Chapter III: The P -scheme algorithm
3.1 Preliminaries
We first review basic notations and facts in algebra. They are standard and can be found in various textbooks, e.g., [Lan02; AM69; Mar77]. Then we discusssplitting of prime idealsin number field extensions. Finally, for the certain rings O¯K, we establish a one-to-one correspondence between their idempotent decompositions and the partitions of certain right coset spaces.
All rings are assumed to be commutative rings with unity.
Ideals. Recall that a subsetI of a ringRis anidealofRif (1)I is a subgroup of the underlying additive abelian group ofR, and (2)R·I ={ra:r ∈R, a∈I} ⊆I. Forx∈R, denote by(x),xRorRxthe ideal{rx :r ∈R}ofRgenerated byx. An ideal of Ris properif it is a proper subset of R. Let I be a proper ideal ofI. We sayI isprimeif I 6= R andab ∈ I impliesa ∈ I orb ∈ I for any a, b ∈ R. AndI ismaximalifI 6=Rand there exists no idealI0 ofRsatisfyingI (I0 (R. A proper idealI is prime (resp. maximal) iff the quotient ringR/I is an integral domain (resp. a field). In particular, maximal ideals are prime. For an ideal I0 of R, the mapI 7→ I/I0 is a one-to-one correspondence between the ideals ofR containingI0 and the ideals ofR/I0, and it preserves primality and maximality.
Ifm1, . . . ,mk andmare maximal ideals of Rand i=1mi ⊆m, thenm = mi for somei∈[k].2 In particular, ifTk
i=1mi = 0, thenm1, . . . ,mkare the only maximal ideals ofR.
Two ideals I, I0 of R are coprimeif I +I0 = R. In particular, distinct maximal ideals are always coprime. For pairwise coprime ideals I1, . . . , Ik, it holds that Tk
i=1Ii =Qk
i=1Ii. We also have
Lemma 3.1(Chinese remainder theorem). SupposeI1, . . . , Ikare pairwise coprime ideals ofR. Then the ring homomorphism
φ:R/
k
\
i=1
Ii →
k
Y
i=1
R/Ii
sendingx+Tk
i=1Iito(x+I1, . . . , x+Ik)is an isomorphism.
Semisimple rings. A (commutative) ring issemisimpleif it is isomorphic to a finite product of fields. The following lemma provides a characterization of semisimple rings.
Lemma 3.2. A ringRis semisimple iff it has finitely many maximal idealsm1, . . . , mk and Tk
i=1mi = 0, in which case R is isomorphic to Qk
i=1R/mi via the map x7→(x+m1, . . . , x+mk).
Proof. SupposeR ∼= Qk
i=1Fi is semisimple where eachFi is a field. Fori ∈ [k], let πi : R → Fi be the ith projection and mi be its kernel. Then R/mi ∼= Fi and hence each mi is a maximal ideal of R. Moreover we have Tk
i=1mi = 0and hencem1, . . . ,mkare the only maximal ideals. Conversely, supposeR has finitely many maximal idealsm1, . . . ,mkandTk
i=1mi = 0. Then by the Chinese remainder theorem, the mapR →Qk
i=1R/misendingx∈Rto(x+m1, . . . , x+mk)is a ring isomorphism. Each direct factorR/mi is a field, and henceRis semisimple.
The semisimple rings considered in this chapter are allsemisimpleFp-algebras, i.e.
semisimple rings that are alsoFp-algebras.
Idempotent elements. An elementxof a ring is anidempotent element(or just an idempotent) ifx2 =x. Two idempotentsx, y areorthogonalifxy= 0. A nonzero idempotentxisprimitiveif it cannot be written as a sum of two nonzero orthogonal
2See [AM69, Proposition 1.11] for a more general statement for prime ideals.
idempotents. As already stated in Definition 3.1, anidempotent decomposition of a ring R is a set I of nonzero mutually orthogonal idempotents of R satisfying P
x∈Ix = 1. We say such an idempotent decomposition is proper if|I| > 1and completeif all idempotents inI are primitive.
Lemma 3.3. LetR be a semisimple ring. For every maximal ideal mofR, there exists a unique primitive idempotentδm ∈Rsatisfyingδm ≡1 (mod m)andδm ≡0 (mod m0)for all maximal idealm0 6=m. Two elements δm andδm0 are orthogonal iffm6=m0. Furthermore
• the mapm7→δmis a one-to-one correspondence between the maximal ideals ofRand the primitive idempotents ofR, and
• the mapB 7→ P
m∈Bδm is a one-to-one correspondence between the sets of maximal ideals ofRand the idempotents ofR.
Proof. This is clear from the isomorphismR ∼= Q
m∈SR/m, whereS denotes the set of all the maximal ideals ofR.
We also need the following lemma.
