§1. Field Theory  Lecture V  Page 1

Lecture V Fundamental Theorem of Algebra

This lecture has primarily mathematical rather than computational goals. Our principal but not sole objective is the Fundamental Theorem of Algebra. We choose a slightly circuitous route, via an investigation of the underlying real field . It is said that the Fundamental Theorem of Algebra depends on two distinct sets of properties: the algebraic properties of a field and the analytic properties of real numbers. But we will see that these “analytic properties” could be formulated purely in algebraic terms. Of course, there is no avoiding some standard construction of the real numbers , such as Dedekind cuts or Cauchy sequences, and verifying that they satisfy our algebraic axioms. But even such constructions can made in a purely algebraic setting. This development originated from Artin and Schreier who gave an affirmative solution to Hilbert’s 17th Problem: if K is a real field and f is a rational function f(X_{1},…, X_{n}) ∈ K(X_{1},…, X_{n}) that is non-negative on all (a_{1},…, a_{n}) ∈ K^{n} whenever f(a_{1},…, a_{n}) is defined, then f is a sum of squares in K(X_{1},…, X_{n}). See [3,6] for a more comprehensive treatment of real fields. More recently, much of this theory has been incorporated into the theory of quadratic forms (see for instance, [4]). For the topology of real algebraic sets, see [1,2].

§1. Elements of Field Theory

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 [6]).

Fields. A field is a commutative ring in which each non-zero element is invertible. This implies that a field is a domain. Often, our field F arises as the quotient field of a domain D. This underlying domain gives F its “arithmetical structure” which is vital for complexity 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 integer p such that \underbrace{1 + 1 + ⋯ + 1}_{p} = 0, then we say the field has characteristic p; otherwise it

has characteristic 0. It is not hard to see that p must be prime.

Extension Fields. If F ⊆ G where G is a field and F is a field under the induced operations of G, then F is a subfield of G, and G an extension field of F. An element θ ∈ G is algebraic over F if p(θ) = 0 for some p(X) ∈ F [X]; otherwise θ is transcendental. G is an algebraic extension of F if every element of G is algebraic over F. If S ⊆ G then F (S), the adjunction of F by S, denotes the smallest subfield of G that contains F ∪ S. In case S = {θ_{1},…, θ_{k}} is a finite set, we write F (θ_{1},…, θ_{k}) for F (S) and call this a finite extension. If k = 1, F (θ_{1}) is called a simple extension. It is easy to see G can be viewed as a vector space over F. We write [G : F] for the dimension of this vector space.

Simple extensions. To study the simple extension F (θ), consider the natural map ϕ : F [X] → G that takes X to θ and which fixes F (this just means ϕ(x) = x for x ∈ F). It is clear that ϕ is a homomorphism. If I is the kernel of ϕ then the image of ϕ is isomorphic to F [X]/I. Furthermore, we have I = (p) for some p(X) ∈ F [X], since I is an ideal and F [X] is a principal ideal domain. Note that p must be irreducible. [Otherwise, p = p_{1}p_{2} for some non-trivial factor p_{1}. Then 0 = ϕ(p) = ϕ(p_{1})ϕ(p_{2}) implies ϕ(p_{1}) = 0 or ϕ(p_{2}) = 0 (since G is a domain). This proves p_{1} or p_{2} is in the kernel I, contradiction.] There are now two possibilities: either p = 0 or p ≠ 0. In the former case, the image is isomorphic to F [X]/(0) = F [X] and θ is a transcendental element; in the latter case, we find that ϕ(p) = p(θ) = 0 so that θ is algebraic.

