Fundamental theorem of algebra

From Wikipedia, the free encyclopedia

In mathematics, the fundamental theorem of algebra states that every non-zero single-variable polynomial, with complex coefficients, has exactly as many complex roots as its degree, if repeated roots are counted up to their multiplicity. Equivalently, the mathematical field of complex numbers is closed under algebraic operations.

In other words, for every complex polynomial p of degree n > 0 the equation p(z) = 0 has exactly n complex solutions, counting multiplicities.

The name of the theorem is now considered a misnomer by many mathematicians, since it is more a theorem in analysis than algebra.

1 History
2 Proofs
3 Corollaries
4 References
5 External links

[edit] History

Peter Rothe (Petrus Roth), in his book Arithmetica Philosophica (published in 1608), wrote that a polynomial equation of degree $n$ (with real coefficients) may have $n$ solutions. Albert Girard, in his book L'invention nouvelle en l'Algèbre (published in 1629), asserted that a polynomial equation of degree $n$ has $n$ solutions, but he did not state that they had to be complex numbers. Furthermore, he added that his assertion holds “unless the equation is incomplete”, by which he meant that no coefficient is equal to $0$ . However, when he explains in detail what he means, it is clear that he actually believes that his assertion is always true; for instance, he shows that the equation $x 4 = 4 x - 3$ , although incomplete, has four solutions (counting multiplicities): $1$ , $1$ , $-1+i\sqrt2$ , and $-1-i\sqrt2$ .

As will be mentioned again below, it follows from the fundamental theorem of algebra that every polynomial with real coefficients and degree greater than $0$ can be written as a product of polynomials with real coefficients whose degree is either $1$ or $2$ . However, in 1702 Leibniz said that no polynomial of the type $x 4 + a 4$ (with $a$ real and distinct from $0$ ) can be written in such a way. Later, Nikolaus Bernoulli made the same assertion concerning the polynomial $x 4 - 4 x 3 + 2 x 2 + 4 x + 4$ , but he got a letter from Euler in 1742 in which he was told that his polynomial happened to be equal to

$(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha),\,$

where $α$ is the square root of $4+2\sqrt7$ , whereas

$x^4+a^4=(x^2+a\sqrt{2}x+a^2)(x^2-a\sqrt{2}x+a^2).\,$

A first attempt at proving the theorem was made by d'Alembert in 1746, but his proof was incomplete. Among other problems, it assumed implicitly a theorem (now known as Puiseux's theorem) which would be proved only more than a century later (and furthermore the proof assumed the fundamental theorem of algebra). Other attempts were made by Euler (1749), de Foncenex (1759), Lagrange (1772), and Laplace (1795). These last four attempts assumed implicitly Girard's assertion; to be more precise, the existence of solutions was assumed and all that remained to be proved was that their form was $a + b i$ for some real numbers $a$ and $b$ . In modern terms, Euler, de Foncenex, Lagrange, and Laplace were assuming the existence of a splitting field of the polynomial $p (z)$ .

At the end of the 18^th century two new proofs were published which did not assume the existence of roots. One of them, due to James Wood and mainly algebraic, was published in 1798 and it was totally ignored. Wood's proof had an algebraic gap. The other one was published by Gauss in 1799 and it was mainly geometric, but it had a topological gap. A rigorous proof was published by Argand in 1806; it was here that, for the first time, the fundamental theorem of algebra was stated for polynomials with complex coefficients, rather than just real coefficients. Gauss produced two other proofs in 1816 and another version of his original proof in 1849.

The first textbook containing a proof of the theorem was Cauchy's Cours d'analyse de l'École Royale Polytechnique (1821). It contained Argand's proof, although Argand is not credited for it.

None of the proofs mentioned so far is constructive. It was Weierstrass who raised for the first time, in 1891, the problem of finding a constructive proof of the fundamental theorem of algebra. Such a proof was obtained by Hellmuth Kneser in 1940 and simplified by his son Martin Kneser in 1981.

[edit] Proofs

All proofs of the fundamental theorem of algebra involve some analysis, at the very least the concept of continuity of real or complex functions. This is unavoidable because the statement of the theorem depends on analysis; the sets of real and complex numbers are analytic objects. Some proofs also use differentiable or even analytic functions.

