Let $F$ be a number field with class number 1. Consider its ring of integers $O_F$. Let $p$ ba a prime number. Is there a totally real field $F$ such that for each prime number $p$, $pO_F$ is also a prime ideal of $O_F$?
We know that this is not true if $F$ is not totally real. Take $F= \mathbb{Q}(i)$. $O_F=\mathbb{Z}[i]$. Take $p=2$. Then $2O_F = (1-i)^2$ and hance 2 is not prime in $O_F=\mathbb{Z}[i]$.
But what if $F$ is totally real??
Any help would be appreciated.
On
For a number field $F$,
The primitive element theorem gives $F=\Bbb{Q}(a)$, we can take $a$ to be an algebraic integer with monic minimal polynomial $f\in \Bbb{Z}[x]$, if $p \nmid Disc(f)$ and $p | f(n)$ then $f$ isn't irreducible in $\Bbb{F}_p[x]$ thus $p$ isn't a prime ideal of $\Bbb{Z}[x]/(f(x))$ thus it is not a prime ideal of $O_F$. Whence there are infinitely many non-inert primes.
If finitely many primes of $\Bbb{Q}$ split completely in $F$ then $$\zeta_F(s) = \prod_p\prod_{P \ prime \ \in O_F,P \ni p}\frac1{1-N(P)^{-s}}$$ converges for $\Re(s) > 1/2$.
It is known to be untrue because for $F$ totally real and $s > 1$
$$\Gamma(s)\zeta_F(s)\ge \Gamma(s)\sum_{a\in (O_F-0)/O_F^\times} N(a)^{-s}=\int_{\Bbb{R}_{>0}^n / U} f(x)(\prod_{j=1}^n x_j)^{s-1}dx_j$$ where $n=[F:\Bbb{Q}]$ $$f(x_1,\ldots,x_n)=\sum_{a\in O_F} \exp(-\sum_{j=1}^n x_j |\sigma_j(a)| x_j)$$ and $$U = \{ (|\sigma_1(u)|,\ldots,|\sigma_n(u)|,u\in O_F^\times\}$$ A more careful analysis shows that $\zeta_F(s)$ has a simple pole at $s=1$ which means that a positive density of the primes split completely in $F$.
If every rational prime were inert in $F$ then there would be unramified extensions of $\Bbb Q$, which don't exist, since the discriminant of any number field larger than $\Bbb Q$ is not equal to $\pm 1$, i.e. it has rational prime divisors, and the ramified primes are exactly those that divide the discriminant.