We briefly review the some basic algebraic properties of field. For a proper treatment, there are many excellent textbooks (including van der Waerden’s classic [209]).
Fields. A field is a commutative ring in which each non-zero element is invertible. This implies that a field is a domain. Often,Farises as the quotient field of a domainD. This underlying domain gives F its “arithmetical structure” which is important for other considerations. For instance, in Lecture III.1, we showed that the concepts of divisibility and unique factorization in a domain extend naturally to its quotient field. If there is a positive integerpsuch that 1 + 1 + · · ·+ 1
p
= 0. If pis chosen as small as possible, we say the field hascharacteristic p; if no such pexists, it has characteristic 0. One verifies thatpmust be prime.
Extension Fields. IfF ⊆G whereGis a field andF is a field under the induced operations of
G, thenF is asubfield ofG, andGanextension field ofF. An element θ∈Gis algebraic overF
if p(θ) = 0 for some p(X)∈ F[X]; otherwise θ is transcendental Gis an algebraic extensionof F
if every element ofG is algebraic overF. If S ⊆Gthen F(S), the adjunction of F by S, denotes the smallest subfield of Gthat contains F ∪S. In caseS = {θ1, . . . , θk} is a finite set, we write
F(θ1, . . . , θk) forF(S) and call this afinite extension. If k= 1,F(θ1) is called asimple extension. It is easy to seeGcan be viewed as a vector space overF. Let [G:F] denote the dimension of this vector space. We call [G:F] thedegreeofGoverF.
Simple extensions. To study the simple extensionF(θ), consider the natural mapφ:F[X]→G
that takesX to θ and which fixes F (this just meansφ(x) =x forx∈F). It is clear thatφ is a homomorphism. IfIis the kernel ofφthen the image ofφis isomorphic toF[X]/I. Furthermore, we haveI= (p) for somep∈F[X], sinceIis an ideal andF[X] is a principal ideal domain. Note thatp
must be irreducible. [Otherwise,p=p1p2for some non-trivial factorp1. Then 0 =φ(p) =φ(p1)φ(p2) implies φ(p1) = 0 or φ(p2) = 0 (since G is a domain). This proves p1 or p2 is in the kernel I, 1LetKbe the field of rational numbers. Hilbert asks if a rational functionf∈K(X1, . . . , Xn) that is non-negatives
at every point (a1, . . . , an)∈Knfor whichf(a1, . . . , an) is defined, is necessarily a sum of squares of rational functions. Artin answered affirmatively in the more general case of any real closed fieldK.
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contradiction.] There are now two possibilities: eitherp= 0 orp= 0. In the former case, the image
is isomorphic toF[X]/(0) =F[X] andθis a transcendental element; in the latter case, we find that
φ(p) =p(θ) = 0 so that θ is algebraic. In casep= 0, it is also easy to see thatF[θ] = F(θ): every non-zero element ofF[θ] has the form q(θ) for some polynomialq(X)∈F[X]. We show thatq(θ) has a multiplicative inverse. By the extended Euclidean algorithm, there existsa(X), b(X)∈F[X] such thatq(X)a(X) +p(X)b(X) = 1. Then p(θ) = 0 implies q(θ)a(θ) = 1,i.e., a(θ) is the inverse ofq(θ).
Splitting fields. Above we started out with a given extension field GofF and ask how we find simple extensions ofF intoG. There is a converse problem: given a fieldF, we want to construct an extension with prescribed properties. In case we want a simple transcendental extension, this is easy: such aGis isomorphic toF(X). IfGis to be an algebraic extension, assume we are given a polynomialp(X)∈F[X] andGis to be the smallest extension such thatp(X) splits into linear factors inG[X]. ThenGis called thesplitting fieldofp(X), and is unique up to isomorphism. We now show such a splitting field may be constructed, proceeding in stages. First let us split off all linear factors
X−α(α∈F) ofp(X). If a non-linear polynomialp1(X) remains after removing the linear factors, let
q1(X) be any irreducible non-linear factor ofp1(X). Then the quotient ringF[X]/(q1) is a domain. But it is in fact a field because (q1) is a maximal ideal. [For, if q∈ (q1) then the irreducibility of
q1 implies GCD(q, q1) = 1, and by the extended Euclidean algorithm F[X] = (1) = (q, q1).] This extension field can be written asF(θ1) where θ1 is the equivalent class ofX in F[X]/(q1). Now in
F(θ1), the polynomialp1/(X−θ1) may split off additional linear factors. If a non-linear polynomial
p2 remains after removing these linear factors, we again pick any irreducible factor q2 of p2, and extendF(θ1) toF(θ1)[X]/(q2), which we write asF(θ1, θ2), etc. This process must eventually stop. The splitting fieldGhas the formF(θ1, . . . , θk) and can be shown to be unique up to isomorphism. We have shown: for any polynomial p(X)∈ F[X] there exists an extension field Gof F in which
p(X)has deg(p)roots.
