Algebraic Background
2.2 Introduction to number theory
24 Ch. 2 Algebraic Background
2.1.4 Vector spaces
In the remainder of this part,Kwill denote a field.
Definition 2.37 Avector spaceV overKis an abelian group for a first operation denoted by+, together with ascalar multiplicationfromK×V intoV, which sends(λ, x)onλxand such that for allx, y∈V, for allλ, µ∈Kwe have
• λ(x+y) =λx+λy
• (λ+µ)x=λx+µx
• (λµ)x=λ(µx)
• 1x=x.
An elementxofV is avectorwhereas an elementλofKis called ascalar.
Definition 2.38 AK-basisof a vector spaceV is a subsetS⊂V which
• islinearly independent over K, i.e., for any finite subset{x1, . . . , xn} ⊂ S and any λ1, . . . , λn∈K, one has that
n i=1
λixi= 0 implies that λi = 0 for alli
• generatesV overK, i.e., for allx∈V there exist finitely many vectorsx1, . . . , xnand scalarsλ1, . . . , λnsuch that
x= n i=1
λixi.
Theorem 2.39 LetV be a vector space overK. IfV is different from{0}thenV has aK-basis.
Definition 2.40 Two bases of a vector spaceV overKhave the same cardinality. This invariant is called theK-dimension of V or simply thedimension of V. Note that the dimension is allowed to be infinite.
Example 2.41 The set of complex numbersCtogether with the usual addition and coefficient wise multiplication with elements ofRis a vector space overR of dimension2. A real basis is for instance{1, i}.
Example 2.42 The setK[X]of polynomials in one variable over a fieldKis an infinite dimensional vector space with the usual addition of polynomials and multiplications with elements fromK. A basis is given by{1, X, X2, . . . , Xn, . . .}
Remark 2.43 When the fieldKis replaced by a ringR, the axioms of Definition 2.37 give rise to a module over the ringR.
§ 2.2 Introduction to number theory 25
2.2.1 Extension of fields
Definition 2.44 LetKandLbe fields, we say thatLis anextension fieldofKif there exists a field homomorphism fromKintoL. Such an extension field is denoted byL/K.
Remark 2.45 As said before, a field homomorphism is always injective, so we shall identifyK with the corresponding subfield ofLwhen consideringL/K.
Example 2.46 LetRbe the field of real numbers with usual addition and multiplication. Obviously, Ris an extension ofQ. Now, let us describe a less trivial example. Consider the element√
2∈R and the subset ofRof the elements of the forma+√
2bwitha, b∈Q. If we put fora+√ 2band a+√
2b
(a+√
2b) + (a+√
2b) =a+a+√
2(b+b) and
(a+√
2b)×(a+√
2b) =aa+ 2bb+√
2(ab+ab), it is easy to see that we obtain a field denoted byQ(√
2), which is an extension ofQ.
Definition 2.47 LetLandLbe two extension fields ofKandσa field isomorphism fromLtoL. One says thatσis aK-isomorphismifσ(x) =xfor allx∈K.
Definition 2.48 LetL/Kbe a field extension thenLcan be considered as aK-vector space. The dimension ofL/Kis called thedegreeofL/Kdenoted by[L:K]ordeg(L/K). If the degree of L/Kis finite then we say that the extensionL/Kisfinite.
The following result is straightforward.
Proposition 2.49 LetK⊂L⊂F be a tower of extension fields then deg(F/K) = deg(F/L) deg(L/K).
Now, letL/Kbe a field extension and letxbe an element ofL. There is a unique ring homomor- phismψ : K[X] → Lsuch thatψ(X) = xand for allz ∈ K,ψ(z) = z. We can consider the kernel of this homomorphism which is either{0}or a maximal idealIofK[X].
Definition 2.50 Suppose thatIas defined above is nonzero. AsK[X]is a principal ideal domain, there exists a unique monic irreducible polynomial
m(X) =Xd+ad−1Xd−1+· · ·+a0
such thatI=m(X)K[X]. We say thatxis analgebraic elementofLof degreedand thatm(X) is theminimal polynomialofx.
Quotienting bykerψ, one sees thatψgives rise to a field inclusionψofK[X]/
m(X)K[X] into L. LetK[x] ={f(x)|f(X)∈K[X]}be the image ofψinL. It is an extension field ofKand the monic polynomialm(X)is an invariant of the extension: in fact, if there existsy∈Lsuch that K[x]K[y]then by constructionxandyhave the same minimal polynomials.
Definition 2.51 If every element ofLis algebraic overK, we say thatLis analgebraic extension ofK.
