Prime Decomposition
Adam Gamzon
1
Introduction
Let K be a number field, let OK be its ring of integers, and let p ∈ Z be a prime. Since
every prime of OK lies over a prime in Z, understanding how primes in Z decompose in OK is
a starting point to understanding the primes of OK . In most cases, the prime decomposition of
p in OK 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 [OK : Z[θ]], then we can decompose pOK as follows. Let
T (X) ≡
g
Y
Ti (X) mod p
i=1
be the decomposition of T into monic irreducible factors in Fp [X]. Then
pOK =
g
Y
pei i ,
i=1
where pi = (p, Ti (θ)).
Proof. See [1, Theorem 4.8.13].
Notice that if there is a power basis, i.e., [OK : Z[θ]] = 1, then Theorem 1 gives a relatively
simple way of computing the prime decomposition of any p ∈ Z. The problem arises when
[OK : Z[θ]] > 1. In this case, Theorem 1 works for all but finitely many primes. If p divides
[OK : Z[θ]], it would be nice if we could choose some θ0 which would give an index [OK : Z[θ0 ]]
relatively prime to p. However, this approach does not work because there are number fields in
which OK does not admit a power basis.
Example 2. Let K = Q(θ), where θ is a root of T (X) = X 3 + X 2 − 2X + 8. This is a classic
example due to Dedekind of a number field in which OK does not admit a power basis. Using
the algorithm that we will introduce, one can show that 2 splits completely in OK . Assuming
that we have computed the splitting of 2 in OK , let’s show that 2 divides [OK : Z[α]] for all
α ∈ OK . Suppose 2 does not divide [OK : Z[α]] for some α ∈ OK . Let F (X) be the minimal
polynomial of α. Then Theorem 1 implies that F (X) factors into three distinct linear terms in
F2 [X] since 2 splits completely in OK . This is a contradiction, however, because there are only
two linear polynomials in F2 [X], namely, X and X + 1.
Primes that divide [OK : Z[θ]] for any θ in OK where K = Q(θ) are called essential discriminantal divisors. So 2 is an essential discriminantal divisor in Example 2. One neat result is
that essential discriminantal divisors must be less than or equal to n − 1 where n Q
= [K : Q]. To
see why this is plausible, consider the case where p splits completely. Then p = ni=1 pi where
the pi are distinct prime ideals of OK . So to have any hope of applying Theorem 1 for some θ,
there had better be at least n distinct linear polynomials in Fp [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 Ip be the radical ideal of pOK . Since OK is a
Q
Dedekind domain, we know pOK = gi=1 pei i for some prime ideals pi ⊂ OK . Furthermore, we
Q
have Ip = gi=1 pi . Set Kj = Ipj + pOK . Then the valuation at pi of Kj is min(ei , j), that is
Kj =
g
Y
min(ei ,j)
pi
.
i=1
Since min(ei , j) ≥ min(ei , j − 1) for all i, Kj−1 divides Kj . In fact
0 , if ei ≤ j − 1,
min(ei , j) − min(ei , j − 1) =
1 , if ei ≥ j.
Hence, if we set Jj := Kj (Kj−1 )−1 , then
Y
Jj =
pi .
ei ≥j
Observe that Jj divides Jj−1 . So, setting Hj := Jj−1 (Jj )−1 , we get
Y
Hj =
pi .
ei =j
The idea is that through this sequence of ideal divisions, we have filtered the primes dividing
pOK by their valuations. Then, letting e = max(ei ),
pOK =
e
Y
Hjj .
j=1
So to get the factorization for pOK , it suffices to decompose each Hj .
We now focus on one ideal Hj , which we abbreviate as H. Consider the Fp -algebra OK /H.
Since H is a product of distinct prime ideals, OK /H is a finite separable algebra over Fp (i.e.,
a finite product of finite extensions of fields of Fp ). The following result about OK /H leads to
an algorithm for splitting H.
Proposition 3. Let A be a finite separable Fp -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 Fp -algebra, we have A ∼
= A1 × · · · × Ak for some k where
each Ai is some finite extension of Fp . So for α ∈ A, write α = (α1 , . . . , αk ) where αi ∈ Ai . Let
ϕ : A → A be defined by ϕ(x) = xp − x and let V := ker ϕ. Note that α ∈ V if and only if
αip − αi = 0 for all i. That is, α ∈ V if and only if αi ∈ Fp for all i where Fp is considered as
embedded into Ai . Therefore, dimFp 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 ε2 = ε and ε 6= 0 and
ε 6= 1. Pick some α = (α1 , . . . , αk ) in V \ Fp (where we now considered Fp as embedded in
A). Compute the minimum polynomial of α, mα (X). In terms of the αi , mα (X) is the least
common multiple of the mαi (X). Furthermore, since αi ∈ Fp for all i, the mαi (X) are degree
2
1 polynomials for all i. Hence mα (X) is a product of at least two distinct linear polynomials
because α ∈ V \ Fp . Write mα (X) = m1 (X)m2 (X) where m1 (X) and m2 (X) are nonconstant
polynomials in Fp [X]. Note that (m1 (X), m2 (X)) = 1 since mα (X) is a square-free product of
polynomials. So we may write
U (X)m1 (X) + V (X)m2 (X) = 1
(3.1)
for some U (X) and V (X) in Fp [X]. Let ε = (U m1 )(α). Then evaluating (3.1) at α and
multiplying by (U m1 )(α) shows that ε2 = ε. Since m1 and m2 are nonconstant and since
(U, m2 ) = 1 and (V, m1 ) = 1 by (3.1), ε 6= 0 and ε 6= 1
Now let’s use Proposition 3 to develop an algorithm for factoring H. Let A = OK /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 OK /H. Let e be any lift of ε. Set H1 := H + eOK and
H2 := H + (1 − e)OK . Note that
e(1 − e) = e − e2
≡ ε − ε2 mod H
≡ 0 mod H,
so H1 H2 ⊂ H. We claim that H1 H2 = H. To see this, pick any x ∈ H. Write x = xe + x(1 − e).
Since
xe ∈ eH ⊂ H1 H2
and
x(1 − e) ∈ (1 − e)H ⊂ H1 H2 ,
x is in H1 H2 . Hence we have factored H as H1 H2 . 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 H1 and H2 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 §2, let’s return to the situation in example 2 and
compute the decomposition of 2OK . In example 2, we had K = Q(θ), where θ is a root of
2
X 3 + +X 2 − 2X + 8. An integral basis is given by {1, θ, θ+θ
2 }. Let v1 = 1, v2 = θ, and
2
v3 = θ+θ
2 . According to our algorithm, the first step is to compute I2 . 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 pj ≥ n, then the nilradical of OK /pOk is
j
equal to the kernel of the map x 7→ xp .
So, in general, if {v1 , . . . , vn } is an integral basis for K/Q then computing the kernel of
j
x 7→ xp is the same as computing the kernel of the matrix (aij ), where the aij are given by
P
j
vjp = ni=1 aij vi and we think of the aij as lying in Fp . In our case, p = 2 and n = 3 so it
suffices to compute the kernel of x 7→ x4 . Computing the aij and reducing modulo 2 gives


