Trivial upper bound for the number of prime ideals $\mathfrak{p} \subset \mathcal{O}_K$ with $N(\mathfrak{p}) \leq x$.

18 Views Asked by At

Let $K$ be a number field of degree $n$ and $\mathcal{O}_K$ denote the ring of integers of $K$. For a given $x > 0$ we define
$$ \theta_K(x):=\sum_{N(\mathfrak{p})\leq x}\log N(\mathfrak{p}), $$ $$ \psi_K(x):=\sum_{N(\mathfrak{p}^m)\leq x}\log N(\mathfrak{p}), $$ where $m$ is a positive integer.

I've shown that $$ \psi_K(x)=\sum_{m\leq \log_2x} \theta_K(x^{1/m}). $$ Now, I am trying to show that $$ \psi_K(x) \sim x \iff \theta_K(x) \sim x. $$ For this purpose we have a trivial inequality for $\theta_K(x)$: $$ \theta_K(x)=\sum_{N(\mathfrak{p})\leq x}\log N(\mathfrak{p}) \leq \sum_{N(\mathfrak{p})\leq x}\log x \leq \ \ ?? $$ That is, I need a trivial upper bound for the number of prime ideals $\mathfrak{p} \subset \mathcal{O}_K$ with $N(\mathfrak{p}) \leq x$.