Chapter IV: Constructing number fields
4.1 Preliminaries
LetKandK0be relative number fields over a number fieldK0. We say an embedding (resp. isomorphism)τ :K →K0is an embedding (resp. isomorphism)overK0 if τ isK0-linear, i.e.,τ(ax) = aτ(x)for alla ∈K0 andx∈K. By choosingx = 1, we see that this is equivalent toτ(a) = afor alla ∈ K0. We writeK ∼=K0 K0 for the statement thatKis isomorphic toK0 overK0.
(K0, g)-subfield systems and the associated subgroup systems. We generalize the notion of(Q, g)-subfield systems (Definition 3.3) and the associated subgroup systems (Definition 3.4) as follows:
Definition 4.1 ((K0, g)-subfield system). Let K0 be a number field. Let g(X) be a polynomial in K0[X] with the splitting field L over K0. LetF be a collection of relative number fields over K0 such that (1) the fields in F are mutually non- isomorphic overK0, and (2) each fieldK0 ∈ F is isomorphic to a subfield ofLover K0. We sayF is a(K0, g)-subfield system.
Definition 4.2. Letg(X)be a polynomial inK0[X]with the splitting field Lover K0. LetF be a(K0, g)-subfield system. DefineP]to be the poset of subfields ofL that includes all the fields isomorphic to those inF overK0:
P] :={K0 ⊆L:K0 ∼=K0 K for someK ∈ F }.
By Galois theory, it corresponds to a posetP of subgroups ofGal(g/K0), given by P :=
H ⊆Gal(g/K0) :LH ∈ P] ,
which is closed under conjugation in Gal(g/K0), and hence is a subgroup system overGal(g/K0). We sayP andP]areassociated withF.
The complexity of a subgroup system. The size of a(K0, g)-subfield systemF is primarily controlled by the total degree of the fields inF overK0, which is the number of coefficients in K0 we need to maintain. We relate this quantity to the complexityof a subgroup system, defined as follows.
Definition 4.3(complexity of a subgroup system). SupposeP is a subgroup system over a finite groupG. ThenGacts onP by conjugation, i.e.,g ∈ GsendsH ∈ P togHg−1 ∈ P. Let P0 ⊆ P be a complete set of representatives of the G-orbits under this action. Define thecomplexityofP to be
c(P) := X
H∈P0
[G:H].
As conjugate subgroups have the same order, the complexityc(P)is well defined.
And we have
Lemma 4.1. For a(K0, g)-subfield systemF, the total degree of the fields inFover K0 equalsc(P), whereP is the subgroup system associated withF.
Proof. Conjugate subgroups correspond to conjugate subfields under the Galois correspondence. So forK ∈ F there exists a unique subgroupH ∈ P0 satisfying LH ∼=K0 K. And the mapK 7→H is a one-to-one correspondence betweenF and P0. Finally note that[K :K0] = [G:H]forHcorresponding toK.
The following lemma bounds the complexity of a system of stabilizers.
Lemma 4.2. Let G be a finite group acting on a finite set S. Let m ∈ N+ and m0 = min{|S|, m}. LetP be the system of stabilizers of depth m0 with respect to the action ofGonS. Then
c(P)≤
m0
X
k=1 k
Y
i=1
(|S| −i) = O
|S|m0 .
Proof. Replacingmwithm0does not changeP. So we may assumem =m0 ≤ |S|. When|S| ≥2, we have
m
X
k=1 k−1
Y
i=0
(|S| −i)≤
m
X
k=1
|S|k =O(|S|m).
The same holds trivially when|S|= 1. Next we provec(P)≤Pm
k=1
Qk
i=1(|S| −i). LetP0 ⊆ P be as in Definition 4.3. It suffices to find an injective map
τ : a
H∈P0
H\G ,→
m
a
k=1
S(k),
since the cardinality of`
H∈P0H\Gisc(P), whereas the cardinality of`m k=1S(k) isPm
k=1
Qk
i=1(|S| −i).
