The Fundamental Theorem of Algebra

Một phần của tài liệu Francis borceux (auth ) an algebraic approach to geometry geometric trilogy ii springer international publishing (2014) (Trang 412 - 415)

Proposition F.2.12 Given two complex numbersx, y∈C,ex+y=exey.

Proof The modulus ofea+bi iseaand its argument isb. The result follows imme-

diately by PropositionF.2.6.

F.3 The Fundamental Theorem of Algebra

We now want to prove that every polynomial over the complex numbers admits a root. Of course, everybody expects at once that:

Lemma F.3.1 The polynomial

aX2+bX+c, a, b, c∈C, a=0 admits the two roots

r1, r2=−b±√

b2−4ac

2a ∈C.

Proof The formula makes sense by PropositionF.2.8and it is straightforward to check that it exhibit roots of the polynomial, whatever the choice of specific square root.

Whenb2−4ac=0, the two roots obtained in this way are distinct, thus they are all the roots of the polynomial (see CorollaryA.6.7).

Whenb2−4ac=0, observe that aX2+bX+c=a

X−−b

2a 2

and the formula in the statement indeed provides the double root of the polyno-

mial.

The crucial step (in our approach) is the following theorem:

Theorem F.3.2 Every non-constant polynomial p(X)∈R[X] admits a complex root.

Proof Letp(X) be a non-constant polynomial over the reals. Since the point is to find a root, we can freely multiply the polynomial by a non-zero real number, namely, the inverse of its leading coefficient. There is thus no loss of generality in assuming thatp(X)has the form

p(X)=Xn+αn−1Xn−1+ ã ã ã +α1X+α0, n >0.

Let us write further the degreenin the form n=2ms

wheresis an odd number. We shall argue by induction on the numbermof factors 2 in the degreen.

Ifm=0, the polynomialp(X)has odd degrees. Therefore

X→−∞lim p(X)= lim

X→−∞Xn= −∞, lim

X→+∞p(X)= lim

X→+∞Xn= +∞. So there exist a real numberr <0 such thatp(r) <0 and a real numbert >0 such thatp(t ) >0. By continuity ofp(X)and the Intermediate Value Theorem, there is a realr < a < t such thatp(a)=0, that is, a roota∈R⊂Cofp(X).

Of course this first argument is not algebraic at all. But it is a matter of fact: the present theorem as well as the fact that the field of complex numbers is algebraically closed (see TheoremF.3.3) depend heavily on the topological completeness of the fieldsRandC, that is, every Cauchy sequence admits a limit. No proof can escape a reference to the topological properties of the real or complex fields.

Suppose now the result has been proved for those polynomial of degree 2m−1s, withs odd, and letp(X)∈R[X] be of degree n=2ms. Of course we can also viewp(X)as a polynomial overCso that by CorollaryA.7.2, there exists a field extension

R⊂C⊆L wherep(X)takes the form

p(X)=(Xa1)ã ã ã(Xan), aiL.

Going back to ExampleE.1.2, we can write the coefficient ofp(X)in terms of the elementary symmetric polynomials

p(X)= n i=0

(−1)iσi(a1, . . . , an)Xni. Fix now an arbitrary real numberrand consider

bij =ai+aj+raiajL, 1≤ijn.

Lettingi take the successive values from 1 ton, we observe that the number of possible values forj—that is, of possible pairs(i, j )—is

n+(n−1)+ ã ã ã +1+0=n(n+1)

2 =2ms(2ms+1) 2

=2m−1s

2ms+1

=2m−1s

F.3 The Fundamental Theorem of Algebra 403 where of course

s=s

2ms+1 remains odd as the product of two odd numbers.

We next consider the following polynomial overL q(X)= $

1≤ijn

(Xbij).

By definition,bijL; but nevertheless, let us prove that the coefficients ofq(X) are real numbers.

• The coefficients ofq(X)are the values of the elementary symmetric polynomials in 2m−1svariables, evaluated at the elementsbijL. They are thus symmetric expressions in thebij’s. Let us recall that all the coefficients of an elementary symmetric polynomial overLare equal to 1 (see ExampleE.1.2) thus, certainly, these coefficients are real numbers.

• Eachbij is itself a symmetric expression, with real coefficients, in terms of ai andaj. Replacing eachbij by this expression, the coefficients ofq(X)now be- come the values of some polynomials innvariables, evaluated at theaiL; as just observed, by construction, these polynomials remain symmetric and with real coefficients.

• By TheoremE.2.1, each coefficient ofq(X)can further be written as r

σ1(a1, . . . , an), . . . , σn(a1, . . . , an)

for some polynomialr(Y1, . . . , Yn)overR, withσi the symmetric polynomials of degreeiinnvariables.

But the elementary symmetric polynomialsσi innvariables, evaluated at the ele- mentsai, are precisely the coefficients ofp(X), which are real numbers by assump- tion. This proves that the coefficients ofq(X)are real numbers.

Sinceq(X)has degree 2m−1s, we can apply the inductive assumption and infer the existence of a complex rootcr ofq(X). But this rootcr ∈C⊆Lis necessarily one of the elementsbij. So there exists a pair of indices, which of course depends on the choice of the real numberr, and that we shall therefore write as(ir, jr), such that

birjr =air+ajr +rairajr =cr ∈C, 1≤irjr < n.

This holds for every real numberr. But since there are infinitely many real numbers rand only finitely many pairs(i, j ), at least one of the pairs(i, j )is obtained as the pair(ir, jr)for infinitely many real numbersr. Consider two such real numbersr andr:

r=r, (ir, jr)=(i, j )=(ir, jr).

We have now found two real numbersr=r and two complex numberscr,cr

such that

ai+aj+raiaj=cr

ai+aj+raiaj=cr. Subtracting these equalities, we find

rr

aiaj=crcr that isaiaj=crcr

rr ∈C.

Next multiply the first equality byr, the second one byr, and then subtract; we get rr

(ai+aj)=rcrrcr that isai+aj=rcrrcr

rr ∈C. This proves that the polynomial

(Xai)(Xaj)=X2−(ai+aj)X+(aiaj)

has complex coefficients. The two rootsai,aj of this polynomial are complex num- bers by LemmaF.3.1. In other words, the two rootsai, ajLofp(X)are in fact

complex numbers.

Now we present the expected fundamental result:

Theorem F.3.3 (Fundamental Theorem of Algebra) Every polynomial over the field of complex numbers has a root.

Proof Letp(X)be a polynomial overC. Writep(X)for the polynomial obtained when replacing the coefficients ofp(X)by their conjugates (see DefinitionF.1.3).

Anticipating PropositionF.4.1, we observe that the polynomialp(X)p(X)has real coefficients. By TheoremF.3.2, this polynomial admits a complex rootc. But

p(c)p(c)=0 =⇒ p(c)=0 orp(c)=0.

Ifp(c)=0, the proof is complete. Ifp(c)=0, anticipating again PropositionF.4.1,

we conclude thatcis a root ofp(X)=p(X).

TheoremF.3.3can thus be rephrased as (see DefinitionA.6.2):

The fieldCof complex numbers is algebraically closed.

Một phần của tài liệu Francis borceux (auth ) an algebraic approach to geometry geometric trilogy ii springer international publishing (2014) (Trang 412 - 415)

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