Splitting fields. Above we started out with a given extension field G of F and ask how we find simple extensions of F into G. There is a converse problem: given a field F, we want to construct an extension with prescribed properties. In case we want a simple transcendental extension, this is easy: such a G is isomorphic to F [X]. If G is to be an algebraic extension, assume we are given a polynomial p(X) ∈ F [X] and G is to be the smallest extension such that p(X) splits into linear factors in G[X]. Then G is called the splitting field of p(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) of p(X). If a non-linear polynomial p_{1}(X) that remains after removing the linear factors, let q_{1}(X) be any irreducible non-linear factor of p_{1}(X). Then the quotient ring F [X]/(q_{1}) is a domain. But it is in fact a field because (q_{1}) is a maximal ideal [For, if q ∉ (q_{1}) then the irreducibility of q_{1} implies GCD(q, q_{1}) = 1, and by the extended Euclidean algorithm F [X] = (1) = (q, q_{1}).] This extension field can be written as F (θ_{1}) where θ_{1} is the equivalent class of X in F [X]/(q_{1}). Now in F (θ_{1}), the polynomial p_{1}/q_{1} may split off additional linear factors. If a non-linear polynomial p_{1} remains, we again pick any irreducible non-linear polynomial q_{2} and extend F (θ_{1}) to F (θ_{1})[X]/(q_{2}), which we write as F (θ_{1}, θ_{2}). This process must eventually stop. The splitting field G has the form F (θ_{1},…, θ_{k}) and can be shown to be unique up to isomorphism. This proves a basic fact: for any polynomial p(X) ∈ F [X] there exists an extension field G of F in which p(X) has deg(p) roots. A field G is said to be a normal extension of F if G is an algebraic extension and every irreducible polynomial p(X) ∈ F [X] is either irreducible in G[X] or has exactly deg(p) linear factors in G[X]. For instance, a quadratic extension F (\sqrt{a}) is normal. It is not hard to show [6, vol. 1,p. 124] that splitting fields of F are normal extensions. A normal extension of a normal extension need not be a normal extension. For example, (^{4}\sqrt{2}) is a normal extension of (\sqrt{2}) but it is not a normal extension of .

Separable extensions. An irreducible polynomial f(X) ∈ F [X] may well have multiple roots α in its splitting field. Such an α is said to be inseparable over F. Fields of characteristic zero cannot have inseparable extensions (such fields are called perfect). In characteristic p > 0, it is easy to verify that inseparable polynomials f(X) have the form ϕ(X^{p^{e}}) where e ≥ 1 and ϕ(Y) is irreducible in F [Y]. If α is the simple root of an irreducible polynomial, then it is separable. An extension G of F is separable over F if if all its elements are separable over F. An extension is Galois if it is normal and separable.

Primitive element. Suppose G = F (θ_{1},…, θ_{k}) is a finite separable extension of F. Then it can be shown that G = F (θ) for some θ. Such an element θ is called a primitive element of G over F. To see the existence of such elements, it is enough to show this when k = 2: consider F (θ_{1} + cθ_{2}) for all c ∈ F. There are only finitely many such fields since they are intermediate between F and G. Suppose for c ≠ c′ we have F (θ_{1}+cθ_{2}) = F (θ_{1}+c′θ_{2}). Letting θ = θ_{1}+cθ_{2}, it is clear that F (θ) ⊆ F (θ_{1}, θ_{2}). To see the converse inclusion, note that (c′−c)θ_{2} = θ−(θ_{1}+c′θ_{2}). Hence θ_{2} ∈ F (θ) and therefore θ_{2} ∈ F (θ).

§2. Ordered Rings  Lecture V  Page 3

Zorn’s Lemma. A powerful principle in mathematical arguments is the Axiom of Choice. This usually appear in algebraic settings as Zorn’s lemma (following Kneser): if P is a partially ordered set such that every chain C in P contains an upper bound in P, then P contains a maximal element. A set C ⊆ P is a chain if for every x, y ∈ C, either x < y or x > y or x = y.

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 \overline{F}, is the smallest algebraically closed field containing F. A theorem of Steinitz says that every field F 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 that is algebraically closed.

Exercise 1.1:

(i) A quadratic extension is a normal extension.

(ii) If α is any root of X^{3} − 2 then (α) is not normal. (iii) (^{4}\sqrt{2}) is not a normal extension of .

§2. Ordered Rings

To study the real field algebraically, we axiomatize one of its distinguishing properties, namely, that it can be ordered. Although it is usual to define ordered fields, we prefer to extend this concept to rings in view of examples such as and [X].

Definition 1 Let R be a commutative ring (as always, with unity). A subset P ⊆ R is called a positive set if it satisfies these properties: (I) For all x ∈ R, either x = 0 or x ∈ P or −x ∈ P, and these are mutually exclusive cases. (II) If x, y ∈ P then x + y and xy ∈ P. We say R is ordered (by P) if R contains a positive set P, and call (R, P) an ordered ring.

As examples, is naturally ordered by the set of positive integers. If R is an ordered ring, we can order R[X] by defining the positive set P to comprise all polynomials whose leading coefficients are positive (in R).