1 0 0
(aij ) =  0 1 0  .
0 0 1
3
This shows that the kernel is trivial, so we conclude that I2 = 2OK . Moreover, this means
H1 = 2OK so we can skip the ideal division part of the algorithm and go straight to computing
the factorization of H1 .
Let A1 = OK /H1 . Let vi be the image of vi in A. Then it’s easy to check that v1 , v2 ,
and v3 form an F2 -basis for A1 . Using this basis, one computes the matrix representation of
the F2 -linear map ϕ : A1 → A1 defined by ϕ(x) = x2 − x to be the zero matrix. Therefore
dim(ker ϕ) = 3 so Proposition 3 implies that there is a nontrivial idempotent. In fact, v2 is
one such idempotent. So, letting H11 = H1 + v2 OK and H12 = H1 + (v2 + 1)OK , we have
H1 = H11 H12 . Computing the multiplication-by-v2 map on OK /H1 with respect to the basis
{v1 , v2 , v3 } gives


0 0 0
 1 1 1 .
0 0 0
Similarly, the matrix corresponding to the multiplication-by-(v2 + 1) map in this basis is


1 0 0
 1 0 1 .
0 0 1
From this, we can conclude that dimF2 OK /H11 = 2 and dimF2 OK /H12 = 1; so H12 is a prime
ideal while the factorization of H11 remains undetermined.
To compute the factorization of H11 , let A11 = OK /H11 . Note that the images of v1 and
v3 in A11 , denoted v1 and v3 , form an F2 -basis for A11 . Computing ϕ : A11 → A11 as before
shows that this is again the zero map. That is, the kernel is the whole space A11 . Since A11 is
2-dimensional over F2 , this means that A11 is not a field so we can split it like before. This time,
we use v3 as the nontrivial idempotent. As before, the minimal polynomial for v3 is X 2 + X.
We may, therefore, factor H11 as H111 H112 , where
H111 = H11 + v3 OK
and
H112 = H11 + (v3 + 1)OK .
Since 2OK can factor into at most three prime ideals, H111 and H112 must be prime without
need for any further calculations. Putting all of this together gives
θ + θ2
θ + θ2
2OK = H12 H111 H112 = (2, θ + 1) 2, θ,
2, θ,
+1 .
2
2
4
References
[1] H. Cohen. A Course in Computational Algebraic Number Theory. Springer-Verlag, 1993.
5