26 Ch. 2 Algebraic Background
It is easy to see thatK[x]/Kis a finite extension andd= deg(m)is equal to its degree: in factψ gives a bijective map from theK-vector space of polynomials with coefficients inKof degree less thandtoK[x]. This bijectionψis called apolynomial representationofK[x]/K.
Not all algebraic extensions are finite, but a finite extension is always algebraic, and ifL/Kis a finite extension then there exists a finite sequence of elementsx1, . . . , xn ∈ L such thatL = K[x1, . . . , xn]. IfL=K[x]we say thatLis amonogenic extensionofK.
Definition 2.52 LetL/Kbe a finite algebraic extension and letx ∈ L. The application of right multiplication byxfromLtoLconsidered as aK-vector space is linear. The trace and the norm of this endomorphism ofLare called respectively thetraceandnormofxand denoted byTrL/K(x) andNL/K(x). We use the notationsTr(x)andN(x)when no confusion is likely to arise.
Ifxis a generating element ofL/Kwith minimal polynomialm(X) =Xd+ad−1Xd−1+· · ·+a0 thenTr(x) =−ad−1andN(x) = (−1)da0.
The trace and norm are both maps ofLtoK. We have the basic properties:
Lemma 2.53 LetL/Kbe a degreedfinite algebraic extension. Forx, y∈Landa∈Kwe have Tr(x+y) = Tr(x) + Tr(y), N(xy) = N(x) N(y)
Tr(a) = da, N(a) = ad
Tr(ax) = aTr(x), N(x) = 0 ⇒ x= 0.
LetK⊂L⊂F be a tower of finite algebraic extensions, letxbe an element ofFthen TrF/K(x) = TrL/K
TrF/L(x)
and NF/K(x) = NL/K
NF/L(x) .
Whenxis not a root of any polynomial equation with coefficients inKone needs a new notion.
Definition 2.54 If the kernel ofψis equal to{0}we say thatxis atranscendental elementofL. If every element ofLis transcendental overK, we say thatLis apure transcendental extensionover K. More generally, if there exists an element ofLwhich is not algebraic overK, thenL/Kis a transcendental extensionofK.
In this case, one can extend ψ to an inclusionψ of the fraction fieldK(X)into L by setting ψ(1/X ) = 1/x. LetK(x)be the image ofψinL. IfLis not an algebraic extension overK(x) then puttingx1=xwe can findx2, a transcendental element ofL/K(x1)which is not inK(x1), and build in the same way an inclusion ofK(X1, X2)intoL. Iterating this process we can find n ∈ N∪ {∞} the maximum number such thatK(x1, . . . , xn)is a subfield of Lisomorphic to K(X1, . . . , Xn). It can be shown thatnis independent of the sequence of transcendental elements x1, . . . , xnofLoverKchosen.
Definition 2.55 The numberndefined above is called thetranscendence degreeofLoverK.
It is quite clear from the previous discussion that every extensionK → L can be written as the compositionK→Ktrans →KalgwhereKtransis a pure transcendental extension ofKandKalgis an algebraic extension overKtrans.
Example 2.56 LetQ(X)be the field of rational functions overQ; then obviouslyQ(X)/Qis a pure transcendental extension ofQof transcendence degree equal to1. Now letQbe the algebraic closure ofQ, then by definitionQ/Qis an algebraic extension but it is not a finite extension. Finally Q(X,√
2)/Qis a transcendental extension that can be written asQ→Q(X)→Q(X,√ 2).
§ 2.2 Introduction to number theory 27
2.2.2 Algebraic closure
LetKbe a field and consider a monogenic algebraic extensionK[x]ofKdefined by a polynomial m(X)irreducible overK. The polynomialm(X)can be written as a product
mi(X)of irre- ducible polynomials overK[x]. As by constructionxis a root ofm(X)inK[x]then(X−x)is an irreducible factor ofm(X)and as a consequence for eachi,degmi <degm. If themi(X)’s are all of degree1then we say thatm(X)splits completelyinK[x].
Ifm(X)does not split completely overK[x]then there existsmi1(X), an irreducible polynomial of degree greater than or equal to2and we can consider the extensionK[x, y]overK[x]defined bymi1(X). Repeating this process, one can build recursively an extension field over whichm(X) splits completely.
Definition 2.57 The smallest extension ofKover whichm(X)completely splits is called thesplit- ting fieldofm(X). It is unique up to aK-isomorphism.
It is well known that every polynomial with coefficients inRsplits completely inC. More generally, ifKis a field, we would like to consider a maximal algebraic extension ofKin which every alge- braic extension ofKcould be embedded. Such an extension has the property that every polynomial ofK[X]splits completely inK. The following theorem [STE 1910] asserts its existence.
Theorem 2.58 (Steinitz) There exists a unique algebraic extension ofKin which every polynomial m(X)∈K[X]splits completely. This extension called thealgebraic closure ofKand denoted by Kis unique up to aK-isomorphism.