Let P ⊆ R be a fixed positive set. We call a non-zero element x positive or negative depending on whether it belongs to P or not. For x, y ∈ R, we say x is 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 to non-negative, non-positive, greater or equal to and less than or equal to, written x ≥ 0, x ≤ 0, x ≥ y and x ≤ y. Define the absolute value |x| of x to be x if x ≥ 0 and −x if x < 0.

We now show that notations are consistent with some familiar properties of these inequality symbols.

§3. Real Rings  Lecture V  Page 4

Lemma 1 Let x, y, z be elements of an ordered ring R.

(i) x > 0 and xy > 0 implies y > 0. (ii) x ≠ 0 implies x^{2} > 0. In particular, 1 > 0. (iii) x > y implies x + z > y + z. (iv) x > y and z > 0 implies xz > yz.

(v) x > 0 implies x^{−1} > 0. (vi) x > y > 0 implies y^{−1} > x^{−1} (provided these are defined). (vii) (transitivity) x > y and y > z implies x > z. (viii) x ≠ 0, y ≠ 0 implies xy ≠ 0. (ix) |xy| = |x| · |y|.

(x) |x + y| ≤ |x| + |y|.

Exercise 2.1: Verify this.

From property (II) in the definition of an ordered ring R, we see that R has characteristic 0 (otherwise if p > 0 is the characteristic of R then \underbrace{1 + 1 + ⋯ + 1}_{p} = 0 is positive, contradiction).

Part (vii) of this lemma says that R is totally ordered by the ‘>’ relation. Part (viii) implies R is a domain.

Ordering is another example of a concept that extends naturally from a domain D to its quotient field Q_{D}. For, we can define an element u/v ∈ Q_{D} to be positive if uv > 0. It is easy to see that this defines an ordering on Q_{D} that extends the ordering of D.

§3. Real Rings

Sum of Squares. In the study of real rings, those elements that can be written as a sum of squares has a special role. Let us define

R^{(2)}

to denote the set of elements of the form ∑_{i=1}^{m} x_{i}^{2}, m ≥ 1, where the x_{i}’s are non-zero elements of R. Since the x_{i}’s are non-zero, it is not automatic that 0 belongs to R^{(2)}.

Lemma 2

(i) 1 ∈ R^{(2)} and R^{(2)} is closed under addition and multiplication. (ii) If x, y ∈ R^{(2)} and y is invertible then x/y ∈ R^{(2)}. (iii) If P ⊆ R is a positive set, then R^{(2)} ⊆ P.

Proof. (i) is easy. To see (ii), note that if y^{−1} exists then x/y = (xy)(y^{−1})^{2} is a product of elements in R^{(2)}, so x/y ∈ R^{(2)}. Finally, (iii) follows from (i) because squares are positive. Q.E.D.

This lemma shows that R^{(2)} has some attributes of a positive set. Under what conditions can R^{(2)} be extended into a positive set? From (iii), we see that 0 ∉ R^{(2)} is a necessary condition. It follows that R has characteristic 0 (otherwise \underbrace{1 + 1 + ⋯ + 1}_{p} = 0 ∈ R^{(2)} where p ≥ 2 is the

characteristic of R).

§4. Constructible Extensions  Lecture V  Page 5

Exercise 3.1: i) If the characteristic of R is not equal to 2 and R is a field then 0 ∈ R^{(2)} implies R = R^{(2)}. ii) If R^{(2)} does not contain 0 then R has no nilpotent elements. iii) Every element in GF (q) is a sum of two squares.

Definition 2 A ring R is (formally) real if x, y ∈ R^{(2)} implies xy ≠ 0.

This notion of “real” is only formal because R need not be a subset of the real numbers . The next section shows such examples. To avoid excess verbiage, we simply say “real” instead of “formally real” in this lecture. The following is immediately from lemma 2(iii):

Corollary 3 If R is ordered then R is real.

To what extent is the converse true? If R is real then 0 ∉ R^{(2)}, and x ∈ R^{(2)} implies −x ∉ R^{(2)}. A real ring R is also a domain: for if xy = 0 and x ≠ 0, y ≠ 0 then 0 ≠ x^{2} ∈ R^{(2)} and 0 ≠ y^{2} ∈ R^{(2)} but x^{2}y^{2} = 0, contradiction. So, R^{(2)} looks even more like a positive set when R is real. We will later show that R^{(2)} can be extended to a positive set of some extension of R.

