calculating the splitting of primes

Let $K|L$ be an extension  of number fields  , with rings of integers  $\mathcal{O}_{K},\mathcal{O}_{L}$. Since this extension is separable (http://planetmath.org/SeparablePolynomial), there exists $\alpha\in K$ with $L(\alpha)=K$ and by multiplying by a suitable integer, we may assume that $\alpha\in\mathcal{O}_{K}$ (we do not require that $\mathcal{O}_{L}[\alpha]=\mathcal{O}_{K}$. There is not, in general, an $\alpha\in\mathcal{O}_{L}$ with this property). Let $f\in\mathcal{O}_{L}[x]$ be the minimal polynomial  of $\alpha$.

Now, let $\mathfrak{p}$ be a prime ideal  of $L$ that does not divide $\Delta(f)\Delta(\mathcal{O}_{K})^{-1}$, and let $\bar{f}\in\mathcal{O}_{L}/\mathfrak{p}\mathcal{O}_{L}[x]$ be the reduction  of $f$ mod $\mathfrak{p}$, and let $\bar{f}=\bar{f}_{1}\cdots\bar{f}_{n}$ be its factorization into irreducible polynomials  . If there are repeated factors, then $p$ splits in $K$ as the product

 $\mathfrak{p}=(\mathfrak{p},f_{1}(\alpha))\cdots(\mathfrak{p},f_{n}(\alpha)),$

where $f_{i}$ is any polynomial    in $\mathcal{O}_{L}[x]$ reducing to $\bar{f}_{i}$. Note that in this case $\mathfrak{p}$ is unramified, since all $f_{i}$ are pairwise coprime mod $\mathfrak{p}$

For example, let $L=\mathbb{Q},K=\mathbb{Q}(\sqrt{d})$ where $d$ is a square-free integer. Then $f=x^{2}-d$. For any prime $\mathfrak{p}$, $f$ is irreducible  mod $\mathfrak{p}$ if and only if it has no roots mod $\mathfrak{p}$, i.e. $d$ is a quadratic non-residue mod $\mathfrak{p}$. Using quadratic reciprocity, we can obtain a congruence     condition mod $4p$ for which primes split and which do not. In general, this is possible for all fields with abelian  Galois groups  , using field .

Furthermore, let $K^{\prime}$ be the splitting field  of $L$. Then $G=\mathrm{Gal}(K^{\prime}|L)$ acts on the roots of $f$, giving a map $G\to S_{m}$, where $m=\deg f$. Given a prime $\mathfrak{p}$ of $\mathcal{O}_{L}$, the Artin symbol  $[\mathfrak{P},K^{\prime}|L]$ for any $\mathfrak{P}$ lying over $\mathfrak{p}$ is determined up to conjugacy by $\mathfrak{p}$. Its in $S_{n}$ is a product of disjoint cycles of length $m_{1},\ldots,m_{n}$ where $m_{i}=\deg f_{i}$. 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 $[\mathfrak{P},K^{\prime}|L]$ for infinitely many primes $\mathfrak{P}$ of $\mathcal{O}_{K^{\prime}}$.

For example, let $f=x^{3}+x^{2}+2\in\mathbb{Z}[x]$. This is irreducible mod 3, and thus irreducible. Galois theory  tells us that $G=\mathrm{Gal}(K^{\prime}|L)$ is a subgroup   of $S_{3}$, and so is isomorphic   to $C_{3}$ or $S_{3}$, but it is not obvious which. But if we consider $p=7$, $f\equiv(x-2)(x^{2}+3x-1)\pmod{7}$, and the quadratic factor is irreducible mod 7. Thus, $G\cong S_{3}$.

Or let $f=x^{4}+ax^{2}+b$ 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 $\bar{f}$ has a root. If $\beta$ is a root of $f$, then so is $-\beta$, and so assuming $p\neq 2$, there are at least two roots, and so a 3-cycle is impossible. Thus $G\cong C_{4}$ or $D_{4}$.

Title calculating the splitting of primes CalculatingTheSplittingOfPrimes 2013-03-22 13:53:24 2013-03-22 13:53:24 mathcam (2727) mathcam (2727) 12 mathcam (2727) Topic msc 11R04 PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ NumberField SplittingAndRamificationInNumberFieldsAndGaloisExtensions