Next, we review some basic properties of algebraic extensions in order to state the main theorem of Galois theory.
2.2.3 Galois theory
For most parts of the book we consider finite algebraic extension fields. Therefore we restrict the discussion of Galois theory to this important case.
Definition 2.59 An extensionLoverKis said to benormalif every irreducible polynomial over Kthat has a root inLsplits completely inL.
As an immediate consequence every field automorphism ofLfixingKleavesLinvariant. LetK be a field,Kits algebraic closure,σan embedding ofKintoKand forx∈K,K[x]an algebraic monogenic extension ofKdefined by a polynomialm(X)of degreed. Letx=x1, x2. . . , xsbe the different roots ofm(X)inK. Then fori = 1, . . . , s, it is possible to define the unique field inclusionσiofK[x]intoKimposing that the restriction ofσionKisσandσi(x) =xi. Theσi’s are all the inclusion homomorphisms ofK[x]inK, the restriction of which is given byσonK.
We remark thatsis always less than or equal tod, the degree ofK[x]/K. This integersis called the degree of separability ofK[x1]overKor the degree of separability ofx1. More generally, we have:
Definition 2.60 LetLbe a finite algebraic extension ofK,Kbe the algebraic closure ofKandσ an inclusion ofKintoK. Then thedegree of separabilityofLoverKdenoted bydegs(L/K)is the numbersof different field inclusionsσi,i= 1, . . . , sofLintoKrestricting toσoverK. If degs(L/K) = deg(L/K), we say thatL/Kisseparable.
Ifx∈L, the elementsσi(x)∈Kare called theconjugatesofx.
An immediate consequence of the definition and the preceding discussion is:
28 Ch. 2 Algebraic Background
Lemma 2.61 A monogenic algebraic extensionK[x]defined by a minimal polynomialm(X)is separable if and only ifm(X)is prime to its derivativem(X).
Concerning the composition of degree of separability we have
Proposition 2.62 LetL/KandF/Kbe a tower of extension fields, then degs(F/K) = degs(F/L) degs(L/K).
We have the basic fact
Fact 2.63 LetF/LandL/K be field extensions and letx ∈ F be separable over K; then it is separable overL.
From the previous proposition, we deduce that an algebraic finite extension is separable if and only if it can be written as the composition of monogenic separable extensions. From this and the preceding fact, we can state
Proposition 2.64 A finite algebraic extensionL/K is separable if and only if every x ∈ L is separable overK.
Then the criterion of Proposition 2.61 tells us that every algebraic extension over a field of charac- teristic0is separable. We have the following definition
Definition 2.65 A field over which every algebraic extension is separable is called aperfect field.
A field is perfect if and only if every irreducible polynomial is prime to its derivative. We saw that every field of characteristic zero is perfect. More generally, we have
Proposition 2.66 A fieldK is a perfect field if and only if one the the following conditions is realized
• char(K) = 0,
• char(K) =p and Kp=K.
As a consequence of this proposition, we shall see that every finite field is perfect. The following theorem shows that every finite algebraic separable extension is in fact monogenic.
Theorem 2.67 IfL/Kis a separable finite algebraic extension ofKthenL/Kis monogenic, i.e., there existsx∈Lsuch thatL=K[x]andxis called adefining element.
Definition 2.68 An extensionL/Kis aGalois extensionif it is normal and separable. We define theGalois group ofLoverKdenoted byGL/K orGal(L/K)to be the group ofK-automorphisms ofL. There is a natural action ofGL/KonLdefined forg∈GL/K andx∈Lbyg·x=g(x). By its very definition, this action leaves the elements ofKinvariant.
IfHis a subgroup ofG, we denote byLH the set of elements ofLinvariant under the action of H. It is easy to see thatLHis a subfield ofL. Moreover,LHis a Galois extension ofKif and only ifHis a normal subgroup ofG.
Obviously, the separability condition of Galois extensions implies that the order of the Galois group ofL/Kis equal to the degree of this extension. Then, we have the following result, which is the starting point of Galois theory
Theorem 2.69 LetL/Kbe a finite Galois extension. Then there is a one-to-one correspondence between the set of subfields ofLcontainingKand the subgroups ofGL/K. To a subgroupH of GL/Kthis correspondence associates the fieldLH.
§ 2.2 Introduction to number theory 29
2.2.4 Number fields
Definition 2.70 Anumber fieldKis an algebraic extension ofQof finite degree. An element ofK is called analgebraic number.
Remarks 2.71
(i) There are number fields of any degreed, since, for instance, the polynomialXd−2is irreducible overQfor every positive integerd.
