Another topological proof of the Fundamental Theorem of Algebra

Elem. Math. 57 (2002) 32 – 37

0013-6018/02/010032-6 $ 1.50+0.20/0 Elemente der Mathematik

Another topological proof of the Fundamental Theorem of Algebra

J.M. Almira, M. Jime´nez, N. Del Toro*

J.M. Almira obtained his Ph.D. on approximation spaces at La Laguna University in 1999. Presently, he is associate professor at the Department of Mathematics of Jae´n University. His main interests are: approximation theory, differential equations, new proofs of classical results, and the history of mathematics.

M. Jime´nez graduated at Granada University in 1989. He also obtained a degree in management sciences at U.N.E.D. University in 2001. Presently, he works with J.M. Almira on his Ph.D. thesis. He is interested in mathematical modeling and optimization techniques.

N. Del Toro graduated at La Laguna University in 1999. Presently, she has a research grant from Junta de Andalucı´a to complete her Ph.D. thesis on approximation theory.

1 Introduction

The Fundamental Theorem of Algebra claims that every polynomialp(z)∈C[z]can be decomposed as a product of linear factors, p(z) =cn

i=1(z−α_i). Now, if we assume thatp(z)is a monic polynomial, thenc=1 and

p(z) =zⁿ−a1(α1, . . . , α_n)zⁿ⁻¹+a2(α1, . . . , α_n)zⁿ⁻²+· · ·+ (−1)ⁿan(α1, . . . , α_n),

.

Allen Lesern wird der Fundamentalsatz der Algebra bekannt sein. Er besagt, dass jedes Polynom P = P(z) u¨ber dem Ko¨rper der komplexen Zahlen mindestens eine komplexe Nullstelle hat. Besitzt P den Gradn, so ergibt sich daraus sofort, dassP (mit Vielfachheiten geza¨hlt) genau n komplexe Nullstellen hat. Die Bestimmung der Nullstellen von Polynomen spielte in der Entwicklung der Algebra eine wichtige Rolle.

Allerdings gelang es erst N.H. Abel zu beweisen, dass die Nullstellen eines Polynoms vom Gradn>4 in der Regel nicht durch Radikale darstellbar sind. Damit musste zum Beweis des Fundamentalsatzes nach neuen Ideen gesucht werden. Neben den Beweisen von C.F. Gauss wird der Fundamentalsatz heute sehr oft als elegante Anwendung aus dem Satz von Liouville in der Funktionentheorie gefolgert. Im vorliegenden Beitrag geben die Autoren einen ebenfalls eleganten Beweis des Fundamentalsatzes, der auf einfachen Ergebnissen der Topologie beruht.

∗) Research partially supported by Junta de Andalucı´a, Grupo de Investigacio´n “Aproximacio´n y Me´todos Nume´ricos” FQM-0178.

where

ak(α1, . . . , α_n) =

1≤i1<i2<···<ik≤n

α_i1α_i2. . . α_ik

are the elementary symmetric functions. Hence the next result holds:

Theorem 1 The following claims are equivalent:

(A) The Fundamental Theorem of Algebra (B) The mapσⁿ:Cⁿ→Cⁿ given by

σⁿ(α1, . . . , α_n) = (a1(α1, . . . , α_n),a2(α1, . . . , α_n), . . . ,a_n(α1, . . . , α_n))

is onto for alln∈N.

Proof.Leta= (a1, . . . ,an)∈Cⁿbe arbitrarily chosen and set pa(z):=zⁿ+

n k=1

(−1)^ka_kz^n−k.

It follows from the Fundamental Theorem of Algebra thatpa(z) =n

i=1(z−α_i0)for a certain choice of complex numbers{α_i0}ⁿi=1⊂C. Hencea=σⁿ(α10, . . . , α_n0)and (A) implies (B). On the other hand, letp(z) =zⁿ+_n

k=1(−1)^kakz^n−kbe arbitrarily chosen.

Then there exists a certain pointα= (α₁₀, . . . , α_n0)∈Cⁿsuch that(a1, . . . ,an) =σⁿ(α).

This implies the identityp(z) =_n

i=1(z−α_i0), which proves (A). 䊐 The main goal of this note is to give an elementary proof of (B) of the theorem above.

