47
$\begingroup$

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...

$\endgroup$
2
  • 1
    $\begingroup$ Seeing how simple the answers from Bjorn and Victor are, this is one case where the "obvious" analytic proof is much, much longer than the elementary one! $\endgroup$ Feb 14, 2010 at 2:30
  • 5
    $\begingroup$ Yes, this is the easy way round. The hard way is showing algebraically that, given an (abelian) extension L/K of number fields, there exists a prime of K that does not split completely in L. This was done by Chevalley when he gave the first purely algebraic proof of the main theorems of class field theory. $\endgroup$
    – JS Milne
    Feb 14, 2010 at 14:16

4 Answers 4

70
$\begingroup$

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 $P$.

$\endgroup$
9
  • $\begingroup$ Excellent! So much simpler than I had anticipated... $\endgroup$ Feb 14, 2010 at 2:13
  • 1
    $\begingroup$ I wrote up a short alternate proof of the lemma Bjorn uses here: qchu.wordpress.com/2009/09/02/… $\endgroup$ Feb 14, 2010 at 7:14
  • 2
    $\begingroup$ Every prime which divides the index $[O_K:Z[\alpha]]$ divides the discriminant of f. So Bjorn's argument is correct as written, I believe. $\endgroup$ Feb 14, 2010 at 15:40
  • 2
    $\begingroup$ There is an elementary proof by Cassels of a result along these lines for any finitely generated field of characteristic 0, showing it embeds into Q_p for infinitely many p with some additional conclusions as well. Find a review of the paper on MathSciNet by searching under Cassels as author and the title is "An Embedding Theorem for Fields" (Bull. Austral. Math. Soc. 1976). $\endgroup$
    – KConrad
    Feb 14, 2010 at 18:37
  • 1
    $\begingroup$ @Albertas: In your example, $p \mid \Delta$, which Bjorn explicitly excludes. Once again, I believe that Bjorn's argument is precisely correct. $\endgroup$ Sep 17, 2015 at 16:47
14
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ Their article 'Chebotarev and his density theorem' might be of interest as well. $\endgroup$
    – user1073
    Feb 14, 2010 at 17:17
12
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Taking $\alpha$ to be a primitive n-th root of 1 in my answer gives your answer for that case! $\endgroup$ Feb 14, 2010 at 2:04
  • $\begingroup$ As I was typing my answer it occurred to me that the same idea gave something like Bjorn's answer, but I wanted to get it in quickly :-). $\endgroup$ Feb 14, 2010 at 2:07
  • $\begingroup$ So, it does bring up the question -- how dense a set of splitting primes can we construct with an elementary argument? The set that "Euclid-like" arguments constructs is awfully sparse. $\endgroup$ Feb 14, 2010 at 2:10
  • 5
    $\begingroup$ I wrote a note some time ago extending Chebyshev's method to number fields which gives an elementary and simple proof that the number of split primes is at least $x^{1/d}/\log x$. jtnb.cedram.org/jtnb-bin/fitem?id=JTNB_2000__12_1_81_0 $\endgroup$ Feb 14, 2010 at 15:51
  • 1
    $\begingroup$ @FelipeVoloch your link above no longer works. Since you didn't mention bibliographic information, here it is for anyone else: "Chebyshev's method for number fields" Journal de Théorie des Nombres de Bordeaux 12 (2000), 81-85. A working link (for now) is numdam.org/article/JTNB_2000__12_1_81_0.pdf. $\endgroup$
    – KConrad
    Jan 15 at 20:25
10
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ For the second paragraph, you still need to show that there are infinitely many distinct primes that divide the values of $f$, and once you have that, it is obvious that infinitely many of them do not divide the discriminant. $\endgroup$ Oct 10, 2021 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.