(ii) As a consequence of Theorem 2.67, for every number fieldK, there is an algebraic numberθ∈Ksuch thatK=Q(θ).
(iii) The degreedofK/Qis equal to the number of field homomorphismsσifromKtoC.
Thus ifK = Q(θ), the degreedis equal to the number of conjugatesσi(θ)ofθ and corresponds to the degree of the minimal polynomial ofθ.
Definition 2.72 Letψbe a homomorphism fromKtoC. If the image ofψis in fact included inR thenψis areal homomorphism. Otherwiseψis called acomplex homomorphism.
Definition 2.73 The numbers of real and complex homomorphisms ofK/Q, respectively denoted byr1and2r2, satisfyd=r1+ 2r2. Ifr2 = 0thenK/Qis said to betotally real. In caser1 = 0 thenK/Qistotally complex. The ordered pair(r1, r2)is called thesignature ofK/Q.
Fact 2.74 It is clear that a Galois extension must be totally real or totally complex.
Remark 2.75 The signature ofK/Qcan be found easily. Let us writeK =Q(θ)and letm(X) be the minimal polynomial ofθ. Thenr1and2r2are respectively the numbers of real and nonreal roots ofm(X).
Example 2.76 The signature of the totally real fieldQ(√
2)of degree2is(2,0).
The extensionQ(i)generated byX2+ 1is totally complex and its signature is(0,1).
The signature ofQ(θ)whereθis the unique real root of the polynomialX3−X−1is(1,1).
Proposition 2.77 LetK/Qbe a number field of degreed, letσ1, . . . , σd be the field homomor- phisms ofKtoCand letαbe an algebraic number inK. Following Definition 2.52, thetraceand normofαare explicitly given by
TrK/Q(α) = d i=1
σi(α) and NK/Q(α) = d i=1
σi(α).
Definition 2.78 LetK/Qbe a number field. An algebraic numberαis calledintegral overZor an algebraic integerifαis a zero of a monic polynomial with coefficients inZ.
The set of all the algebraic integers ofKunder the addition and the multiplication ofKis a ring, called theinteger ring ofKand is denoted byOK.
Remarks 2.79
(i) IfK =Q(θ)is of degreedthen the ringOK is aZ-module having anintegral basis, that is a set of integral elements{α1, . . . , αd}such that everyα∈ OKcan be written as
α=a1α1+· · ·+adαd for someai∈Z.
(ii) The ringZ[θ] = {f(θ) |f(X)∈Z[X]}is a subring ofOK. In general it is different fromOK.
30 Ch. 2 Algebraic Background
Example 2.80 Letαbe an algebraic integer such thatα2=nwithn∈Zdifferent from 0, 1, and squarefree. ThenOKis explicitly determined.
• Ifn≡1 (mod 4)thenOK=Z1+α
2
·
• Ifn≡2,3 (mod 4)thenOK=Z[α].
Definition 2.81 LetK be a number field of degreed. Anorder ofKis a subring ofOK of finite index which contains an integral basis of lengthd. The ringOK is itself an order known as the maximal order ofK.
Theorem 2.82 The ringOKis aDedekind ring, in other words
• it isNoetherian, i.e., every idealaofOKis finitely generated
• it isintegrally closed, that is, the set of all the roots of polynomials with coefficients in OKis equal toOK itself
• every nonzero prime ideal is maximal inOK.
In a Dedekind ring, an element has not necessarily a unique factorization. For instance, in the field Q(i√
5)the ring of integers isZ[i√
5]and one has 21 = 3×7
= (1 + 2i√
5)(1−2i√ 5).
However, if we consider ideals instead, we have unique factorization. First, we define theproduct of two idealsaandbby
ab=
i
aibi|ai∈a, bi∈b .
Theorem 2.83 LetObe a Dedekind ring andabe an ideal ofOdifferent from(0)and(1). Thena admits a factorization
a=p1. . .pr
where thepi’s are nonzero prime ideals. The factorization is unique up to the order of the factors.
Definition 2.84 LetKbe a number field and let an orderObe a Dedekind ring. Afractional ideal ofKis a submodule ofKoverO.
Remark 2.85 AnO-submoduleais a fractional ideal ofKif and only if there existsc ∈ Osuch thatca⊂ O.
The fractional ideals form a groupJKunder the product defined above and theinverse of a fractional idealainJK is the fractional ideal
a−1={x∈K|xa⊂ O}.
The fractional principal ideals, i.e., fractional ideals of the formaOfora ∈K∗form a subgroup PK ofJK.
Definition 2.86 LetKbe a field. Theclass groupofOKis the quotient groupClK =JK/PK. Theorem 2.87 For any number fieldK, the class groupClKis a finite abelian group. Its cardinality, called theclass numberis denoted byh(K).