calculating the splitting of primes

Let K|L be an extensionPlanetmathPlanetmath of number fieldsMathworldPlanetmath, with rings of integersMathworldPlanetmath 𝒪K,𝒪L. Since this extension is separable (, there exists αK with L(α)=K and by multiplying by a suitable integer, we may assume that α𝒪K (we do not require that 𝒪L[α]=𝒪K. There is not, in general, an α𝒪L with this property). Let f𝒪L[x] be the minimal polynomialPlanetmathPlanetmath of α.

Now, let 𝔭 be a prime idealPlanetmathPlanetmath of L that does not divide Δ(f)Δ(𝒪K)-1, and let f¯𝒪L/𝔭𝒪L[x] be the reductionPlanetmathPlanetmath of f mod 𝔭, and let f¯=f¯1f¯n be its factorization into irreducible polynomialsMathworldPlanetmath. If there are repeated factors, then p splits in K as the product


where fi is any polynomialMathworldPlanetmathPlanetmathPlanetmath in 𝒪L[x] reducing to f¯i. Note that in this case 𝔭 is unramified, since all fi are pairwise coprime mod 𝔭

For example, let L=,K=(d) where d is a square-free integer. Then f=x2-d. For any prime 𝔭, f is irreduciblePlanetmathPlanetmath mod 𝔭 if and only if it has no roots mod 𝔭, i.e. d is a quadratic non-residue mod 𝔭. Using quadratic reciprocity, we can obtain a congruenceMathworldPlanetmathPlanetmathPlanetmathPlanetmath condition mod 4p for which primes split and which do not. In general, this is possible for all fields with abelianMathworldPlanetmath Galois groupsMathworldPlanetmath, using field .

Furthermore, let K be the splitting fieldMathworldPlanetmath of L. Then G=Gal(K|L) acts on the roots of f, giving a map GSm, where m=degf. Given a prime 𝔭 of 𝒪L, the Artin symbolMathworldPlanetmath [𝔓,K|L] for any 𝔓 lying over 𝔭 is determined up to conjugacy by 𝔭. Its in Sn is a product of disjoint cycles of length m1,,mn where mi=degfi. This is useful not just for prime splitting, but also for the calculation of Galois groups.

Another useful fact is the Frobenius theorem, which that every element of G is [𝔓,K|L] for infinitely many primes 𝔓 of 𝒪K.

For example, let f=x3+x2+2[x]. This is irreducible mod 3, and thus irreducible. Galois theoryMathworldPlanetmath tells us that G=Gal(K|L) is a subgroupMathworldPlanetmathPlanetmath of S3, and so is isomorphicPlanetmathPlanetmathPlanetmath to C3 or S3, but it is not obvious which. But if we consider p=7, f(x-2)(x2+3x-1)(mod7), and the quadratic factor is irreducible mod 7. Thus, GS3.

Or let f=x4+ax2+b for some integers a,b and is irreducible. For a prime p, consider the factorization of f. Either it remains irreducible (G contains a 4-cycle), splits as the product of irreducible quadratics (G contains a cycle of the form (12)(34)) or f¯ has a root. If β is a root of f, then so is -β, and so assuming p2, there are at least two roots, and so a 3-cycle is impossible. Thus GC4 or D4.

Title calculating the splitting of primes
Canonical name CalculatingTheSplittingOfPrimes
Date of creation 2013-03-22 13:53:24
Last modified on 2013-03-22 13:53:24
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 12
Author mathcam (2727)
Entry type Topic
Classification msc 11R04
Related topic PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ
Related topic PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ
Related topic NumberField
Related topic SplittingAndRamificationInNumberFieldsAndGaloisExtensions