For each k ∈ [m], the group G acts diagonally on S(k). For each H ∈ P0, we pickk = k(H)≤ m andx =x(H) ∈ S(k) such thatH = Gx with respect to the diagonal action. By Lemma 2.1, we have an injective map H\G → S(k) whose image is the G-orbit of x. These maps altogether give the map τ. To show τ is injective, it suffices to show that for different H, H0 ∈ P0, the coset spacesH\G andH0\G are mapped to differentG-orbits. Assume to the contrary that they are mapped to the the sameG-orbitO. Sox(H), x(H0)∈O. Thenk(H) =k(H0)and x(H0) =g(x(H))for someg ∈G. But then we have
H0 =Gx(H0)=Ggx(H)=gGx(H)g−1 =gHg−1, which is a contradiction to the choice ofP0. Soτ is injective.
Algebraic numbers. The fields in a (K0, g)-subfield system F are encoded by polynomials in K0[X]. So to bound the size of F, we also need to bound the size of the coefficients of these polynomials, which are algebraic numbers inK0. This is closely related to the following definition, introduced in [WR76].
Definition 4.4. For an algebraic numberα, definekαkto be the greatest absolute value ofi(α)∈Cwhereiranges over the embeddings ofQ(α)inC.2
For algebraic numbersα, β, we clearly havekα+βk ≤ kαk+kβkandkα·βk ≤ kαk · kβk.
The following lemma relates the size of an algebraic number α ∈ K0 (i.e., the number of bits used to encodeαinK0) tokαk.
Lemma 4.3. SupposeK0is a number field encoded by a polynomialh(X)∈Q[X]
irreducible overQof degreenand sizes0. Let αbe an algebraic number inK0 of sizes. LetDbe the smallest positive integer such thatDα is an algebraic integer.
Thensis polynomial inlogkαk, logDands0. Conversely,logkαkandlogDare polynomial insands0.
Proof. Suppose h(X) = Pn
i=0ciXi where n = deg(h) and ci ∈ Q for all i. By substituting X with X/k for some large enough k ∈ N+ and clearing the denominators, we may assumeh(X) ∈Z[X]andcn = 1. Both the encoding ofh and that ofαuse at leastncoefficients inQ. So we haves, s0 ≥n.
The algebraic number α ∈ K0 is encoded by the constants d0, . . . , dn−1 ∈ Q satisfying
α=
n−1
X
i=0
diβi, (4.1)
whereβ is a root of hinK0. So we havekαk ≤ Pn−1
i=0 |di|kβki. It was shown in [WR76] thatkβk ≤Pn−1
i=0 |ci|. And we clearly havelog|ci| ≤ s0 andlog|di| ≤ s for0≤i≤n−1. It follows thatlogkαkis polynomial insands0.
LetD0 ∈ N+be the least common multiple of the denominators ofdi. Ash(X)∈ Z[X]andcn = 1, we knowβis an algebraic integer. ThenD0αis also an algebraic integer by (4.1). So Dis bounded byD0. It follows thatlogDis polynomial in s ands0. Then the second claim of the lemma is proved.
For the first claim, it suffices to show that the size of eachdiis polynomial inlogkαk, logDands0. This follows from [WR76, Section 7 and Lemma 8.3].
The following lemma relates the size of the minimal polynomial of an algebraic numberαover a number fieldK0 tokαk.
2kαkis called the size ofαin [WR76]. We reserve the termsize(of an object) for the number of bits used to encode an object in an algorithm.
Lemma 4.4. Suppose K0 is a number field encoded by a rational polynomial irreducible overQof sizes0 (lets0 = 1ifK0 =Q). Letαbe an algebraic number, and letDbe the smallest positive integer such thatDαis an algebraic integer. Let h(X) ∈ K0[X]be the minimal polynomial of α whose size iss and degree is n.
Thens is polynomial in logkαk, logD, s0 andn. Conversely, logkαkandlogD are polynomial insands0.
Proof. We clearly have n ≤ s. Supposeh(X) = Pn
i=0ciXi, whereci ∈ K0 and cn = 1. It was as shown in [WR76] that kαk ≤ Pn−1
i=0 kcik. It follows from Lemma 4.3 thatlogkαkis polynomial insands0.
Note that for sufficiently largek ∈N+that is polynomial insands0, the coefficients of the polynomial knh(X/k) are all algebraic integers. It follows that kα is an algebraic integer (cf. [AM69, Corollary 5.4]). SoDis bounded bykand hence is polynomial insands0. Then the second claim of the lemma is proved.
