PlanetMath: calculating the splitting of primes

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calculating the splitting of primes

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Let $K\vert L$ be an extension of number fields, with rings of integers $\O _K,\O _L$ . Since this extension is separable, there exists $\alpha\in K$ with $L(\alpha)=K$ and by multiplying by a suitable integer, we may assume that $\alpha\in\O _K$ (we do not require that $\O _L[\alpha]=\O _K$ . There is not, in general, an $\alpha\in\O _L$ with this property). Let $f\in \O _L[x]$ be the minimal polynomial of $\alpha$ .

Now, let $\mathfrak{p}$ be a prime ideal of that does not divide $\Delta(f)\Delta(\O _K)^{-1}$ , and let $\bar{f}\in \O _L/\mathfrak{p}\O _L[x]$ be the reduction of 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 splits in as the product

$\displaystyle \mathfrak{p}=(\mathfrak{p},f_1(\alpha))\cdots(\mathfrak{p},f_n(\alpha)),$

where

is any polynomial in $\O _L[x]$ reducing to $\bar{f}_i$ . Note that in this case $\mathfrak{p}$ is unramified, since all

are pairwise coprime mod $\mathfrak{p}$

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

Furthermore, let be the splitting field of . Then $G=\mathrm{Gal}(K'\vert L)$ acts on the roots of , giving a map $G\to S_m$ , where $m=\deg f$ . Given a prime $\mathfrak{p}$ of $\O _L$ , the Artin symbol $[\P ,K'\vert L]$ for any $\P$ lying over $\mathfrak{p}$ is determined up to conjugacy by $\mathfrak{p}$ . Its image in is a product of disjoint cycles of length $m_1,\ldots,m_n$ where $m_i=\deg f_i$ . This information is useful not just for prime splitting, but also for the calculation of Galois groups.

Another useful fact is the Frobenius density theorem, which states that every element of is $[\P ,K'\vert L]$ for infinitely many primes $\P$ of $\O _{K'}$ .

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 $G=\mathrm{Gal}(K'\vert L)$ is a subgroup of , and so is isomorphic to or , but it is not obvious which. But if we consider , $f\equiv (x-2)(x^2+3x-1)\pmod 7$ , and the quadratic factor is irreducible mod 7. Thus, $G\cong S_3$ .

Or let for some integers and is irreducible. For a prime , consider the factorization of . Either it remains irreducible ( contains a 4-cycle), splits as the product of irreducible quadratics ( contains a cycle of the form ) or $\bar{f}$ has a root. If $\beta$ is a root of , 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 .