Prime Decomposition. Adam Gamzon. 1 Introduction

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1 Prime Decomposition Adam Gamzon 1 Introduction Let K be a number field, let O K be its ring of integers, and let p Z be a prime. Since every prime of O K lies over a prime in Z, understanding how primes in Z decompose in O K is a starting point to understanding the primes of O K. In most cases, the prime decomposition of p in O K is straightforward to compute by using the following theorem. Theorem 1. Let K = Q(θ) where θ is an algebraic integer whose minimal polynomial is denoted by T (X). If p does not divide [O K : Z[θ]], then we can decompose po K as follows. Let T (X) T i (X) mod p be the decomposition of T into monic irreducible factors in F p [X]. Then po K = p e i i, where p i = (p, T i (θ)). Proof. See [1, Theorem ]. Notice that if there is a power basis, i.e., [O K : Z[θ]] = 1, then Theorem 1 gives a relatively simple way of computing the prime decomposition of any p Z. The problem arises when [O K : Z[θ]] > 1. In this case, Theorem 1 works for all but finitely many primes. If p divides [O K : Z[θ]], it would be nice if we could choose some θ which would give an index [O K : Z[θ ]] relatively prime to p. However, this approach does not work because there are number fields in which O K does not admit a power basis. Example. Let K = Q(θ), where θ is a root of T (X) = X 3 + X X + 8. This is a classic example due to Dedekind of a number field in which O K does not admit a power basis. Using the algorithm that we will introduce, one can show that splits completely in O K. Assuming that we have computed the splitting of in O K, let s show that divides [O K : Z[α]] for all α O K. Suppose does not divide [O K : Z[α]] for some α O K. Let F (X) be the minimal polynomial of α. Then Theorem 1 implies that F (X) factors into three distinct linear terms in F [X] since splits completely in O K. This is a contradiction, however, because there are only two linear polynomials in F [X], namely, X and X + 1. Primes that divide [O K : Z[θ]] for any θ in O K where K = Q(θ) are called essential discriminantal divisors. So is an essential discriminantal divisor in Example. One neat result is that essential discriminantal divisors must be less than or equal to n 1 where n = [K : Q]. To see why this is plausible, consider the case where p splits completely. Then p = n p i where the p i are distinct prime ideals of O K. So to have any hope of applying Theorem 1 for some θ, there had better be at least n distinct linear polynomials in F p [X]. That is, we need p n. 1

2 Factoring Essential Discriminantal Divisors In order to properly handle these essential discriminantal divisors, we will employ the following method of Buchman and Lenstra. Let I p be the radical ideal of po K. Since O K is a Dedekind domain, we know po K = g pe i i for some prime ideals p i O K. Furthermore, we have I p = g p i. Set K j = Ip j + po K. Then the valuation at p i of K j is min(e i, j), that is K j = p min(e i,j) i. Since min(e i, j) min(e i, j 1) for all i, K j 1 divides K j. In fact { 0, if ei j 1, min(e i, j) min(e i, j 1) = 1, if e i j. Hence, if we set J j := K j (K j 1 ) 1, then J j = e i j p i. Observe that J j divides J j 1. So, setting H j := J j 1 (J j ) 1, we get H j = e i =j p i. The idea is that through this sequence of ideal divisions, we have filtered the primes dividing po K by their valuations. Then, letting e = max(e i ), po K = e H j j. So to get the factorization for po K, it suffices to decompose each H j. We now focus on one ideal H j, which we abbreviate as H. Consider the F p -algebra O K /H. Since H is a product of distinct prime ideals, O K /H is a finite separable algebra over F p (i.e., a finite product of finite extensions of fields of F p ). The following result about O K /H leads to an algorithm for splitting H. Proposition 3. Let A be a finite separable F p -algebra. Then there exists an efficient algorithm that either shows that A is a field or finds a nontrivial idempotent in A. Proof. Since A is a finite separable F p -algebra, we have A = A 1 A k for some k where each A i is some finite extension of F p. So for α A, write α = (α 1,..., α k ) where α i A i. Let ϕ : A A be defined by ϕ(x) = x p x and let V := ker ϕ. Note that α V if and only if α p i α i = 0 for all i. That is, α V if and only if α i F p for all i where F p is considered as embedded into A i. Therefore, dim Fp V = k. In particular, dim V = 1 if and only if A is a field. Suppose dim V > 1. We will show that there is some ε such that ε = ε and ε 0 and ε 1. Pick some α = (α 1,..., α k ) in V \ F p (where we now considered F p as embedded in A). Compute the minimum polynomial of α, m α (X). In terms of the α i, m α (X) is the least common multiple of the m αi (X). Furthermore, since α i F p for all i, the m αi (X) are degree j=1