Normal extensions. A fieldGis said to be anormal extensionofF (or,normal overF) ifGis an algebraic extension and for every irreducible polynomialp(X)∈F[X], either Ghas no roots of
p(X) orGcontains the splitting field of p(X). We can equivalently characterize normal extensions as follows: two elements of G are conjugates of each otherover F if they have the same minimal polynomial inF[X]. ThenG is a normal extension of F iff G is closed under conjugates overF, i.e., if a ∈ G then G contains all the conjugates of a over F. If G is also a finite extension of
F, it can be shown that G must be a splitting field of some polynomial over F. For instance, a quadratic extensionF(√a) is normal overF. On the other hand,Q(a1/3)⊆Ris not normal overQ for any positive integerathat is square-free. To see this, note that by Eisenstein’s criterion (§III.1,
Exercise),X3−ais irreducible overZand hence overQ. But
X3−a= (X−a1/3)(X−ρa1/3)(X−ρ2a1/3)
whereρ, ρ2= (−1±√−3)/2 are the two primitive cube-roots of unity. IfQ(a1/3) were normal over
QthenQ(a1/3) would contain a non-real elementρa1/3, which is impossible. It is not hard to show that splitting fields ofF are normal extensions. A normal extension of a normal extension ofF need not be a normal extension ofF (Exercise).
Separable extensions. An irreducible polynomialf(X)∈F[X] may well have multiple rootsα
in its splitting field. Such anαis said to be inseparableoverF. But ifαis a multiple root off(X), then it is a common root off(X) anddf(X)/dX =f(X). Sincef(X) is irreducible, this implies
f(X) is identically zero. Clearly this is impossible ifF has characteristic zero (in general, such fields are calledperfect). In characteristicp >0, it is easy to verify thatf(X)≡0 impliesf(X) =φ(Xpe) for somee≥1 andφ(Y) is irreducible in F[Y]. Ifαis a simple root of an irreducible polynomial,
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then it isseparable. An extensionGofF isseparableoverF if if all its elements are separable over
F. An extension isGaloisif it is normal and separable.
Galois theory. IfE is an extension field ofF, let Γ(E/F) denote the group of automorphisms of
E that fixesF. We call g∈Γ(E/F) an automorphism ofE overF. We claim: g must map each
θ∈Eto a conjugate elementg(θ) overF. [In proof, note that ifp(X)∈F[X] then
g(p(θ)) =p(g(θ)).
The claim follows if we let p(X) be the minimal polynomial ofθ, wherebyp(θ) = 0 and g(p(θ)) =
g(0) = 0, so thatp(g(θ)) = 0.] In consequence, the group Γ(E/F) is finite whenEis a splitting field of some polynomialp(X) over F. To see this, note that our claim implies that each g ∈ Γ(E/F) determines a permutationπ of α1, . . . , αn. Conversely, each permutationπcan extend to at most one g ∈Γ(E/F) since g is completely determined by its action on the roots ofp(X) becauseE is generated by the roots ofp(X) overF.
IfG is any subgroup of Γ(E/F), then thefixed field of G is the set of elementsx∈E such that
g(x) = xfor all g ∈ G. Galois theory relates subgroups of Γ(E/F) to the subfields of E over F. Two subfieldsK, K ofEoverF areconjugateif there is an automorphismσofEoverF such that
σ(K) =K.
