Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 27 September 2014

Lecture 32: Revision: The Fundamental Theorem of Algebra

(II) Week 1 §2: Show that the field of complex numbers is algebraically closed.

To show: If f=a0+a1t+⋯+aℓtℓ∈ℂ[t] then there exists β1,β2,…,βℓ∈ℂ such that a0+a1t+⋯+aℓtℓ =(t-β1)⋯ (t-βℓ). Proof by induction on ℓ:

If f=a0+a1t+⋯+aℓtℓ∈ℂ[t] then there exist β∈ℂ and f1=b0+b1t+⋯+bℓ-1tℓ-1 such that f=(t-β)f1.

Another version of the Lemma is

If f=a0+a1t+⋯+aℓtℓ∈ℂ[t] then there exist β∈ℂ such that f(β)=0.

ℝ is not algebraically closed: t2+1 does not factor in ℝ[t], even though t2+1=(t+i)(t-i) factors in ℂ[t].

If f=a0+a1t+⋯+aℓtℓ∈ℝ[t] then there exist p1,…,pr with deg(pi) equal to 1 or 2 such that f=p1·p2⋯pr.

This follows from the fundamental theorem of algebra and the fact that If f∈ℝ[t] and β∈ℂ such that f(β)=0 then f(β‾)=0. So f = (t-γ1)⋯(t-γr) (t-β1)(t-β1‾)⋯ (t-βs)(t-βs‾) = (t-γ1)⋯(t-γr) (t2-(β1+β1‾+β1β1‾))⋯ (t2-(βs+βs‾)+βsβs‾) with γ1,…,γr∈ℝ and β1,…,βs∈ℂ with β1,…,βs∉ℝ. Note that βj+βj‾∈ℝ and βjβj‾∈ℝ.

This theorem is called the fundamental theorem of algebra. It was first proved by d'Alembert, after which Gauss studied the theorem intensively providing 14 proofs.

Further references: Wikipedia - Fundamental Theorem of Algebra, Math Overflow - Fundamental Theorem of Algebra, Article of Harm Derksen.

ℂ is algebraically closed.

Proof (d'Alembert-Gauss [Bou, Top. Ch VIII §1, no. 1, Theorem 1]).

Bourbaki defines ℂ as ℝ[X](X2+1).

To show:
(a) If a∈ℝ≥0 then there exists a∈ℝ.
(b) If p(t)∈ℝ[t] and deg p is odd then there exists α∈ℝ such that p(α)=0.
(b) Assume p(t)=anXn+an-1Xn-1+⋯+a0 with n odd and an≠0.
If x∈ℝ and x≠0 then p(x)=anxng(x), where g(x)=1+an-1anx+⋯+a0anxn.
limx→∞g(x)=1 andlimx→-∞g(x)=1. So there exists a∈ℝ≥0 such that sign(an)= sign(f(a))and sign(-an)= sign(f(-a)). Thus, by Bolzano's theorem, [Bou, Top IV §6, no. 1, Theorem 2], there exists α∈[-a,a]ℝ such that f(α)=0.

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Proof 2 [Bou, Top. Ch VIII §2. Exercise 2].

Let f(t)∈ℂ[t] such that f(t)≠0.
To show: There exists r∈ℝ≥0 such that if z∈ℂ and |z|≥r then |f(z)|>|f(0)|. Use [Exercise 1] and Weierstrass' theorem [Bou, Top. ChIV §6 no. 1, Theorem 1] to show ℂ is algebraically closed.

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[Bou Top. Ch. VIII §2, Exercise 1]
Let a∈ℂ, a≠0 and n∈ℤ>0.

To show:
(a) If r∈ℝ>0 such that rn≤|a| then there exists z∈ℂ such that |z|=r and |a+zn|= |a|-rn.
(b) If f(z)∈ℂ[t] and deg(f)>0 and z0∈ℂ, with f(z0)≠0 then there exists z∈Bε(z0) such that |f(z0)|> |f(z)|.

Notes and References

These are a typed copy of Lecture 32 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 19, 2011.

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