For the proof we will need to use Brouwer’s Theorem of Invariance of Domain (see [2, Theorem 36.5, p. 207]) and a certain separation property ofRⁿ :

Theorem 2 (Brouwer) Let us assume that f :Ω⊂Rⁿ→Rⁿ is a continuous injective map, andΩis an open subset ofRⁿ. Then f(Ω)is an open subset ofRⁿ.

Theorem 3 Let M ⊂Rⁿ be an embedded submanifold of Rⁿ of dimension ≤n−2, and let∆ be a subset of M. Then ∆⊂Rⁿ does not separateRⁿ (we say that the set

∆⊂Rⁿ separatesRⁿifRⁿ\∆is not arcwise connected).

Proof. Recall that the k-manifoldM ⊂Rⁿ is embedded in Rⁿ if and only if for each x ∈ M there exists a bounded neighborhood Ux of x in Rⁿ and a homeomorphism ϕx :C_n→Ux such that ϕx(0_n) =xandM∩Ux =ϕx(C_n^k), whereC_n = [−1,1]ⁿ,0_n denotes the origin of coordinates ofRⁿandC_n^k= [−1,1]^k× {0_n−k}. If k ≤n−2 then C_n\C_n^k is arcwise connected, so that

ϕx(C_n\C_n^k) =Ux\(M∩Ux)

is arcwise connected. It follows from the fact thatRⁿhas a numerable dense subset that there exists a sequence{x_n}^∞_n=1⊂M such that

(Ux_n\(M∩Ux_n))∩(Ux_n+1\(M∩Ux_n+1))=∅

for alln, andM⊂_∞

n=1Ux_n. Hence T(M) =

∞ n=1

(Ux_n\(M∩Ux_n))

is arcwise connected.

We may assume without loss of generality that∆ =M, since ifMdoes not separateRⁿ then∆cannot separateRⁿ. Leta,b∈Rⁿ\Mbe arbitrarily chosen and letα:[0,1]→Rⁿ be a path with end points {a,b}. If α([0,1])∩M =∅then we will find another path β : [0,1] → Rⁿ withβ([0,1])∩M = ∅. With this idea in mind, let us consider the intersection

α([0,1])∩T(M).

It follows from compactness of α([0,1]) that there exists t0,t1 ∈ (0,1), t0 <t1 such thatα(t)∈/T(M)for allt∈[0,1]\[t0,t1]andα(ti)∈T(M)fori=0,1. On the other hand, there exists a path η : [t0,t1] → T(M)such that η(ti) = α(ti), i = 0,1 (since T(M)is arcwise connected). It follows that

β(t) =

η(t) ift∈[t0,t1], α(t) otherwise

is a pathβ :[0,1]→Rⁿwithβ([0,1])∩M=∅and end points{a,b}. 䊐 We believe that there are several advantages of our focus with respect to other proofs of the Fundamental Theorem of Algebra based in algebraic topology: Firstly, the starting point is very clear and no trick is used (you know what you must do from the very beginning) and, as a consequence, the proof is very intuitive. On the other hand, this proof needs the same background than others which are based on the knowledge of the homology of spheres. There are also analytical proofs of the Fundamental Theorem of Algebra (e.g., the one based on Rouche´’s theorem) which only require some topological background.

2 Proof of the main result

In order to prove thatσⁿ is onto, we need firstly to state several technical results:

Lemma 4 Let us assume thatσⁿ(A)is a bounded subset ofCⁿ. ThenA is a bounded subset ofCⁿ.

Proof. Let us assume that sup_α∈Aσⁿ(α)_∞ := sup_α∈Amax_k≤n|a_k(α)| ≤ C, and let α∈Abe arbitrarily chosen. Then

αⁿ_k+

n i=1

(−1)ⁱai(α)αⁿ⁻ⁱ_k

=0 for all k ≤n. Hence |α_k|ⁿ≤ n

i=1

|ai(α)||α_k|ⁿ⁻ⁱ.

If|α_k| ≤1 we do nothing. Otherwise

|α_k| ≤ n

i=1

|a_i(α)||α_k|¹⁻ⁱ≤ n

i=1

|a_i(α)| ≤nC. Hence sup

α∈Amax

k≤n|α_k| ≤max{nC,1}.

This ends the proof. 䊐

Corollary 5 The map σⁿis closed(i.e, it sends closed sets to closed sets).

