Is there an "elementary" proof of the infinitude of completely split primes?

Question

Let $K$ be a Galois extension of the rationals with degree $n$. The Chebotarev Density Theorem guarantees that the rational primes that split completely in $K$ have density $1/n$ and thus there are infinitely many such primes. As Kevin Buzzard pointed out to me in a comment, there is a simpler way to see that there are infinitely many rational primes that split completely in $K$, namely that the Dedekind zeta-function $\zeta_K(s)$ has a simple pole at $s = 1$. While this result is certainly much easier to prove than Chebotarev's Theorem, it is still not an elementary proof.

Is there a known elementary proof of the fact that there are infinitely many rational primes that split completely in $K$?

Selberg's elementary proof of Dirichlet's Theorem for primes in arithmetic progressions handles the case where $\text{Gal}(K/\mathbb{Q})$ is Abelian. I don't know anything about the general case. Since Dirichlet's Theorem is stronger than required, it is possible that an simpler proof exists even in the Abelian case.

Remarks on the meaning of elementary. I am aware that there is no uniformly recognized definition of "elementary proof" in number theory. While I am not opposed to alternate definitions, my personal definition is a proof which can be carried out in first-order arithmetic, i.e. without quantification over real numbers or higher-type objects. Obviously, I don't require it to be explicitly formulated in that way — even logicians don't do that! Odds are that whatever you believe is elementary is also elementary in my sense.

Kurt Gödel observed that proofs of (first-order) arithmetical facts can be much, much shorter in second-order arithmetic than in first-order arithmetic. This observation explains some of the effectiveness of analytic number theory, which is implicitly second-order. In view of Gödel's observation, it is possible that we have encountered arithmetical facts with a reasonably short second-order proof (i.e. could be found in an analytic number theory textbook) but no reasonable first-order proof (i.e. the production of any such proof would necessarily exhaust all of our natural resources). The above is unlikely to be such, but it is interesting to know that beasts of this type could exist...

Answer 1

By the primitive element theorem, $K=\mathbb{Q}(\alpha)$ for some nonzero $\alpha \in K$, and we may assume that the minimal polynomial $f(x)$ of $\alpha$ has integer coefficients. Let $\Delta$ be the discriminant of $f$. Since $K/\mathbb{Q}$ is Galois, a prime $p \nmid \Delta$ splits completely in $K$ if and only if there is a degree $1$ prime above $p$, which is if and only if $p | f(n)$ for some $n \in \mathbb{Z}$. Suppose that the set $P$ of such primes is finite. Enlarge $P$ to include the primes dividing $\Delta$. Let $t$ be a positive integer such that $\operatorname{ord}_p t> \operatorname{ord}_p f(0)$ for all $p \in P$. For any integer $m$, we have $f(mt) \equiv f(0) \;(\bmod \; t)$, so $\operatorname{ord}_p f(mt) = \operatorname{ord}_p f(0)$ for all $p \in P$. But $f(mt) \to \infty$ as $m \to \infty$, so eventually it must have a prime factor outside $P$, contradicting the definition of $P$.

Answer 2

I don't think that Lenstra's and Stevenhagen's article

Primes of degree one and algebraic cases of Čebotarev's theorem, L'Enseign. Math. 37 (1991), 17-30

has been mentioned yet. It is available online here.

Answer 3

There's an old easy proof of the fact that there are infinitely many primes $p$, $p \equiv 1 \bmod n$: Let $\Phi_n(X)$ be the $n$-th cyclotomic polynomial. Show that $\Phi_n(X)$ has a root in $\mathbb{F}_p$ if and only if $p \equiv 1 \bmod n$. Do as in Euclid's proof of the infinitude of primes: if $p_1, \dots, p_r $ are primes $\equiv 1 \bmod n$ then consider $\Phi_n(n p_1 \dots p_r)$. It's bigger than 1 and not divisible by any of the $p_i$ or any prime dividing $n$. So this shows the statement for cyclotomic fields.

Answer 4

If $p \mid f(n)$ precisely, then one of the ideals $(n - \sigma(\alpha))\mathcal{O}_K$ must have a factor of $p\mathcal{O}_K$ in its factorization and hence all of them do. Therefore $p\mathcal{O}_K$ splits into a product of $\deg(f)$ ideals and therefore they must all have degree one.

One could see quite elementarily that there exist infinitely many primes that divide some value of $f$ precisely: if $p$ is not such a prime, then if $p^2 \mid f(n)$, then $p^2 \mid f(n + p) = f(n) + pf'(n) + p^2A$ for some integer $A$ and therefore $p \mid f'(n)$. Thus, $n$ is a common zero of both $f, f'$ in $\mathbb{F}_p$, which cannot happen for $p$ large enough, as $\mathrm{Res}(f, f')$ does not vanish.