§4. Constructible Extensions

We may form the quotient field Q_{R} for any real ring R. In general, if R is a domain, it is not hard to see that R is real if and only if Q_{R} is real. Henceforth, we might as well assume that we are dealing with some real field F. It is easy to show that for a field F,

F is real iff − 1 ∉ F^{(2)}.

For a real field F, we call a field extension of the form

G = F (\sqrt{a_{1}}, \sqrt{a_{2}},…, \sqrt{a_{n}}), a_{i} ∈ F,

a finite constructible extension of F provided G is real. G is a simple constructible extension if n = 1.

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) constructible if it is equal to the distance between two constructed 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 a constructed curve is 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 the constructible numbers in this sense coincide precisely with our notion the set of constructible real numbers over .

Exercise 4.1: In this exercise, constructible means “ruler-and-compass constructible”. i) Show that if S ⊆ is a set of constructible numbers, so is are the positive elements in the smallest field F ⊆ containing S. [In particular, the positive elements in are constructible.] ii) Show that if the positive elements in a field F ⊆ are constructible, so are the positive elements in F (\sqrt{a}), for any positive a ∈ F. [In view of i), it suffices to construct \sqrt{a}.] iii) Show that if x is any number constructible from elements of F ⊆ then x is in F (\sqrt{a_{1}},…, \sqrt{a_{k}}) for some positive numbers a_{i} ∈ F, k ≥ 0.

We prove a fundamental lemma:

Lemma 4 If F is a real field, a ∈ F and F (\sqrt{a}) is not real then a ∉ F^{(2)} and −a ∈ F^{(2)}.

Proof. Since F (\sqrt{a}) is a domain, F (\sqrt{a}) is not real is equivalent to 0 ∈ F (\sqrt{a})^{(2)}. Hence

0 = ∑_{i}(b_{i} + c_{i}\sqrt{a})^{2},  (b_{i}, c_{i} ∈ F)
= ∑_{i}(b_{i}^{2} + c_{i}^{2}a) + 2\sqrt{a} ∑_{i} b_{i}c_{i}
= u + v\sqrt{a},

where the last equation defines u and v. If u ≠ 0 then v ≠ 0; hence \sqrt{a} = −u/v ∈ F and F (\sqrt{a}) = F is real, contradiction. Hence we may assume u = 0. If a ∈ F^{(2)} then u which is defined as ∑_{i}(b_{i}^{2} + c_{i}^{2}a) also belongs to F^{(2)}. This give the contradiction 0 = u ∈ F^{(2)}. This proves a ∉ F^{(2)}, as required. We also see from the definition of u that

−a = (∑_{i} b_{i}^{2})/(∑_{i} c_{i}^{2}) ∈ F^{(2)},

as required. Q.E.D.

Corollary 5

(i) a ∈ F^{(2)} implies F (\sqrt{a}) is real. (ii) a ∈ F implies F (\sqrt{a}) or F (\sqrt{−a}) is real.

Constructible closure. Let H be any extension of F. Call x ∈ H a constructible element of H over F provided x ∈ G ⊆ H where G is any finite constructible extension of F. Call H a constructible extension of F if every element in H is constructible. A field F is constructible closed if for any a ∈ F, F (\sqrt{a}) = F. We call F real constructible closed if F is real and for any a ∈ F^{(2)}, F (\sqrt{a}) = F. Beware that if F is constructible closed then it cannot be real because \sqrt{−1} ∈ F; in particular, F is not real constructible closed. We define the real constructible closure F̃ of a real field F, to be the smallest real constructible closed extension of F such that for any field G, F ⊆ G ⊂ F̃, G is not constructible closed. Applying the last corollary by repeatedly extending F with squareroots of elements in F^{(2)}, Zorn’s lemma implies:

Corollary 6 Every real field F has a real constructible closure F̃.

§5. Real Closed Fields  Lecture V  Page 7

Lemma 7 If F is real constructible closed then it is uniquely ordered by F^{(2)}.

Proof. It is sufficient to show that F^{(2)} is a positive set. We already know F^{(2)} is closed under addition and multiplication, and 0 ∉ F^{(2)}. We must show that for every non-zero element x, either x or −x belongs to F^{(2)}. But this is a consequence of corollary 5 since either \sqrt{x} or \sqrt{−x} is in F. Q.E.D.