Some proofs of the theorem only prove that any polynomial with real coefficients has some complex root. This is enough to establish the theorem in the general case because, given a polynomial $p (z)$ with complex coefficients, the polynomial $q(z)=p(z)\overline{p(\overline{z})}$ has only real coefficients and, if $z$ is a zero of $q (z)$ , then either $z$ or its conjugate is a root of $p (z)$ .

A large number of non-algebraic proofs of the theorem use the fact (sometimes called “growth lemma”) that an $n$ ^th degree polynomial function $p (z)$ behaves like $z n$ when $| z |$ is large enough. A more precise statement is: there is some positive real number $R$ such that

$\frac{1}{2}|z^n|<|p(z)|<\frac{3}{2}|z^n|$

when $| z | > R$ .

[edit] Complex-analytic proofs

Find a closed disk $D$ of radius $r$ centered at the origin such that $| p (z) | > | p (0) |$ whenever $| z |$ ≥ $r$ . The minimum of $| p (z) |$ on $D$ , which must exist since $D$ is compact, is therefore achieved at some point $z 0$ in the interior of $D$ , but not at any point of its boundary. The minimum modulus principle implies then that $p (z 0) = 0$ . In other words, $z 0$ is a zero of $p (z)$ .

Another analytic proof can be obtained along this line of thought observing that, since $| p (z) | > | p (0) |$ outside $D$ , the minimum of $| p (z) |$ on the whole complex plane is achieved at $z 0$ . If $| p (z 0) | > 0$ , then $1 / p$ is a bounded holomorphic function in the entire complex plane since, for each complex number $z$ , $| 1 / p (z) |$ ≤ $| 1 / p (z 0) |$ . Applying Liouville's theorem, which states that a bounded entire function must be constant, this would imply that $1 / p$ is constant and therefore that $p$ is constant. This gives a contradiction, and hence $p (z 0) = 0$

Yet another analytic proof uses argument principle. Let $R$ be a positive real number large enough so that every root of $p (z)$ has absolute value smaller than $R$ ; such a number must exist because every non-constant polynomial function of degree $n$ has at most $n$ zeros. For each $r > R$ , consider the number

$\frac{1}{2\pi i}\int_{c(r)}\frac{p'(z)}{p(z)}\,dz,$

where $c (r)$ is the circle centered at $0$ with radius $r$ oriented counterclockwise; then the argument principle says that this number is the number $N$ of zeros of $p (z)$ in the open ball centered at $0$ with radius $r$ , which, since $r > R$ is the total number of zeros of $p (z)$ . On the other hand, the integral of $n / z$ along $c (r)$ divided by $2π i$ is equal to $n$ . But the difference between the two numbers is

$\frac{1}{2\pi i}\int_{c(r)}\frac{p'(z)}{p(z)}-\frac{n}{z}\,dz=\frac{1}{2\pi i}\int_{c(r)}\frac{zp'(z)-np(z)}{zp(z)}\,dz.$

The numerator of the rational expression being integrated has degree at most $n - 1$ and the degree of the denominator is $n + 1$ . Therefore, the number above tends to $0$ as $r$ tends to +∞. But the number is also equal to $N - n$ and so $N = n$ .

[edit] Topological proofs

As an alternative to the use of Liouville's theorem in the previous proof, we can write $p (z)$ as a polynomial in $z - z 0$ : there is some natural number $k$ and there are some complex numbers $c k$ , $c k + 1$ ,…, $c n$ such that $c k$ ≠ $0$ and that

$p(z)=p(z_0)+c_k(z-z_0)^k+c_{k+1}(z-z_0)^{k+1}+ \cdots +c_n(z-z_0)^n$ .

It follows that if $a$ is a $k$ ^th root of $- p (z 0) / c k$ and if $t$ is positive and sufficiently small, then $| p (z 0 + t a) | < | p (z 0) |$ , which is impossible, since $| p (z 0) |$ is the minimum of $| p |$ on $D$ .