The Fundamental theorem of Galois theory says this. Supposep(X)∈F[X] is separable overF and
E is the splitting field ofp(X).
(i) There is a one-one correspondence between subfields ofEoverFand the subgroups of Γ(E/F): a subfieldK corresponds to a subgroupH iff the fixed field ofH is equal toK.
(ii) IfK is another subfield that corresponds toH⊆Γ(E/F) thenK⊆K iffH ⊆H. (iii) IfK andK are conjugate subfields thenH andH are conjugate subgroups.
Primitive element. SupposeG=F(θ1, . . . , θk) is a finite separable extension of an infinite field
F. Then it can be shown that G = F(θ) for some θ. Such an element θ is called a primitive elementfield!primitive element ofGoverF. The existence of such elements is easy to show provided we accept the fact2 that are only finitely many fields that are intermediate betweenF andG: it is enough to show this whenk= 2. Consider F(θ1+cθ2) for all c∈F. Since there are only finitely many such fields (being intermediate between F and G), suppose F(θ1+cθ2) = F(θ1+cθ2) for somec=c. Letting θ =θ1+cθ2, it is clear thatF(θ)⊆F(θ1, θ2). To see the converse inclusion, note that (c−c)θ2=θ−(θ1+cθ2). Henceθ2∈F(θ) and alsoθ1∈F(θ).
Zorn’s Lemma. A powerful principle in mathematical arguments is the Axiom of Choice. This usually appears in algebraic settings as Zorn’s lemma (following Kneser): ifP is a partially ordered set such that every chain C inP has an upper bound inP, thenP contains a maximal element. A set C⊆P is a chain if for everyx, y∈C, eitherx < yor x > yor x=y. A typical application is this: letP be a collection of fields, partially ordered by set inclusion. IfC is a chain inP, we note that itsunion∪Cis also a field defined in the natural way: ifx, y∈ ∪C then there is a fieldF ∈C
that containsx, yand we definex+yandxyas if they are elements inF. Assume thatP is closed under unions of chains. Then Zorn’s lemma implies thatP contains a maximal field.
Algebraic Closure. If every non-linear polynomial in F[X] is reducible then we say that F is algebraically closed. The algebraic closure of F, denoted F, is a smallest algebraically closed field
§2. Ordered Rings Lecture V Page 127
containingF. A theorem of Steinitz says that every fieldF has an algebraic closure, and this closure is unique up to isomorphism. The proof uses Zorn’s lemma. But the existence of algebraic closures is intuitively clear: we simply iterate the splitting field construction for each polynomial, using transfinite induction. The Fundamental Theorem of Algebra is the assertion thatCis algebraically closed.
Exercises
Exercise 1.1:
(i) A quadratic extension is a normal extension.
(ii) Letabe a positive square-free integer. Ifαis any root ofX3−athenQ(α) is not normal. (This is more general than stated in the text.)
(iii) Q(√42) is not a normal extension of Q. Thus, a normal extension of a normal extension
need not be a normal extension. HINT: X4−2 = (X2−√2)(X2+√2). 2
Exercise 1.2: The splitting fieldEoff(X)∈F[X] overF has index [E:F]≤n! wheren= deg(f).
HINT: use induction onn. 2
Exercise 1.3:
(i) Compute a basis ofEoverF =Qin the following cases ofE: E=Q(√2,√3),Q(√2,√−2),
Q(√2,√32),Q(√2, ω) whereω= (1 +√−3)/2,Q(√32, ω),Q(√2,√32,√35).
(ii) Compute the group Γ(E/F), represented as a subgroup of the permutations on the previ-
ously computed basis. Which of these extensions are normal? 2
§2. Ordered Rings
To study the real field Ralgebraically, we axiomatize one of its distinguishing properties, namely, that it can3be ordered.
LetR be a commutative ring (as always, with unity). A subsetP ⊆R is called apositive set if it satisfies these properties:
(I) For allx∈R, eitherx= 0 orx∈P or−x∈P, and these are mutually exclusive cases. (II) Ifx, y∈P thenx+yandxy∈P.