We now investigate the consequence of adding i = \sqrt{−1} to a real constructible closed field. This is analogous to the extension from to = (i).

Theorem 8 If F is real constructible closed then F (i) is constructible closed.

Proof. Let a + bi where a, b ∈ F, ab ≠ 0. We must show that there exist c, d ∈ F such that (c + di)^{2} = a + bi. Such values c, d satisfy

c^{2} − d^{2} = a,  2cd = b.  (1)

Hence (c^{2} + d^{2})^{2} = a^{2} + b^{2} ∈ F^{(2)}. It follows that c^{2} + d^{2} = e where e = \sqrt{a^{2} + b^{2}} ∈ F. As e is a sum of squares, it must be positive in F. Note that c, d must next satisfy

c^{2} = \frac{e − a}{2},  d^{2} = \frac{e + a}{2}.  (2)

But e ≥ |a| (since e^{2} = a^{2} +b^{2} ≥ |a|^{2}) implies that both (e−a)/2 and (e+a)/2 are non-negative. This proves there exist c, d ∈ F that satisfies equation (2). We must still choose the sign of c and d so that equation (1) holds. In fact, it is sufficient to choose the signs of c and d to agree iff b ≥ 0. Q.E.D.

§5. Real Closed Fields

A field is real closed if it is real and all algebraic proper extension is non-real. We will show that this property essentially identifies the real numbers, . First let us remark that if F is real closed then it must be real constructible closed: for if a ∈ F^{(2)} then F (\sqrt{a}) is real and hence \sqrt{a} ∈ F.

Lemma 9 If F is real and p(X) ∈ F [X] is irreducible of odd degree n then F (α) is real for any root α of f(X).

Proof. The result is true for n = 1. Suppose inductively that the result holds for all smaller odd values of n. If F (α) is not real then

−1 = ∑_{i∈I} q_{i}(α)^{2}

for some finite index set I and q_{i}(X) ∈ F [X], deg q_{i} ≤ n − 1. Since p(X) is irreducible, F (α) is isomorphic to F [X]/(p(X)). Thus we get

−1 = ∑_{i∈I} q_{i}(X)^{2} + r(X)p(X)

for some r(X) ∈ F [X]. But ∑_{i∈I} q_{i}(X)^{2} has a positive even degree at most 2n − 2. Hence, in order for this equation to hold, the degree of r(X)p(X) must equal that of ∑_{i∈I} q_{i}(X)^{2}. Then r(X) has odd degree at most n − 2. If r′(X) is any irreducible factor of r(X) of odd degree and β is a root of r′(X) then we get

−1 = ∑_{i∈I} q_{i}(β)^{2}.

This proves F (β) is not real, constradicting our inductive assumption. Q.E.D.

For instance, X^{3} − 2 is irreducible in [X] (cf. Lecture III.1, using Eisenstein’s criteria). Thus (α) is formally real for α any cube-root of 2. If we choose α ∉ , then (α) is not a subset of .

Theorem 10 (Characterization of real closed fields) The following statements are equivalent.

(i) F is real closed. (ii) F is real constructible closed and every polynomial in F [X] of odd degree has a root in F. (iii) F is not algebraically closed but F (i) is.

Proof.

(i) implies (ii): this follows from the previous lemma and the remark at the beginning of this section.

(ii) implies (iii): clearly F is real implies it is not algebraically closed since X^{2} + 1 has no solution in F. To see that F (i) is algebraically closed, we claim it suffices to prove that nonconstant polynomials in F [X] has roots in F (i). Write \overline{f(X)} for the conjugate of f(X) ∈ F (i)[X], obtained by conjugating each coefficient (the conjugate of x + yi is x − yi). It is not hard to verify that

g(X) = f(X)\overline{f}(X) ∈ F [X]

where f(X) ∈ F (i)[X]. Moreover, if g(X) has a root α ∈ F (i) implies f(α) = 0 or \overline{f}(α) = 0. But the latter is equivalent to f(\overline{α}) = 0. So g(X) has a root in F (i) iff f(X) has a root in F (i), establising our claim. Hence we may assume that we are given g ∈ F [X]. We now use induction on n = deg g to show that g has a root in F (i). Let n = 2^{i}q where q is odd and i ≥ 0. We already know that F (i) is constructible closed (theorem 8). Thus the result is true if i = 0. We now assume i ≥ 1. Let α_{1},…, α_{n} be the roots of g in an algebraic extension of F. We may assume these roots are distinct since otherwise we can easily show g already has a root in F (i). Consider the set of values

