A simple proof of the Fundamental Theorem of Algebra
aa r X i v : . [ m a t h . C V ] J a n A SIMPLE PROOF OF THE FUNDAMENTAL THEOREM OF ALGEBRA
RICARDO P´EREZ-MARCO
Abstract.
We present a simple short proof of the Fundamental Theorem of Algebra, withoutcomplex analysis and with a minimal use of topology. It can be taught in a first year calculusclass. Statement.
Theorem 1.1.
A non constant polynomial P ( z ) ∈ C [ z ] with complex coefficients has a root. The proof is based only on the following elementary facts: • A polynomial has at most a finite number of roots. • The Implicit Function Theorem. • Removing from C a finite number of points leaves an open connected space.2. The proof.
It is enough to consider a monic polynomial P . We denote by C = ( P ′ ) − (0) the finite set ofcritical points of P , and by D = P ( C ) the finite set of critical values of P . • Let R = { c ∈ C ; the polynomial P ( z ) − c has at least a simple root and no double roots } . • R ⊂ C − D . This is because if c ∈ D , then c = P ( z ) for some critical point z ∈ C , hence P ′ ( z ) = 0 and P ( z ) − c = 0 has a double root at z . Note that C − D is open and connected ( D being finite). • R is open. This is an application of the Implicit Function Theorem. Let c ∈ R ⊂ C − D ,and z ∈ C be a root of P ( z ) − c . We apply the Implicit Function Theorem to the equation F ( z, c ) = P ( z ) − c = 0. Since ∂F∂z ( z , c ) = P ′ ( z ) = 0, there is a neighborhood U of c such that for c ∈ U we have a root z ( c ) of P ( z ) − c . Taking U small enough, by continuity of P ′ and c z ( c ),we have P ′ ( z ( c )) = 0 and the root z ( c ) is simple. Since C − D is open we can take U ⊂ C − D and P ( z ) − c does not have any double root, thus U ⊂ R . • R is closed in C − D . Because P is monic, if c is uniformly bounded then any root of P ( z ) − c isuniformly bounded (since P ( z ) /z n → z → ∞ , if n is the degree). We can take asubsequence of c n → c ∞ ∈ C − D and a converging subsequence of roots of P ( z ) − c n . By continuity,the limit is a root of P ( z ) − c ∞ , so this polynomial has roots. Moreover, all roots of P ( z ) − c ∞ aresimple since c ∞ ∈ C − D . Mathematics Subject Classification.
Key words and phrases.
Roots, complex polynomials, fundamental theorem of algebra. • R is non-empty. For any a ∈ C we have that for c = P ( a ), P ( z ) − c has at least z = a asroot. If we choose a ∈ C − P − ( D ), then for any root z of P ( z ) − c with c = P ( a ), we have P ( z ) = P ( a ) / ∈ D , so z / ∈ P − ( D ), but C ⊂ P − ( D ), and z / ∈ C , and the root z is simple.The above proves that R = C − D . Now, if 0 ∈ D , then 0 = P ( z ) for a critical point z of P that is also a root of P . If 0 / ∈ D , then 0 ∈ R = C − D and the equation P ( z ) − P has a root. ⋄ Comment.
The above proof is inspired from a beautiful proof by Daniel Litt [1]. He works in the globalspace of monic polynomials of degree n ≥ C n ), and removes the algebraic locus D n , defined by the discriminant, of polynomials with a double root. He uses that the complementof an algebraic variety in C n is connected. Essentially the proof above achieves the same goal ina more elementary way working with n = 1. In particular, we only need the simpler fact that thecomplement of a finite set in the plane is connected (which for n = 1 is the same as the connectednessof the complement of an algebraic variety in C n ). We also avoid the use of discriminants. Acknowledgment.
I am grateful to my friends Marie-Claude Arnaud, Kingshook Biswas, AlainChenciner and Yann Levagnini for their comments and suggestions to improve the presentation. Inparticular, to Kingshook that proposed a simplification of a first draft.
References [1] LITT, D.;
Yet another proof of the Fundamental Theorem of Algebra
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