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I have a conjecture that the primes that split completely in a finite abelian extension $K/Q$ are determined by a congruence relation to modulus $\Delta$, the discriminant of the extension. I believe that Quadratic Reciprocity implies this for quadratic extensions. Until recently I was under the impression that Artin Reciprocity implied this for all abelian extensions of the rationals, but after reading the full statement of the theorem I no longer think this is true: it seems to me that Artin's law implies that there is some modulus for which this is true, and we may take this modulus to be divisible only by the primes ramified in the extension, but this leaves open the possibility that the best possible modulus does not divide the discriminant. This doesn't seem computationally useful: how can I use Artin reciprocity to determine the primes that split if I don't know an acceptable modulus, only that such a modulus exists?

For example, consider the extension $K := Q(\beta)$ where $\beta$ is a root of the polynomial $x^3 - 7x +7$. This is an abelian extension with discriminant $49$; its Galois group $G$ is cyclic of order $3$. I want to show that a rational prime $p$ splits completely in this extension iff $p \equiv \pm 1 \ (7)$. $K$ is totally real so the infinite prime of $Q$ doesn't ramify, so that $7$ is the only ramified prime. Artin reciprocity now tells us (I think) that there is some $r$ such that the Artin map gives a homomorphism $$ (\textbf{Z}/7^r\textbf{Z})^* \rightarrow G$$ such that a prime $p$ splits iff the image of (the residue class of) $p$ under this map is the identity element. If I can take $r = 2$ (ie, if the discriminant is an acceptable modulus) then I can show that any such map factors through $(\textbf{Z}/7\textbf{Z})^*$ and immediately deduce the splitting law for the extension. Artin reciprocity on its own, however, only seems to tell me that some $r$ works, which is no good. How do I get round this?

My questions: 1.) Is my conjecture (discriminant always works) true?

2.) If so, is this an easy consequence of Artin Reciprocity or a much deeper result?

3.) If not, is there a way round this difficulty, so that we can still use Artin to work out the splitting in concrete cases, as above?

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  • $\begingroup$ At first : The conductor is the least $n$ such that $K \subset \mathbb{Q}(\zeta_n)$. Since the splitting of prime ideals above $p$ in $\mathbb{Q}(\zeta_n)$ only depends on $p \bmod n$ and $Gal(\mathbb{Q}(\zeta_N)/\mathbb{Q}) =Gal(\mathbb{Q}(\zeta_N)/K)\times Gal(K/\mathbb{Q})$, it is the same in $K$. Then you look at class field theory and Artin reciprocity for generalizing this. $\endgroup$ – reuns Dec 28 '16 at 7:08
  • $\begingroup$ I got two good answers to my question below. I will mention for completeness that I realised soon after posting that in the specific example I gave the splitting can be calculated without knowing the exact value of r, only that such r exists: the point is both $(Z/7^rZ)^*$ and $G \equiv C_3$ are cyclic, and it is easy to deduce that a prime is in the kernel of the Artin map iff it is a cube mod $7^r$. An integer is a cube mod $7^r$ iff it is a cube $7$ and $\pm 1$ are the only cubes mod $7$. $\endgroup$ – S. Davey Dec 28 '16 at 20:32
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The smallest possible modulus is called the conductor of $K/Q.$ It is true that the conductor always divides the discriminant of $K/Q;$ this follows from the "conductor-discriminant formula" which writes the discriminant explicitly in terms of the conductor. The conductor-discriminant formula is typically proved with $L$ functions.

I do not know whether you can prove that the conductor divides the discriminant using only Artin reciprocity.

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  • $\begingroup$ The Artin reciprocity should be enough for finding the factorization of $\zeta_{K}(s)$ in $\prod_{\chi \in X} L(s,\chi)$ $\endgroup$ – reuns Dec 28 '16 at 7:00
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First, to avoid possible confusion, let me recall that for a given number field $K$ and a finite abelian extension $L/K$, the conductor $\mathcal F_{L/K}$ is the gcd of the $K$-modulii $\mathcal M$ such that $L$ is contained in the ray class field mod $\mathcal M$ of $K$ . To simplify, let us consider only the ramification of non archimedean primes. Then a prime $\mathcal P$ of $K$ ramifies in $L$ iff $\mathcal P$ divides $\mathcal F_{L/K}$ , iff $\mathcal P$ divides $\mathcal D_{L/K}$, the discriminant ideal. Note that in general $\mathcal F_{L/K}$ and $\mathcal D_{L/K}$ are not the same: let $X(L/K)$ be the group of characters of $Gal(L/K)$, and for each $\chi \in X(L/K)$, let $\mathcal F_{\chi}$ be the conductor of the extension of $L$ cut out by $\chi$ ; then $\mathcal F_{L/K}$ is the lowest common multiple of the $\mathcal F_{\chi}$'s, whereas $\mathcal D_{L/K}$ is the product of their finite parts.

A prime $\mathcal P$ splits totally in $L/K$ iff $\mathcal P$ does not divide $\mathcal F_{L/K}$ and the Artin symbol of $\mathcal P$ relative to $L/K$ is trivial. This implies that the set $Spl(L/K)$ (obvious notation) can be characterized by congruences mod $\mathcal F_{L/K}$ in the following sense: there exists a finite number of fractional ideals $\mathcal A_i$ of $K$, coprime to $\mathcal F_{L/K}$, such that $\mathcal P \in Spl(L/K)$ iff $\mathcal P\equiv \mathcal A_i$ mod $\mathcal F_{L/K}$ for all $i$.

If $K = Q$, the conductor $f_L$ is the smallest $n$ such that, by Kronecker-Weber, $L$ is contained in the cyclotomic field $Q(\zeta_n)$, and the discriminant $d_L$ is the usual one (up to signs). If $L$ is a quadratic field, they happen to coincide (as you noticed), but not in general. Here is a concrete example taken from D. Garbanati, Rocky Mountain J.of Math., 11, 12 (1981) : $L=Q(\sqrt 5, \sqrt -3), d_L=\pm 5^2.3^2 , f_L=5.3$, $Spl(L/Q)$={primes congruent to 1 or 4 mod 15} .

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