We sayR isordered(byP) ifRcontains a positive setP, and call (R, P) anordered ring.
As examples,Zis naturally ordered by the set of positive integers. If R is an ordered ring, we can extend this ordering to the polynomial ring R[X], by defining the positive set P to comprise all polynomials whose leading coefficients are positive (inR).
LetP ⊆R be a fixed positive set. We call a non-zero elementxpositive ornegative depending on whether it belongs to P or not. For x, y ∈ R, we say xis less than y, written “x < y”, if y−x
is positive. Similarly, x is greater than y if x−y ∈ P, written “x > y”. In particular, positive and negative elements are denoted x > 0 and x < 0, respectively. We extend in the usual way the terminology tonon-negative,non-positive, greater or equal toandless than or equal to, written
x≥0,x≤0,x≥yand x≤y. Define theabsolute value|x|ofxto bexifx≥0 and−xifx <0. 3It is conventional to define “ordered fields”. But the usual concept applies to rings directly. Moreover, we are
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We now show that notations are consistent with some familiar properties of these inequality symbols.
Lemma 1 Let x, y, z be elements of an ordered ringR. (i)x >0and xy >0 impliesy >0.
(ii) x= 0 impliesx2>0. In particular, 1>0. (iii)x > y impliesx+z > y+z.
(iv)x > y andz >0 impliesxz > yz. (v)x >0 impliesx−1>0.
(vi)x > y >0 impliesy−1> x−1 (provided these are defined). (vii) (transitivity) x > yandy > z impliesx > z.
(viii)x= 0, y= 0implies xy= 0.
(ix)|xy|=|x| · |y|.
(x)|x+y| ≤ |x|+|y|.
(xi)x2> y2 implies|x|>|y|.
The proof is left as an exercise.
From property (II) in the definition of an ordered ringR, we see thatRhas characteristic 0 (otherwise ifp >0 is the characteristic ofRthen 1 + 1 + · · ·+ 1
p
= 0 is positive, contradiction). Parts (ii) and (iii) of the lemma implies that 0 < 1 < 2 < · · ·. Part (vii) of this lemma says that R is totally ordered by the ‘>’ relation. Part (viii) implies Ris a domain.
An ordered domain (or field) is an ordered ring that happens to be a domain (or field). If D is an ordered domain, then its quotient field QD is also ordered: define an element u/v ∈QD to be positive ifuvis positive inD. It is easy to verify that this defines an ordering onQD that extends the ordering onD.
Exercises
Exercise 2.1: Verify lemma 1. 2
Exercise 2.2: In an ordered fieldF, the polynomialXn−chas at most one positive root, denoted n
√
c. Ifnis odd, it cannot have more than one root; ifnis even, it has at most two roots (one
is the negative of the other). 2
Exercise 2.3: If the ordering ofQDpreserves the ordering ofD, then this ordering ofQDis unique. 2
§3. Formally Real Rings
Sum of Squares. In the study of ordered rings, those elements that can be written as sums of squares have a special role. At least for the real fields, these are necessarily positive elements. Are they necessarily positive in an ordered ringR? To investigate this question, let us define
§4. Constructible Extensions Lecture V Page 129
to denote the set of elements of the formmi=1x2i,m≥1, where thexi’s are non-zero elements of
R. Thexi’s here are not necessarily distinct. But since the xi’s are non-zero, it is not automatic that 0 belongs toR(2). Indeed, whether 0 belongs toR(2) is critical in our investigations.
Lemma 2
(i)1∈R(2) andR(2) is closed under addition and multiplication. (ii) Ifx, y ∈R(2) andy is invertible then x/y∈R(2).
(iii) IfP ⊆Ris a positive set, thenR(2)⊆P.
Proof. (i) is easy. To see (ii), note that ify−1exists thenx/y= (xy)(y−1)2 is a product of elements inR(2), sox/y∈R(2). Finally, (iii) follows from (i) because squares are positive. Q.E.D.
This lemma shows thatR(2) has some attributes of a positive set. Under what conditions canR(2)
be extended into a positive set? From (iii), we see that 0 ∈ R(2) is a necessary condition. This further implies thatR has characteristicp= 0 (otherwise 1 + 1 + · · ·+ 1
p
= 0∈R(2)).
