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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.

Arturo Magidin
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Poitou-Tate
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1 Answers1

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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.

Mathmo123
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  • Could you tell me why $p$ splits(ramify) in $KL$ is equivalent to $p$ splits(ramify) in $K$ and $L$ ? – Poitou-Tate Jun 13 '23 at 19:48
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    See https://math.stackexchange.com/questions/78091/splitting-of-primes-in-the-compositum-of-fields. But be careful... replace splits completely with totally ramifies and the result is false. – Mathmo123 Jun 13 '23 at 20:38