Finding field extensions in which a given prime is inert

Question

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 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 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).

Answer 1

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.