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Let $K$ be a number field, and $p$ be a rational prime. Then the Chebotarev Density Theorem implies we can find primes $v$ and $w$ of $K$ of degree 1 which are split and nonsplit respectively in $K[\sqrt{p}]$. What is the best known effective (upper) bound for the norms of the least such primes (not assuming GRH)? In particular, is there a bound which is asymptotically strictly less than $\sqrt{p}$ (times a constant coming from the field $K$)?

EDIT: I'd like to clarify, in response to the comments below. The situation I'm wondering about is when we fix K, and let p vary. So when K is a cyclotomic field (adjoin, say, the qth root of unity for a prime q), I'm asking about the least prime which is a quadratic nonresidue (resp. residue) mod p, which is 1 mod some fixed prime q, and I'm hoping that there is a bound of the form $\sqrt{p}$ times (something in terms of q). Under GRH this is true --- in fact under GRH, we can get a bound of the shape $(\log p)^2$ times constants coming from K.

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(Prompted by the comments in GH's answer) Do you really want $\sqrt p$, not square root of discriminant? See edit to my answer. –  Felipe Voloch Jan 17 '11 at 15:29
    
See my edit above to the question which addresses your example below. –  Eric Larson Jan 17 '11 at 19:58
    
I still think that $\sqrt{p}$ is pretty hopeless with present day technology, but at least for $K$ abelian I see room for improving upon what I wrote originally. For example, if $q>1$ is fixed and you are looking for an $\ell$ which is $1$ mod $q$ and of a given quadratic character mod $p$, then Linnik's theorem (see Wikipedia) guarantees $\ell\ll p^c$ with some constant $c$ (the current record is $c=5.2$). Linnik's theorem is pretty deep though, and the way I applied it is rather dull. –  GH from MO Jan 17 '11 at 21:50
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2 Answers 2

I assume $\sqrt{p}$ is not contained in $K$, then the bound you are looking for is available.

Let $\chi$ be the ray class character attached to the quadratic extension $K(\sqrt{p})/K$, then the $L$-function $L(s,\chi)$ has conductor essentially $p$. By a recent result of Venkatesh (Theorem 6.1 in Annals of Math. 172 (2010), 989-1094) we have the subconvex bound $L(s,\chi)\ll |s|^N p^{1/4-1/200}$ on the criticial line $\Re s=1/2$, where $N>0$ is a constant. It follows, by a simple Mellin transformation technique, that for any fixed smooth function $V:(0,\infty)\to\mathbb{C}$ of compact support we have

$$\sum_{\mathfrak{m}\subset\mathcal{O}_K}\chi(\mathfrak{m})V(N\mathfrak{m}/X)\ll p^{1/4-1/200} X^{1/2}.$$

Therefore the absolute value of the left hand side is smaller than $X$ for some $X\gg p^{1/2-1/100}$, where the implied constant depends only on $K$ and $V$. This implies that $\chi$ takes both values $\pm 1$ on prime ideals with norm $\ll p^{1/2-1/100}$.

Perhaps one can complement this with Vinogradov's trick, see Corollary 9.19 in Montgomery-Vaughan: Multiplicative number theory I.

EDIT: As the OP pointed out, all the $p$'s above should be replaced by $p^{(K:\mathbb{Q})}$.

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I'm looking at Venkatesh's Theorem 6.1, and it looks like this gives a bound of $p^{(1/2 - 1/100) n}$, where $n$ is the degree of $K$, as opposed to $p^{1/2 - 1/100}$. Hopefully, I'm misunderstanding either your argument above or the statement of the theorem. But it looks like the bound depends on the norm (from K to Q) of the conductor, which would be $p^n$, as opposed to $p$? –  Eric Larson Jan 17 '11 at 1:06
    
Eric, I am afraid you are right. My $p$ is really the norm of the relative discriminant of the field extension which, as you say, is $p^n$ instead of $p$. I'd think that in this case $p^{n/2}$ is the natural "easy" bound for the norms of your $v$ and $w$, not $p^{1/2}$. This is certainly the case for the sum in my response which is a natural analytic device to study your question. The number you can put in front of $X^{1/2}$ on the right is really the quantity that Venkatesh studied in part of his paper. So personally I'd be surprised if my response did not contain the state of the art. –  GH from MO Jan 17 '11 at 5:48
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The paper:

J. C. Lagarias, H. L. Montgomery and A. M. Odlyzko, A bound for the least prime ideal in the Chebotarev density theorem, Invent. Math. 54 (1979) 271-296

gives a bound of the form $c \sqrt p$ for some unspecified $c$.

My paper with J. Vaaler:

The least nonsplit prime in Galois extensions of Q, J. Number Theory, 85 (2000), 320-335.

gives an effective constant (when $K=\mathbb Q$, but the argument should generalize). Actually, the quadratic field case of our argument is already in Gauss. Our paper has a bunch of other references, including what you can get with GRH. Improving the square root bound without GRH is a big open problem.

The paper also gives me Erdos number 2 :-)

EDIT: As in GH's answer, the natural quantity for the bounds is the discriminant, so $p$ needs to be replaced by $p^n, n=[K:\mathbb Q]$, in the case of $K(\sqrt p)$. Here is an example where this will make a big difference. Take $K$ to be the cyclotomic field of $p$-th roots of unity where $p \equiv 3 \mod 4$, so the quadratic extension is non trivial. The OP asks for degree one primes, these are primes above rational primes $l \equiv 1 \mod p$, so they have norm $l > p$ and you can't expect a $\sqrt p$ bound.

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The square root bound has been improved (for abelian extensions), see my response sent a minute ago. –  GH from MO Jan 16 '11 at 14:00
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@GH: Thanks. Your comment also reminded me of this recent paper: arxiv.org/abs/1003.5718 which has improvements for extensions of degree at least three. –  Felipe Voloch Jan 16 '11 at 15:56
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Thanks for the reference! I keep collecting arxiv preprints, but I am always behind with naming and looking at them properly. So although I had Li's paper on my hard disk, I was not really aware of it. BTW I edited some exponents in my original response. –  GH from MO Jan 16 '11 at 17:38
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