Lemma 3.4. Supposeφ :R0 →Ris a ring homomorphism between two semisimple ringsR, R0. Letδ, δ0be idempotents ofRandR0 respectively satisfyingφ(δ0)δ=δ.
Then φ induces a ring homomorphism from R0/(1−δ0) to R/(1− δ) sending x+ (1−δ0)toφ(x) + (1−δ)forx∈R0.
Proof. It suffices to show thatφ(1−δ0)is in the ideal(1−δ)ofR, which holds since(1−φ(δ0))(1−δ) = 1−φ(δ0)−δ+φ(δ0)δ= 1−φ(δ0) =φ(1−δ0).
Finitely generated modules and free modules. A subset S of an R-moduleM generates M if P
x∈SRx = M. And M is finitely generatedif it is generated by a finite subset S. A basis of M over R, or an R-basis of M is a subset S ⊆ M generatingM for which the sumM =P
x∈SRxis a direct sum. We sayM isfree (overR) if it admits anR-basis. Therankof a finitely generated free module over Ris the cardinality of anyR-basis of it, which is finite and depends only onR.
Number fields. Elements in the algebraic closure Q¯ of Q are called algebraic numbers. An algebraic number isintegralor analgebraic integerif it is a root of a monic polynomial inZ[X]. The set of algebraic integers is a subring ofQ, denoted¯ byA. Anumber fieldis a finite degree field extension ofQinQ. For a number field¯ K, the subringOK :=A∩K is called thering of integersofK. It is embedded in theQ-vector spaceK as a lattice of rank[K :Q].
SupposeK/K0 is a number field extension. We sayα ∈K is aprimitive element ofK overK0 ifK =K0(α). Primitive elements always exist for any number field extension by theprimitive element theorem.
Galois theory. Let K/K0 be a field extension. The set of automorphisms of K fixingK0 is a group, called theautomorphism groupofKoverK0, and is denoted by Aut(K/K0). We say K is Galois over K0 if |Aut(K/K0)| = [K : K0], in which caseAut(K/K0)is also called theGalois groupofK overK0 and denoted byGal(K/K0).
Theorem 3.3 (fundamental theorem of Galois theory). Let K/K0 be a Galois extension. Then for any intermediate fieldK0 ⊆E ⊆K, the extensionK/Eis also a Galois extension. Furthermore, the mapE 7→Gal(K/E)is an inclusion-reversing one-to-one correspondence between the poset of intermediate fieldsK0 ⊆E ⊆ K and the poset of subgroups ofGal(K/K0), with the inverse mapH 7→KH.
Given a Galois extension K/K0, two subfieldsE, E0 between K andK0 arecon- jugate over K0 if there exists an isomorphism τ0 : E → E0 fixing K0. Such an isomorphism always extends to an automorphism τ ∈ Gal(K/K0) of K. The corresponding Galois groups Gal(K/E) and Gal(K/E0) satisfy Gal(K/E0) = τGal(K/E)τ−1. So conjugate subfields of K over K0 correspond to conjugate subgroups inGal(K/K0).
Now we restrict to number field extensions. LetK/K0be a number field extension.
There exists a unique minimal number field that containsK and is Galois overK0, called theGalois closure of K/K0. For a polynomial f(X) ∈ K0[X]with roots α1, . . . , αk ∈Q, the number field¯ K0 =K0(α1, . . . , αk)is called thesplitting field offoverK0and is Galois overK0. We also writeGal(f /K0)for the corresponding Galois groupGal(K0/K0), called the Galois group offoverK0. Iffis the minimal polynomial of a primitive element ofK overK0, the splitting field off overK0 is exactly the Galois closure ofK/K0.
Suppose K/K0 is a Galois extension with the Galois group G. If x ∈ K is an algebraic integer, so isgxfor anyg ∈GsinceZ⊆K0 is fixed byG. So the action ofGonK restricts to an action onOK.
Splitting of prime ideals. The ring of integers of a number field is an example of aDedekind domain[AM69; Mar77]. An ideal of a Dedekind domain is a nonzero prime ideal iff it is a maximal ideal, and hence these two notions are interchangeable.
By convention, we use the notion of (nonzero) prime ideals instead of maximal ideals.
LetKbe a number field. It follows from the theory of Dedekind domains [Mar77]
that the idealpOKofOKsplits uniquely (up to the ordering) into a product of prime ideals ofOK:
pOK =
k
Y
i=1
Pi.
Fori ∈ [k], the quotient ring OK/Pi is a finite field extension of degreedi ∈ N+ overFp, and Pk
i=1di = [K : Q]. We sayP1, . . . ,Pk are the prime ideals ofOK lying overp. IfP1. . . ,Pkare distinct andOK/Pi ∼=Fp for alli∈[k](and hence k = [K :Q]), we saypsplits completelyinK. It is known that ifpsplits completely in K, then it also splits completely in any subfield of the Galois closure of K/Q.
