Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

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++at[t] then there exists β1,β2,,β such that a0+a1t++at =(t-β1) (t-β). Proof by induction on :

If f=a0+a1t++at[t] then there exist β and f1=b0+b1t++b-1t-1 such that f=(t-β)f1.

Another version of the Lemma is

If f=a0+a1t++at[t] then there exist β such that f(β)=0.

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

If f=a0+a1t++at[t] then there exist p1,,pr with deg(pi) equal to 1 or 2 such that f=p1·p2pr.

This follows from the fundamental theorem of algebra and the fact that If f[t] and β 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 and β1,,βs with β1,,βs. Note that βj+βj and βjβj.

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.

is algebraically closed.

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

Bourbaki defines as [X](X2+1).

To show:
(a) If a0 then there exists a.
(b) If p(t)[t] and degp is odd then there exists α such that p(α)=0.
(b) Assume p(t)=anXn+an-1Xn-1++a0 with n odd and an0.
If x and x0 then p(x)=anxng(x), where g(x)=1+an-1anx++a0anxn.
limxg(x)=1 andlimx-g(x)=1. So there exists a0 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] such that f(α)=0.

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

Let f(t)[t] such that f(t)0.
To show: There exists r0 such that if z and |z|r then |f(z)|>|f(0)|. Use [Exercise 1] and Weierstrass' theorem [Bou, Top. ChIV §6 no. 1, Theorem 1] to show is algebraically closed.

[Bou Top. Ch. VIII §2, Exercise 1]
Let a, a0 and n>0.

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