A ring Ris formally realif 0∈R(2). This notion of “real” is only formal becauseR need not be a subset of the real numbersR(Exercise). The following is immediately from lemma 2(iii):
Corollary 3 IfR is ordered then Ris formally real.
To what extent is the converse true? If R is formally real, then 0 ∈ R(2), and x ∈ R(2) implies −x∈R(2). So,R(2) has some of the attributes of a positive set. In the next section, we show that ifRis a formally real domain thenR(2) can be extended to a positive set of some extension of R.
Exercises
Exercise 3.1: (a) If the characteristic ofR is not equal to 2 andR is a field then 0∈R(2) implies
R=R(2).
(b) IfR(2) does not contain 0 thenR has no nilpotent elements.
(c) Every element inGF(q) is a sum of two squares. 2
Exercise 3.2:
(a) Letα∈Cbe any root ofX3−2. ThenQ(α) is formally real (but not necessarily real). (b) Let Q(α) be an algebraic number field andf(X) is the minimal polynomial ofα. Then
Q(α) is formally real ifff(X) has a root inR. 2
Exercise 3.3: LetK be a field.
(a) Let G2(K) denote the set {x∈K\ {0} :x =a2+b2, a, b∈ K}. Show thatG2(K) is a group under multiplication. HINT: consider the identity|zz|=|z|·|z|wherez, zare complex numbers.
(b) LetG4(K) denote the set{x∈K\ {0}:x=a2+b2+c2+d2, a, b, c, d∈K}. Show that G4(K) is a group under multiplication. HINT: consider the identity|qq|=|q| · |q|whereq, q
§4. Constructible Extensions Lecture V Page 130
§4. Constructible Extensions
Let F be a formally real field. For instance, if D be a domain, then its quotient field F =QD is formally real iffD is formally real. It is immediate that
F is formally real iff −1∈F(2).
We call a field extension of the form
G=F(√a1,√a2, . . . ,√an), ai∈F,
a finite constructible extension of F provided G is formally real. If n = 1, we call G a simple constructible extension. Note that “F(√a)” is just a convenient notation for the splitting field of the polynomialX2−aoverF (that is, we do not know if√acan be uniquely specified as the “positive square root ofa” at this point).
Ruler and compass constructions. Our “constructible” terminology comes from the classical problem of ruler-and-compass constructions. More precisely, a number is(ruler-and-compass) con- structible if it is equal to the distance between twoconstructed points. By definition, constructible numbers are positive. Initially, we are given two points (regarded as constructed) that are unit distance apart. Subsequent points can be constructed as an intersection point of two constructed curves where aconstructed curveis either a line through two constructed points or a circle centered at a constructed point with radius equal to a constructed number. [Thus, our ruler is only used as a “straight-edge” and our compass is used to transfer the distance between two constructed points as well as to draw circles.] The following exercise shows that “constructible numbers” in this sense coincides with our abstract notion of constructible real numbers overQ.
Exercise 4.1: In this exercise, constructible means “ruler-and-compass constructible”.
i) Show that if S ⊆ R is a set of constructible numbers, so are the positive elements in the smallest fieldF⊆RcontainingS. [In particular, the positive elements inQare constructible.] ii) Show that if the positive elements in a field F ⊆ Rare constructible, so are the positive elements in F(√a), for any positivea∈F. [In view of i), it suffices to construct√a.]
iii) Show that if x is any number constructible from elements of F ⊆ R then x is in
F(√a1, . . . ,√ak) for some positive numbersai∈F,k≥0. 2
Lemma 4 If F is a formally real field, a∈F and F(√a) is not formally real then a∈ F(2) and −a∈F(2).
Proof. F(√a) is not formally real is equivalent to 0∈F(√a)(2). Hence
0 = i (bi+ci√a)2, (bi, ci∈F) = i (b2i +c2ia) + 2√a i bici = u+v√a,
where the last equation definesuandv. Ifu= 0 thenv= 0; hence√a=−u/v∈F andF(√a) =F
§4. Constructible Extensions Lecture V Page 131