α_{j}α_{k} + c(α_{ℓ} − α_{m})

where 1 ≤ j < k ≤ n and 1 ≤ ℓ < m ≤ n. There are O(n^{4}) values of c for which these are not distinct because each coincidence of values uniquely determines a c. Since F is infinite, we may pick c so that these are all distinct. Next consider the set

B = {α_{j}α_{k} + c(α_{j} − α_{k}) : 1 ≤ j < k ≤ n}

§6. Fundamental Theorem  Lecture V  Page 9

of N = \binom{n}{2} values. Let s_{i} denote ith elementary symmetric function of the elements in B (thus s_{1} = ∑_{x∈B} x). But s_{i} is also symmetric in α_{1},…, α_{n}. Hence the s_{i} are rational integral polynomials in the elementary symmetric functions σ_{0},…, σ_{n} on α_{1},…, α_{n}. But these σ_{i}’s are precisely the coefficients of g(X). Thus the polynomial G(X) = ∑_{i=0}^{N} s_{i}X^{i} belongs to F [X] and its roots are precisely the elements of B. Notice the degree of G is N = 2^{m−1}q′ for some odd q′. By induction hypothesis, G has a root in F (i). Without loss of generality, let this root be

ϕ = α_{1}α_{2} + c(α_{1} − α_{2}) ∈ F (i).

Similar to the construction of G(X), let u(X) ∈ F [X] be the polynomial whose roots are precisely the set {α_{j}α_{k} : 1 ≤ j < k ≤ n}, and likewise v(X) ∈ F [X] be the polynomial whose roots are {α_{j} − α_{k} : 1 ≤ j < k ≤ n}. We note that

u(α_{1}α_{2}) = 0, v(\frac{ϕ − α_{1}α_{2}}{c}) = v(α_{1} − α_{2}) = 0.

Moreover, for any α_{j}α_{k} where (j, k) ≠ (1, 2), v(\frac{ϕ−α_{j}α_{k}}{c}) ≠ 0 because our choice of c implies \frac{ϕ−α_{j}α_{k}}{c} ≠ α_{ℓ} − α_{m} for any 1 ≤ ℓ < m ≤ n. This means that the polynomials

u(X), v(\frac{ϕ − X}{c}) ∈ F (ϕ)[X] ⊆ F (i)[X]

have α_{1}α_{2} as their only common root. Their GCD is thus X − α_{1}α_{2}, which must be an element of F (i)[X]. This proves that α_{1}α_{2} ∈ F (i) and therefore α_{1} − α_{2} ∈ F (i). We can determine α_{1} and α_{2} by solving a quadratic equation in F (i). This proves g(X) has solutions α_{1}, α_{2} in F (i).

(iii) implies (i): we must show that F is real. We first observe that an irreducible polynomial f(X) in F [X] must have degree 1 or 2 because of the inequality

2 = [F (i) : F] ≥ [E : F] = deg f

where E ⊆ F (i) is the splitting field of f over F. Next we see that it is sufficient to show that every sum a^{2} + b^{2} of squares (a, b ∈ F) is a square in F. For, by induction, this would prove that every element in F^{(2)} is a square, and the formal reality of F then follows because −1 is not a square in F. To show a^{2} +b^{2} is a square, we may assume ab ≠ 0. Consider the polynomial f(X) = (X^{2} − a)^{2} + b^{2} ∈ F [X]. It factors as

(X^{2} − a − bi)(X^{2} − a + bi)

over F (i). Since F (i) is algebraically closed, there are c, d ∈ F (i) such that

c^{2} = a + bi, d^{2} = a − bi.

This gives f(X) = (X −c)(X +c)(X −d)(X +d). Note that ±a±bi are not elements in F. Thus f(X) has no linear factors in F [X]. It must therefore split into quadratic factors. Consider the factor that contains X − c. This cannot be (X − c)(X + c) = X^{2} − c^{2}. Hence it must be (X − c)(X ± d). In either case, we notice on multiplying out that \sqrt{a^{2} + b^{2}} ∈ F, as we wanted to show. Q.E.D.

Exercise 5.1: Show that f(X)\overline{f}(X) ∈ F [X] if f(X) ∈ F (i)[X].

