Let $K$ be a number field which does not contain $\Bbb{Q}(i)$.
Then, I want to prove there are infinitely many prime number $p$ such that $p≡3\pmod4$ and $p$ splits completely in $K$.
To prove this kind of theorem, Chebotarev's density theorem should work well.
What I know is following.
・In Galois extension $K/\Bbb{Q}$ of degree $n$, $1/n$ percent of primes spilts completely in $K$(Chebotarev's density theorem)
・Since $K$ does not contain $\Bbb{Q}(i)$, $K\cap \Bbb{Q}(i)=\Bbb{Q}$.
・$p$ such that $p≡3\pmod4$ does not spilt in $\Bbb{Q}(i)/\Bbb{Q}$.
How can I combine these facts to prove the statement ?
Thank you for your help.
Best Answer
By assumption, $K(i)$ is the compositum of $K$ and $\mathbb Q(i)$. Hence, a prime $p$ splits in $K(i)$ if and only if it splits in $K$ and in $\mathbb Q(i)$. It follows that $$\{p: p\text{ splits in }K(i)\} = \{p\equiv 1\pmod 4 : p\text{ splits in }K\}.$$
Suppose that almost all of the primes that split in $K$ are $1\pmod 4$. Then the densities of the sets $$\{p : p\text{ splits in }K(i)\},\ \{p\equiv 1\pmod 4 : p\text{ splits in }K\},\text{ and }\{p : p\text{ splits in }K\}$$ are equal, contradicting the Chebotarev density theorem.