For the first claim, we may assumeαis an algebraic integer by replacingαwithDα andci withDn−ici. Then any conjugateα0 ofαoverQis also an algebraic integer, andkα0k=kαk. For0≤i≤n−1, the coefficientciofhis (up to sign) given by theith elementary symmetric polynomial in a subset of conjugates ofα overQ. It follows from Lemma4.3that the size of eachci is polynomial inlogkαk,logD,s0 andn. Sosis polynomial inlogkαk,logD,s0 andnas well.
Finding a primitive element over Q. Suppose K0 = Q(α) is a number field encoded by the minimal polynomial of a primitive elementαoverQ, andK =K0(β) is a relative number field overK0, encoded by the minimal polynomial of a primitive elementβoverK0. We would like to representKdirectly in the formQ(γ), encoded by the minimal polynomial of a primitive element γ over Q. The first step is to find such an elementγ, which can be achieved using a constructive version of the primitive element theorem (see, e.g., [Wae91]). For completeness, we give the details as follows.
Lemma 4.5. SupposeK0 is a number field and α, β are algebraic numbers. Let d = [K0(α, β) : K0]. Thenkα+β is a primitive element ofK0(α, β)overK0 for some integerk∈[1, d+ 1].
Proof. Consider a “bad” nonzero integerkfor whichK0(kα+β)is a proper subfield ofK0(α, β). LetLbe the Galois closure ofK0(α, β)/K0. Then by the fundamental
theorem of Galois theory, there exists an automorphismφofLfixingK0(kα+β) but not K0(α, β). Then either φ(α) 6= α or φ(β) 6= β. As φ fixes kα+β, we have kφ(α) +φ(β) = φ(kα+β) = kα+β, from which we see that actually φ(α) 6= α and φ(β) 6= β both hold. Then k is determined by φ(α) andφ(β)via k = (φ(β)−β)/(α−φ(α)). So the number of bad choices of k is bounded by the number of(φ(α), φ(β))whereφranges over the automorphisms ofLfixingK0. The later is the cardinality of the orbit of(α, β)under the action ofGal(L/K0). By the orbit-stabilizer theorem, it equals
[Gal(L/K0) : Gal(L/K0(α, β))] = [K0(α, β) :K0] =d.
So there are at mostdbad choices ofk. The lemma follows since[1, d+ 1]contains more thandintegers.
This gives an efficient algorithm of finding a primitive element overQ:
Lemma 4.6. There exists a polynomial-time algorithm that given a number fieldK0
and a relative number fieldK overK0, find a primitive elementγ ofK overQand its minimal polynomialh(X)∈Q[X]overQ.
Proof. SupposeK0 is encoded by a polynomialg(X)∈ Q[X]irreducible overQ, andK is encoded by a polynomialg0(X) ∈ K0[X]irreducible overK0. Then we are explicitly given a rootαofg(X)and a rootβofg0(X)inK, andK =Q(α, β). Enumerate the integersk ∈[1, d+ 1], whered= [K :Q]. For eachk, we compute γ =kα+β ∈K, and then compute its minimal polynomialh(X)∈Q[X]overQ by solving linear equations overQ. This step runs in polynomial time by Lemma 4.4.
Outputγandhwheneverdeg(h) = [K :Q]. By Lemma 4.5, a primitive elementγ is guaranteed to be found.
By computing a primitive element overQ, we can efficiently turn a relative number field into an ordinary number field:
Corollary 4.1. There exists a polynomial-time algorithm that given a number field K0 and a relative number fieldK overK0, computes an ordinary number fieldK0, a Q-basisB of K, and an isomorphismφ : K → K0 encoded by φ(x) ∈ K0 for x∈B.
Proof. Find a primitive element γ of K over Q and its minimal polynomial h(X) ∈ Q[X] over Q using Lemma 4.6. Compute K0 := Q[X]/(h(X)) and B = {1, γ, γ2, . . . , γd−1}, where d = [K : Q]. Then compute the isomorphism φ:K →K0, which sendsγi toXi+ (h(X))fori= 0,1, . . . , d−1.
As an application, we generalize Lemma 3.10 to obtain an efficient algorithm that computes embeddings of relative number fields over a given number field.
Lemma 4.7. There exists a polynomial-time algorithm ComputeRelEmbeddings that given a number fieldK0and relative number fieldsKandK0overK0, computes all the embeddings ofK inK0overK0.
Proof. IdentifyKandK0with ordinary number fields using Corollary 4.1. Run the algorithmComputeEmbeddingsin Lemma 3.10 to compute all the embeddings of K inK0, and ignore those not fixingK0.