It's given a finite extension $F$ of $\mathbb{Q}$ and suppose it's ring of integers is in the form $\mathbb{Z}[x]$ for some integrer $x$. For each prime $p$ in $\mathbb{Z}$ proof there are at most $p$ prime ideals $P_i$ in $\mathbb{Z}[x]$ such that $P_i\cap\mathbb{Z}=p\mathbb{Z}$ and $dim_{\frac{\mathbb{Z}}{p\mathbb{Z}}}(\frac{\mathbb{Z}[x]}{P_i})=1$ (for $i=1,\ldots,p$).
Appreciated even ideas and suggestions cause I have no ideas on how start this proof.
Suppose that $f(t)\in {\mathbb Z}[t] $ is the irreducible monic polynomial corresponding to your generator $x$ of the ring of integers of $F$. By Dedekind's theorem on the factorisation of rational primes, the distinct primes above $p$ correspond to the distinct irreducible factors of $\bar f ( t )$ mod $p$. The prime factors $ P$ above $p$ of degree $1$ (i.e., matching your dimension criterion above) are in bijection with the linear factors of $\bar f$. These are of the form $(t - a)$, for $a\in {\mathbb F}_p$, so there can be no more than $p$ such factors, and thus no more than $p$ such primes $P$.
