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 $p\eq 2$ , there are at least two roots, and so a 3-cycle is impossible. Thus $G\\cong C_{4}$ or $D_{4}$ .