See, e.g., [Mar77, Chapter 4]. We also need the following result that identifies the set of prime ideals lying over p with a right coset space in the case that p splits completely in a Galois extension containingK.
Theorem 3.4. LetLbe a Galois extension ofQsuch thatpsplits completely inL, and letG= Gal(L/Q). Fix a prime idealQ0ofOLlying overp. For any subgroup H ⊆ G and its fixed field K = LH, the map Hg 7→ gQ0 ∩ OK is a one-to-one correspondence between the right cosets inH\Gand the prime ideals ofOK lying overp.3
See, e.g., [Mar77, Theorem 33]. As the prime ideals of OK lying over p are exactly those containingpOK, we get the following correspondence by passing to the quotient ringO¯K :=OK/pOK.
Corollary 3.1. Let L, G, Q0 be as in Theorem 3.4. For any subgroup H ⊆ G and its fixed field K = LH, the map Hg 7→ (gQ0 ∩ OK)/pOK is a one-to-one
3Note that this map is well defined: for another representativehg∈GofHgwhereh∈H, we havehgQ0∩ OK =h(gQ0∩ OK) =gQ0∩ OKsinceOK is fixed byH.
correspondence between the right cosets in H\G and the prime (and maximal) ideals ofO¯K.
Idempotent decompositions vs. partitions of a right coset space. Suppose p splits completely into a product of prime idealsP1, . . . ,Pk in a number fieldK. Then P1/pOK, . . . ,Pk/pOK are the prime (and maximal) ideals of O¯K. As the intersection of these ideals equalspOK/pOK = 0, the ring O¯K is semisimple by Lemma 3.2. The prime idealsPi/pOK correspond to the primitive idempotents of O¯Kby Lemma 3.3 and also to the cosets in a right coset space by Corollary 3.1. We combine them and establish a correspondence between the idempotent decomposi- tions ofO¯K and the partitions of a certain right coset space.
For a number field extensionL/K, the inclusionOK ,→ OLinduces a map iK,L: ¯OK →O¯L
with the kernel(pOL∩ OK)/pOK. AspOL∩ OK =pOK,4 this map is injective, which identifies O¯K with a subring of O¯L. Also note that if L/Q is a Galois extension with the Galois group G, the action of G on OL induces an action on O¯L and permutes the maximal ideals of O¯L. These observations are used in Definition 3.2 below.
Fix the following notations: letLbe a Galois extension ofQwithGal(L/Q) =G and suppose p splits completely in L. For a nonzero prime ideal Q of OL lying overp, defineQ¯ :=Q/pOLwhich is a prime (and hence maximal) ideal ofO¯L, and letδQ¯ be the primitive idempotent ofO¯LsatisfyingδQ¯ ≡ 1 (mod ¯Q)andδQ¯ ≡ 0 (mod ¯Q0)for all maximal idealQ¯0 6= ¯QofO¯L(cf. Lemma 3.3).
Definition 3.2. SupposeHis a subgroup ofGandK =LH. Fix a prime idealQ0 ofOLlying overp. Then
• for an idempotent decompositionI ofO¯K, defineP(I)to be the partition of H\G such that Hg, Hg0 ∈ H\G are in the same block iff g−1(iK,L(δ)) ≡
g0−1
(iK,L(δ)) (mod ¯Q0)holds for allδ ∈I, and
• for a partition P ofH\G, define I(P) to be the idempotent decomposition ofO¯K consisting of the idempotentsδB :=i−1K,L
P
g∈G:Hg∈B gδQ¯0
, whereB ranges over the blocks inP.5
4To see this, note that ifx∈pOL∩ OK, thenx/p∈ OL∩K=OK.
5We show in the proof of Lemma 3.5 thatP
g∈G:Hg∈B
gδQ¯0does lie in the image ofiK,L, and henceδBis well defined.
We have the following lemmas.
Lemma 3.5. The partitionsP(I)and the idempotent decompositionsI(P)are well defined. And for any idempotent decomposition I of O¯K, the idempotents δ ∈ I correspond one-to-one to the blocks ofP(I)via the mapδ7→Bδ :={Hg ∈H\G:
g−1
(iK,L(δ))≡1 (mod ¯Q0)}with the inverse mapB 7→δB.
Lemma 3.6. The map I 7→ P(I) is a one-to-one correspondence between the idempotent decompositions ofO¯K and the partitions ofH\G, with the inverse map P 7→I(P).
The proofs of Lemma 3.5 and Lemma 3.6 are routine and can be found in Ap- pendix B. In particular, Lemma 3.6 establishes a one-to-one correspondence be- tween the idempotent decompositions ofO¯K and the partitions ofH\G.