Natural density implies Dirichlet density

Here is a proof when $K=Q$, for general number field proof will be analogous. $\# \{ p\in M :p\leq x\}=M(x)$. Let's define $a_n=1$ when $n=p$ prime, and $0$ otherwise. So, the sum

$\displaystyle\sum_{p\in M,\\p\leq x}\frac{1}{p^s}=\sum_{n\leq x}\frac{a_n}{n^s}=\frac{M(x)}{x^s}+s\int_{1}^{x}\frac{M(t)}{t^{s+1}}dt$.

Now suppose $\delta(M)= \lim_{x\rightarrow \infty} \frac{ \# \{ p \in M : p \leq x\}}{ \# \{ p : p \leq x\}}$ exists and equals to $R$. That means by the Prime Number Theorem

$\lim_{x\to\infty}\frac{M(x)\log x}{x}=R$ or $M(x)=\frac{Rx}{\log x}+o(\frac{x}{\log x})$

Now plug in this formula of $M(x)$ into above summation. We have,

$d(M)=\lim_{s\to1+}\frac{\lim_{x\to\infty}\frac{M(x)}{x^s}+s\int_{1}^{x}\frac{M(t)}{t^{s+1}}dt}{-\log (s-1)}$,

as, $\sum_{p}\frac{1}{p^s}=-\log (s-1) + O(1)$.

Now using L'hospital rule we can have that,

$d(M)=\lim_{s\to1+}(s-1)\int_{1}^{\infty}\frac{M(t)\log t}{t^{s+1}}dt$

So, $\displaystyle\\d(M)=\lim_{s\to1+}(s-1)(R\int_{1}^{\infty}\frac{dt}{t^{s}}+o(\int_{1}^{\infty}\frac{dt}{t^{s}}))$

We know that, $\lim_{s\to1+}(s-1)\int_{1}^{\infty}\frac{dt}{t^{s}}=1$ as $\epsilon\to0$.

so $d(M)=R+o(1)=R$

For general case just "switch all $p$'s by $N(P)$'s", and use Prime Ideal Theorem instead of PNT.