Proof.Let us assume that ybelongs to the closure ofσⁿ(M)andM is a closed subset of Cⁿ. Then there exists a sequence of points{x_n} ⊂Msuch that σⁿ(x_n)→y. Hence σⁿ({x_n}^∞_n=1)is a bounded subset ofCⁿand it follows from Theorem 2 that{x_n} ⊂Cⁿ is bounded. Hence there exists a convergent subsequence {x_nk} →x∈M (sinceM is closed) and

σⁿ(x) = lim

k→∞σⁿ(x_nk) = lim

n→∞σⁿ(x_n) =y.

This proves thaty∈σⁿ(M). 䊐

Lemma 6 Let us assume that α_i = α_j for all i = j. Then there exists an open set U=U^◦ ⊂Cⁿsuch thatσ_|Uⁿ is one-to-one and(α₁, . . . , α_n)∈U.

Proof. It is easy to prove that under these hypotheses, there exists a neighborhood of α= (α1, . . . , α_n)such that

(β1, . . . , β_n)∈U⇒(β_θ(1), . . . , β_θ(n))∈/U for all θ∈Σ_n\ {id}.

Now, assume that σⁿ(β1, . . . , β_n) = σⁿ(β₁^∗, . . . , β^∗_n) and (β1, . . . , β_n) = (β₁^∗, . . . , β_n^∗) belong both to U. Then p(z) := n

i=1(z−β_i) = n

i=1(z−β_i^∗) is a polynomial of degreenwhich vanishes on the set {β_i}ⁿ_i=1∪ {β_i^∗}ⁿ_i=1. This implies that(β₁^∗, . . . , β^∗_n) = (β_θ(1), . . . , β_θ(n)) for a certain θ ∈ Σ_n\ {id}, because of the divisibility properties of polynomials (which are proved as a consequence of the division algorithm of Euclid), a

contradiction. 䊐

Corollary 7 Let us assume thatα_i =α_j for alli=j andi,j∈ {1, . . . ,n}. Then there exists an open setU=U^◦ ⊂Cⁿ such that(α1, . . . , α_n)∈U, σⁿ(U)is an open subset ofCⁿ, andσⁿ_|U :U→σⁿ(U)is a homeomorphism.

Proof.If we takeUas in the lemma above, thenσ_|Uⁿ :U→σⁿ(U)is continuous, closed and bijective. This obviously implies that it is also open, hence it is a homeomorphism.

Furthermore, the Theorem of Invariance of Domain claims thatσⁿ(U)is an open subset

ofCⁿ. 䊐

Lemma 8 LetH_i,j ={(z1,z2, . . . ,zn)∈Cⁿ:zi−zj =0}. Then Γ:=σⁿ(H_i,j) =σⁿ

t<s

H_t,s

for all1≤i<j≤n.

Proof. Let 1≤t<s≤n andα∈H_t,s be arbitrarily chosen. Let θ∈Σ_n be such that θ(t) =i andθ(s) =j. Then

σⁿ(α) =σⁿ(α1, . . . , α_n) =σⁿ(α_θ(1), . . . , α_θ(n))∈σⁿ(H_i,j).

This means that σⁿ(H_t,s) ⊂ σⁿ(H_i,j) for all t <s. Hence σⁿ

t<sH_t,s

⊂ σⁿ(H_i,j),

and the proof follows. 䊐

Lemma 9 Cⁿ\

t<sH_t,s andCⁿ\Γare both open connected sets. Furthermore, σⁿ

Cⁿ\

t<s

H_t,s ⊆Cⁿ\Γ. (1)

Proof. It is clear that both sets are open, since

t<sH_t,s is closed, Γ=σⁿ

t<sH_t,s

andσⁿis a closed map. If we write the equations which defineH_t,sas a linear subspace of R²ⁿ ≡ Cⁿ (where we identify z_i = x_i+iy_i with the pair (x_i,y_i)), we have that H_t,s=kerL_(t,s), where

L(t,s)(x1,y1,x2,y2, . . . ,xn,yn) = (xt−xs,yt−ys).

Hence dimH_t,s=2n−2 for allt<s, and this implies that

t<sH_t,s does not separate R²ⁿ. HenceCⁿ\

t<sH_t,s is a connected set.