Exercise 5.2: Let a, b ∈ F and f(X) ∈ F [X] where F is real closed. If f(a) < 0 and f(b) > 0 then there exists c between a and b such that f(c) = 0. HINT: suffices to show this for deg f = 2.

§6. Fundamental Theorem of Algebra

D’Alembert (1746) published the first formulation and proof of the Fundamental Theorem of Algebra. Gauss is credited with the first proof that is acceptable by modern standards^{1}. For this section, “real” refers to the real numbers . We note two analytic properties of real numbers:

1. The reals are ordered (and hence formally real).

2. (Weierstrass’ Nullstellensatz) If f(X) is a real function continuous in an interval [a, b], and f(a)f(b) < 0 then f(c) = 0 for some c between a and b.

Recall that f(X) is continuous at a point X = a if for all ϵ > 0 there is a δ > 0 such that |f(a + d) − f(a)| < ϵ whenever |d| < δ. It is not hard to show that the constant functions, the identity function, the sums and products of continuous functions are all continuous. In particular, polynomials are continous. It is not hard to conclude from Weierstrass’ Nullstellensatz:

a) A positive real number c has a positive real square root, \sqrt{c}.

b) A real polynomial f(X) of odd degree has a real root. It follows from theorem 10 that is real closed. Since is defined to be (\sqrt{−1}), we obtain:

Theorem 11 (Fundamental Theorem of Algbera) is algebraically closed.

Exercise 6.1: Prove properties a) and b).

Cantor’s construction of the real closure. Let us briefly recall a construction of the reals. If F is a field and F̂ is the smallest real closed field containing F, then we call F̂ a real closure of F. The real closure of any real field F exists, by Zorn’s lemma. But we now outline a “constructive” method to obtain a real closure of F assuming F is ordered. This just mirrors the construction of from using Cauchy sequences. A Cauchy or fundamental sequence of F is a = (a_{1}, a_{2},…) such that for all positive ϵ there exists n = n(ϵ) such that |a_{i} − a_{j}| < ϵ for all i, j > n. We may add and multiply such sequences in a componentwise manner. In particular, if b = (b_{1}, b_{2},…) then ab = (a_{1}b_{1}, a_{2}b_{2},…). It is easy to check that Cauchy sequences form a commutative ring. We define a sequence (a_{1}, a_{2},…) to be null if for all ϵ > 0 there exists n = n(ϵ) where a_{i} > ϵ for i > n. Similarly, we define the sequence to be positive if there is an ϵ > 0 and n such that for all i > n, a_{i} > ϵ. The set of null sequences form a maximal ideal in the ring of fundamental sequences; hence the ring of Cauchy sequences modulo this null ideal is a field F̂ that extends F in a canonical way. If F = then F̂ is, by definition, equal to . Now F̂ is an ordered field because the set of positive Cauchy sequences correspond to the positive

elements of F̂. This construction, if repeated on F̂ yields nothing new: we can show that ˆ F̂ is isomorphic to F̂. An ordering of R is Archimedean if for all a ∈ R, there is an n ∈ such that a < n. If F is archimedean ordered, we have a canonical isomorphism between F̂ and .

1Incidentally, one of Gauss’ proofs was in turn found wanting, presumably also by modern standards only!

Apropos of a footnote in (§VI.1) concerning the second-class status of complex numbers, Gauss personally marked the transition to the modern view of complex numbers: in his 1799 dissertation on the Fundamental Theorem of Algebra, he deliberately avoided imaginary numbers (by factoring polynomials only up to linear or quadratic factors). In 1849, he returned to give his fourth and last proof of the theorem, using imaginaries. See [5, p. 116,122].

References

[1] S. Akbulut and H. King. Topology of Real Algebraic Sets. Mathematical Sciences Research Institute Publications. Springer-Verlag, 1992.

[2] R. Benedetti and J.-J. Risler. Real algebraic and semi-algebraic sets. Actualités Mathématiques. Hermann, Paris, 1990.

[3] Serge Lang. Algebra. Addison-Wesley, 1971.

[4] Winfried Scharlau. Quadratic and Hermitian Forms. Grundlehren der mathematischen Wissenschaften. Springer-Verlag, Berlin, 1985.

[5] D.J. Struik, editor. A source book in Mathematics, 1200-1800. Princeton University Press, 1986.

[6] B.L. van der Waerden. Algebra, volume 1 & 2. Frederick Ungar Publishing Co., 1970.