Natural density implies Dirichlet density

Question

Let $M$ be a set of prime ideals of a number field $K$ . The limit $$d(M)= \lim_{s\rightarrow 1^+} \frac{ \sum_{P \in M} N(P)^{-s} }{ \sum_{P} N(P)^{-s}}$$ Where $P$ is a prime of $K$ and $N$ denotes the norm of $K/ \mathbb{Q}$ is called Dirichlet Density of $M$. Also, the Natural density is the limit $$ \delta(M)= \lim_{x\rightarrow \infty} \frac{ \# \{ P \in M : N(P) \leq x\}}{ \# \{ P : N(P) \leq x\}} $$

Show that the Natural denstity implies the Dirichhlet density. Also, $d(M)=\delta(M)$.

Any help? I'm trying for days to prove this exercise.

Answer 1

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.

Answer 2

It suffices to prove the theorem for $K=Q$. For general number field proof will be analogous. Let $M(x)=\# \{ p\in M :p\leq x\}$. Now suppose $\delta(M)= \lim_{x\rightarrow \infty} \frac{ \# \{ p \in M : p \leq x\}}{ \# \{ p : p \leq x\}}=\frac{M(x)}{\pi(x)}$ exists and equals to $R$. Since $\lim_{x\to\infty}\frac{M(x)}{x}=0=\lim_{x\to\infty}\frac{\pi(x)}{x}$, the Dirichlet series $$\sum_{p\in M}\frac{1}{p^s}=s\int_1^\infty\frac{M(x)}{x^{s+1}}dx,\quad\sum_p\frac{1}{p^s}=s\int_1^\infty\frac{\pi(x)}{x^{s+1}}dx,\quad s>1$$

In order to prove our theorem, we need a lemma.

lemma Let $A(x),B(x)$ be non-negative integrable function and the limit $\lim_{x\to\infty}\frac{A(x)}{B(x)}=R$. If $$\lim_{s\to 1^+}\int_1^\infty\frac{B(x)}{x^{s+1}}dx=+\infty$$ then we have that $$\lim_{s\to1^+}\frac{\int_1^\infty\frac{A(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R.$$ proof

Since $\lim_{x\to\infty}\frac{A(x)}{B(x)}=R$, for every $\epsilon>0$, there exists a constant $C=C(\epsilon)$ such that $|R(x):=A(x)-RB(x)|\leq C+\epsilon B(x)$ for all $x\geq1$.

Hence $$\frac{\int_1^\infty\frac{A(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R+ \frac{\int_1^\infty\frac{A(x)-RB(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R+ \frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}.$$ Hence, it suffices to prove $$\lim_{s\to 1^+}\frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=0$$ Since $$0\leq \left|\int_1^\infty\frac{R(x)}{x^{s+1}}dx\right|\leq \int_1^\infty\frac{|R(x)|}{x^{s+1}}dx\leq \int_1^\infty\frac{C+\epsilon B(x)}{x^{s+1}}dx=\frac{C}{s}+\epsilon\int_1^\infty\frac{B(x)}{x^{s+1}}dx$$ We have that $$\limsup_{s\to1^+}\frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}\leq \epsilon$$ Hence $$\lim_{s\to 1^+}\frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=0$$ and $$\lim_{s\to 1^+}\frac{\int_1^\infty\frac{A(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R$$ Now, the proof is complete.