In Dirichlet's theorem for number fields, I asked about an analogue of Dirichlet's theorem (or I guess I should call it the Prime Number Theorem for Arithmetic Progressions) for number fields. There I received a confirmation of the guess that primes are also "essentially equally distributed" among the residue classes of $K/qK$. On this note, Theorem 1 of the paper "On the Theorem of Barban and Davenport-Halberstam in Algebraic Number Fields" (Journal of Number Theory 13, 463-484, 1981) implies that for $\mathfrak{q}$ an integral domain of a number field $K$ and $\beta \in K$ an integer with $(\beta, \mathfrak{q}) = 1$, the number of primes in $K$ congruent to $\beta \pmod{\mathfrak{q}}$ with norm less than $X$ is

$ \frac{1}{\phi(\mathfrak{q})}I(1) + o(I(1)), $

where $I(1)$ is the analogue of the logarithmic integral.

My next question: each prime in $K$ has a number of associates. Roughly, does each associate appear equally often?

More precisely, my work is in $\mathbb{Z}[\omega]$ ($\omega$ a primitive cube root of unity) where there are $6$ associates of each prime. I can specify a specific associate by requiring it to be primary, i.e. $2 \pmod{3}$, and in the upper half plane. Now the above theorems imply that $\frac{1}{\phi (Nq)}(1+o(1))$ of the primes in $\mathbb{Z}[\omega]$ are congruent to a particular $\beta \pmod{q}$. Is it also true that one sixth of these are primary and in the upper half plane? Supposing that $3|q$ and $\beta \equiv 2 \pmod{3}$ so that such primes are automatically primary -- is it true that one half of primes in this congruence class are in the upper half plane?


In the case you're interested in the answer is yes: just replace the modulus $q$ by $q\pi^3$, where $\pi$ generates the prime ideal with norm $3$. In general number fields, the question is not meaningful as it stands because the unit group has infinite order except when $K$ is complex quadratic.

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  • $\begingroup$ Sorry for being dense, but how does this guarantee that the resulting primes will be in the upper half-plane? $\endgroup$ – Tony Jun 24 '10 at 0:13

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