For a number field $K$, does there exist totally splitting prime? Let $K$ be a number field.
Then does there exist a rational prime number $l$ which splits completely in $K$?
I think this follows from Cebotarev density theorem.
But I think there exists more elementary proof.
 A: I don't know if this counts as elementary. Let $L$ be the Galois closure
of $K$ over $\Bbb Q$. If $l$ splits completely over $L$ then it splits
completely over $\Bbb Q$. The ideals of prime norm in $L$ are, apart from
the (finitely many) primes that ramify over $\Bbb Q$, the prime ideal factors
of the rational primes that split completely in $L$ over $\Bbb Q$. But, from the
Dedekind zeta function having a pole at $s=1$, there must be infinitely many
prime ideals of prime norm in $L$. So infinitely many rational primes split 
completely in $L$, and so also in $K$.
A: Wojowu gave a link to an elementary proof that the Galois closure has infinitely many primes splitting completely, because the minimal polynomial of its primitive element as a funtion on the integers has a polynomial growth rate, giving infinitely many $p$ with $f(m) \equiv 0 \bmod p$ so those $p$ split completely in the Galois closure as well as in $K$.
Now an interesting alternative is to look at the Dedekind zeta function containing all the information about those things :
Let $\sigma_1,\ldots,\sigma_n$ be the embeddings $K \to \mathbb{C}$, counting for only one each pair of complex embeddings. Let $\nu_j = 2$ for a complex embedding, $=1$ for a real-embedding.
If there are finitely many primes splitting completely then   $$\zeta_K(s) =  \prod_p \prod_{j=1}^{g(p)} \frac{1}{1-p^{-s f_j(p)}}$$ is analytic at $s=1$ since $g(p) \le N$ and for almost every $p$, $f_j(p) \ge 2$
We know it isn't the case because $$\Gamma(s) \zeta_K(s) \ge \Gamma(s)\sum_{a \in O_K^*/O_K^\times}N(a)^{-s} = \int_{\mathbb{R}^n / \log \iota(O_K^\times)} (\Theta(e^x)-1) |e^x|^sd^n x$$ with  $e^x = (e^{x_1},\ldots,e^{x_n}), |e^x| = \prod_{j=1}^n |e^x_j|$ and  $$\log\iota(a) = (\log\sigma_1(a),\ldots,\log\sigma_n(a)), \qquad \log \iota(O_K^\times) \text{ a lattice of rank } n-m$$
and $$\Theta(x)= \sum_{a \in O_K}\exp(-\sum_{j=1}^n x_j  |\sigma_j(a)|^{ \nu_j})$$
and $\Theta(x)-1 \to \infty $ as $x \to 0$