3 1 polynomials for all i. Hence m α (X) is a product of at least two distinct linear polynomials because α V \ F p. Write m α (X) = m 1 (X)m (X) where m 1 (X) and m (X) are nonconstant polynomials in F p [X]. Note that (m 1 (X), m (X)) = 1 since m α (X) is a square-free product of polynomials. So we may write U(X)m 1 (X) + V (X)m (X) = 1 (3.1) for some U(X) and V (X) in F p [X]. Let ε = (Um 1 )(α). Then evaluating (3.1) at α and multiplying by (Um 1 )(α) shows that ε = ε. Since m 1 and m are nonconstant and since (U, m ) = 1 and (V, m 1 ) = 1 by (3.1), ε 0 and ε 1 Now let s use Proposition 3 to develop an algorithm for factoring H. Let A = O K /H. If A is a field then we re done because that means H is a prime ideal. Otherwise, by Proposition 3, there is some nontrivial idempotent ε in O K /H. Let e be any lift of ε. Set H 1 := H + eo K and H := H + (1 e)o K. Note that e(1 e) = e e ε ε mod H 0 mod H, so H 1 H H. We claim that H 1 H = H. To see this, pick any x H. Write x = xe + x(1 e). Since xe eh H 1 H and x(1 e) (1 e)h H 1 H, x is in H 1 H. Hence we have factored H as H 1 H. Moreover, this factorization is nontrivial since otherwise e 1 or 0 mod H. To get the complete decomposition of H, just repeat this process with H 1 and H in place of H. It s easy to see that this procedure will stop after k steps where k is the number of prime factors of H. 3 Application As an application of the algorithm from, let s return to the situation in example and compute the decomposition of O K. In example, we had K = Q(θ), where θ is a root of X 3 + +X X + 8. An integral basis is given by {1, θ, θ+θ }. Let v 1 = 1, v = θ, and v 3 = θ+θ. According to our algorithm, the first step is to compute I. To do this, we use the next result found in [1, Lemma 6.1.6]. Lemma 4. If n = [K : Q] and if j 1 is such that p j n, then the nilradical of O K /po k is equal to the kernel of the map x x pj. So, in general, if {v 1,..., v n } is an integral basis for K/Q then computing the kernel of x x pj is the same as computing the kernel of the matrix (a ij ), where the a ij are given by v pj j = n a ijv i and we think of the a ij as lying in F p. In our case, p = and n = 3 so it suffices to compute the kernel of x x 4. Computing the a ij and reducing modulo gives (a ij ) =

4 This shows that the kernel is trivial, so we conclude that I = O K. Moreover, this means H 1 = O K so we can skip the ideal division part of the algorithm and go straight to computing the factorization of H 1. Let A 1 = O K /H 1. Let v i be the image of v i in A. Then it s easy to check that v 1, v, and v 3 form an F -basis for A 1. Using this basis, one computes the matrix representation of the F -linear map ϕ : A 1 A 1 defined by ϕ(x) = x x to be the zero matrix. Therefore dim(ker ϕ) = 3 so Proposition 3 implies that there is a nontrivial idempotent. In fact, v is one such idempotent. So, letting H 11 = H 1 + v O K and H 1 = H 1 + (v + 1)O K, we have H 1 = H 11 H 1. Computing the multiplication-by-v map on O K /H 1 with respect to the basis {v 1, v, v 3 } gives Similarly, the matrix corresponding to the multiplication-by-(v + 1) map in this basis is From this, we can conclude that dim F O K /H 11 = and dim F O K /H 1 = 1; so H 1 is a prime ideal while the factorization of H 11 remains undetermined. To compute the factorization of H 11, let A 11 = O K /H 11. Note that the images of v 1 and v 3 in A 11, denoted v 1 and v 3, form an F -basis for A 11. Computing ϕ : A 11 A 11 as before shows that this is again the zero map. That is, the kernel is the whole space A 11. Since A 11 is -dimensional over F, this means that A 11 is not a field so we can split it like before. This time, we use v 3 as the nontrivial idempotent. As before, the minimal polynomial for v 3 is X + X. We may, therefore, factor H 11 as H 111 H 11, where and. H 111 = H 11 + v 3 O K H 11 = H 11 + (v 3 + 1)O K. Since O K can factor into at most three prime ideals, H 111 and H 11 must be prime without need for any further calculations. Putting all of this together gives O K = H 1 H 111 H 11 = (, θ + 1) (, θ, θ + ) (, θ θ, θ + ) θ

5 References [1] H. Cohen. A Course in Computational Algebraic Number Theory. Springer-Verlag,