Dirichlet density for number fields $K$?

Question

Let $K$ be a number field. Let $P$ be a subset of the set of nonzero prime ideals in $K$. For $\mathfrak{p} \in P$, let $N(\mathfrak{p})$ be its absolute norm, so $N(\mathfrak{p}) = \#\mathcal{O}_K/\mathfrak{p}$. The Dirichlet density of $P$ is defined as$$\lim_{s \to 1^+} {{\sum_{\mathfrak{p} \in P} N(\mathfrak{p})^{-s}}\over{\log {1\over{s-1}}}}.$$I have two questions.

1. What is the Dirichlet density of the set of all prime ideals?
2. Let $P$ be the set of prime ideals in $\mathbb{Q}(i)$ that lie above primes congruent to $3$ mod $4$. What is the Dirichlet density of $P$?

Answer 1

I'll answer the questions in reverse order, just because it's convenient to have (2) for (1).

(2) The density is $0$. Any such prime ideal is inert, hence the numerator is $\sum_{p \equiv 3 \pmod 4} p^{-2s}$, which converges to a finite value when $s = 1$. More generally, in any number field $K$ the set of prime ideals of degree $\ge 2$ has zero Dirichlet density.

(1) The density is $1$,. You can see this in this particular case $K = \mathbb Q(i)$ as follows: by (1), we only need to consider those prime ideals $\mathfrak p$ above primes $p \equiv 1 \pmod 4$ (ignoring the ramified prime $2$). Each such $p$ factors as $\mathfrak p_1 \mathfrak p_2$, with inertial degree $1$. Thus from $N(\mathfrak p_1) + N(\mathfrak p_2) = 2p$ we see the Dirichlet density is $$ \lim_{s \to 1} \frac{\sum_{p \equiv 1 \pmod 4} 2p^{-s}}{-\log(s-1)}.$$ This is twice the $\mathbb Q$-Dirichlet density of the $1 \pmod 4$ primes, which is $2 \cdot \frac12 = 1$.

The proof in the general case is much the same, see Lemma 8.5 in Lemmermeyer for example.

Answer 2

1. The Dirichlet analytic class number formula shows that $\zeta_K(s)$, extended to the region $\text{Re}\,s > 1 - 1/[K: \mathbb{Q}]$, satisfies$$\zeta_K(s) = {1\over{s-1}}\,f(s)$$for some function $f(s)$ that is holomorphic and nonvanishing in a neighborhood of $1$. Thus, if $\log$ is the principal branch of the complex logarithm, then$$\log \zeta_K(s) = \log {1\over{s-1}} + O(1)$$as $s \to 1^+$. On the other hand, the Euler product for $\zeta_K(s)$ leads to$$\log \zeta_K(s) = \sum_\mathfrak{p} - \log\left(1 - N(\mathfrak{p})^{-s}\right) = \sum_\mathfrak{p}\left(N(\mathfrak{p})^{-s} + O\left(N(\mathfrak{p})^{-2s}\right)\right),$$and $\sum_\mathfrak{p} N(\mathfrak{p})^{-2s}$ converges for $\text{Re}\,s > 1/2$, since the Euler product for $\zeta_K(s)$ converges for $\text{Re}\,s > 1$. Comparing the two formulas shows that as $s \to 1^+$, we have$$\sum_\mathfrak{p} N(\mathfrak{p})^{-s} = \log {1\over{s - 1}} + O(1),$$but $\log {1\over{s-1}} \to \infty$, so the ratio defining the Dirichlet density tends to $1$.
2. If $p \equiv 3 \text{ }(\text{mod }4)$, then $-1$ is not a square $\text{mod }p$, so $x^2 + 1 \text{ mod }p$ is irreducible, so $p$ is inert in $\mathbb{Q}(i)/\mathbb{Q}$, so there is just one prime ideal in $\mathbb{Q}(i)$ above $p$, and its norm is $p^2$. Hence$$\sum_{\mathfrak{p} \in P} N(\mathfrak{p})^{-s} = \sum_{p \,\equiv\, 3 \text{ }(\text{mod }4)} p^{-2s} \le \sum_{n \ge 1} n^{-2s} = O(1)$$ as $s \to 1^+$, so the Dirichlet density of $P$ is $0$.

Edit. Whoops, I did not see E Chen's answer before writing this. Oh well.