\_K$ and $S$ $=$ $\mathfrak{O}\_F$ (or overrings thereof in $K$ resp. $F$), and we can argue as follows. Let $p\mathbb{Z}$ = $\mathfrak{p}\cap\mathbb{Z}$. Let $q\_1,\cdots,q\_s$ be the rational primes that ramify in $F$. If not on the list, include $p$ as well. Take a prime number $d$ $>$ $[F:\mathbb{Q}]$. Then $d$ $>$ $n$ and, with $q$ $:=$ $q\_i$ and $\mathfrak{q}$ an $\mathfrak{O}\_K$-prime lying over $q$, $d$ is larger than the residue class degree $r$ of $\mathfrak{q}$ over $q$. The smallest field containing $\mathbb{F}\_{q^d}$ and $\mathbb{F}\_{q^r}$ is $\mathbb{F}\_{q^{dr}}$, which is of degree $d$ over $\mathbb{F}\_{q^r}$. So when $\mathbb{Q}\subseteq D$ is any extension of degree $d$ in which all $q\_i$ are inert, the $\mathfrak{q}$ will be inert, and in particular unramified in the compositum $DK$. This goes in particular for $\mathfrak{q}$ $=$ $\mathfrak{p}$, and so $\widetilde{K}$ $:=$ $DK$ meets the requirements.


Now for $1\leq i\leq s$, pick a monic polynomial $f\_i$ $\in$ $\mathbb{Z}[X]$ of degree $d$ and irreducible mod $q\_i$. By the Chinese Remainder theorem, there is a monic $f$ in $\mathbb{Z}[X]$ of degree $d$ that is congruent to $f\_i$ mod $q\_i$ for every $i$. So each of the $q\_i$ is inert in $D$ $:=$ $\mathbb{Q}[X]/(f)$.


**Note.** This answer replaces two earlier attempts, which were flawed. Apologies to anyone who may have wasted time on them.