On the other hand, it follows from the identitiesΓ=σⁿ(H1,2)and

(z+α)²(zⁿ⁻²+b1zⁿ⁻³+· · ·+b_n−2)

=zⁿ+ (b1+2α)zⁿ⁻¹+ (b2+2αb1+α²)zⁿ⁻²

+ n−2

k=3

(bk+2αb_k−1+α²b_k−2)z^n−k+2αb_n−2z+α²b_n−2,

thatΓ is a subset ofM={A(α)·b:α∈C,b= (b1, . . . ,b_n−2)∈Cⁿ⁻²}, where

A(α) =















−1 0 0 · · · 0 0

2α 1 0 ... ...

−α² −2α −1 · · · 0 α² 2α · · ·

0 0 −α² · · · (−1)ⁿ⁻³ 0

0 0 0 2(−1)ⁿ⁻²α (−1)ⁿ⁻²

... ... ... (−1)ⁿ⁻¹α² 2(−1)ⁿ⁻¹α 0 0 0 · · · 0 (−1)ⁿα²















∈M_n×(n−2)(C)

for allα∈C. Now, rank(A(α)) =n−2 (as a complex matrix) for allα∈C. HenceM is a ruled submanifold ofCⁿ≡R²ⁿ of complex dimension equal ton−1, so that it has real dimension 2n−2. HenceR²ⁿ\∆is arcwise connected for all∆⊆M. This means thatCⁿ\Γis arcwise connected.

Now, we prove the inclusion formula (1). Let us assume thatσⁿ(α)∈Γ. Then there exists a certainβ∈H_1,2 such thatσⁿ(α) =σⁿ(β). But this implies thatα= (β_θ(1), . . . , β_θ(n)) for a certainθ∈Σ_n(as in the proof of Lemma 4). Henceα∈

t<sH_t,s. This ends the

proof. 䊐

Now we are able to prove the main result of this paper:

Theorem 10 The mapσⁿ:Cⁿ→Cⁿ is onto for alln∈N. Proof.It is clear that

σⁿ(Cⁿ) =σⁿ

Cⁿ\

t<s

H_t,s ∪Γ.

Hence we only need to prove thatσⁿ(Cⁿ\

t<sH_t,s) =Cⁿ\Γ. Now, it follows from Corollary 5 and 7 thatσⁿ(Cⁿ\

t<sH_t,s)is a connected open subset ofCⁿ\Γ. On the other hand, ify∈Cⁿ\Γis a point of the closure ofσⁿ(Cⁿ\

t<sH_t,s)inCⁿ\Γ, then (as it was proved in Corollary 5), there exists a convergent sequence{x_nk} ⊂Cⁿ\

t<sH_t,s, such that{x_nk} →x∈Cⁿ\

t<sH_t,s, σⁿ(x) = lim

k→∞σⁿ(x_nk) =y.

Buty∈/Γimplies thatx∈Cⁿ\

t<sH_t,s. Henceσⁿ(Cⁿ\

t<sH_t,s)is a closed subset of Cⁿ\Γ. Now we use thatCⁿ\Γis connected to obtain thatσⁿ(Cⁿ\

t<sH_t,s) =Cⁿ\Γ,

which is what we wish to prove. 䊐

Remark 11 The proof uses that we are dealing with complex polynomials since oth- erwise the sets H_i,j would be hyperspaces in Rⁿ, so thatRⁿ\

t<sH_t,s could not be a connected set.

Remark 12 With the use of a little of complex analysis there are several more elementary proofs of the Fundamental Theorem of Algebra (see [1], [3]). The standard focus is to use Liouville’s theorem. Another point of view (that usually does not appear in textbooks), very near to our proof, is as follows: first prove the open mapping theorem (i.e., that non-constant holomorphic functions are open maps), then prove by similar arguments to those given in Lemma 4 and Corollary 5, thatp(z) =zⁿ+n

k=1akz^n−kis a closed map for all choices of coefficientsa1, . . . ,an∈C. Finally, you use thatCis connected.

References

[1] Benjamin, F., Rosenberger, G.:The Fundamental Theorem of Algebra, Undergraduate texts in Math.

Springer, 1997.

[2] Munkres, J.R.:Elements of Algebraic Topology, Addison-Wesley, 1984.

[3] Rudin, W.:Real and Complex Analysis, McGraw-Hill, 1974.

J.M. Almira, M. Jime´nez, N. Del Toro Departamento de Matema´ticas

Universidad de Jae´n E.U.P. Linares

23700 Linares (Jae´n), Spain e-mail:jmalmira@ujaen.es