For another topological proof by contradiction, suppose that $p (z)$ has no zeros. Choose a large positive number $R$ such that, for $| z | = R$ , the leading term $z n$ of $p (z)$ dominates all other terms combined; in other words, such that $| z | n > | a n - 1 z n - 1 +$ ··· $+ a 0 |$ . As $z$ traverses the circle $| z | = R$ once counter-clockwise, $p (z)$ , like $z n$ , winds $n$ times counter-clockwise around $0$ . At the other extreme, with $| z | = 0$ , the “curve” $p (z)$ is simply the single (nonzero) point $p (0)$ , whose winding number is clearly $0$ . If the loop followed by $z$ is continuously deformed between these extremes, the path of $p (z)$ also deforms continuously. Since $p (z)$ has no zeros, the path can never cross over $0$ as it deforms, and hence its winding number with respect to $0$ will never change. However, given that the winding number started as $n$ and ended as $0$ , this is absurd. Therefore, $p (z)$ has at least one zero.

[edit] Algebraic proofs

These proofs use two facts about real numbers that require only a small amount of analysis (more precisely, the intermediate value theorem):

every polynomial with odd degree and real coefficients has some real root;
every non-negative real number has a square root.

The second fact, together with the quadratic formula, implies the theorem for real quadratic polynomials.

As mentioned above, it suffices to check that every polynomial $p (z)$ with real coefficients has a complex root. The theorem can be proved by induction on the greatest non-negative integer k such that $2 k$ divides the degree n of $p (z)$ . Let $F$ be a splitting field of $p (z)$ (seen as a polynomial with complex coefficients); in other words, the field $F$ contains C and there are elements $z 1$ , $z 2$ , …, $z n$ in $F$ such that

$p(z)=(z-z_1)(z-z_2) \cdots (z-z_n).$

If $k = 0$ , then $n$ is odd, and therefore $p (z)$ has a real root. Now, suppose that $n = 2 k m$ (with $m$ odd and $k > 0$ ) and that the theorem is already proved when the degree of the polynomial has the form $2 k - 1 m'$ with $m'$ odd. For a real number $t$ , define:

$q_t(z)=\prod_{1\le i<j\le n}\left(z-z_i-z_j-tz_iz_j\right).\,$

Then the coefficients of $q t (z)$ are symmetric polynomials in the $z i$ 's with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the elementary symmetric polynomials, that is, in $- a 1$ , $a 2$ , …, $( - 1) n a n$ . So $q t$ has in fact real coefficients. Furthermore, the degree of $q t$ is $n (n - 1) / 2 = 2 k - 1 m (n - 1)$ , and $m (n - 1)$ is an odd number. So, using the induction hypothesis, $q t$ has at least one complex root; in other words, $z i + z j + t z i z j$ is complex for two distinct elements $i$ and $j$ from { $1$ ,…, $n$ }. Since there are more real numbers than pairs $(i, j)$ , one can find distinct real numbers $t$ and $s$ such that $z i + z j + t z i z j$ and $z i + z j + s z i z j$ are complex (for the same $i$ and $j$ ). So, both $z i + z j$ and $z i z j$ are complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree 2 has a complex root by the quadratic formula. It follows that $z i$ and $z j$ are complex numbers, since they are roots of the quadratic polynomial $z 2 - (z i + z j) z + z i z j$ .

Another algebraic proof of the fundamental theorem can be given using Galois theory. It suffices to show that C has no proper finite field extension. Let K/C be a finite extension, and without loss of generality assume that K is a normal extension of R. Let G be the Galois group of this extension, and let H be a Sylow 2-group of G, so that the order of H is a power of 2, and the index of H in G is odd. By the fundamental theorem of Galois theory, there exists a subextension L of K/R such that Gal(K/L)=H. As [L:R]=[G:H] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have L = R, thus [K:R] and [K:C] are powers of 2. Assuming for contradiction [K:C] > 1, the 2-group Gal(K/C) contains a subgroup of index 2, thus there exists a subextension M of K/C of degree 2. However, C has no extension of degree 2, because every quadratic complex polynomial has a complex root, as mentioned above.

[edit] Corollaries

Since the fundamental theorem of algebra can be seen as the statement that the field of complex numbers is algebraically closed, it follows that any theorem concerning algebraically closed fields applies to the field of complex numbers. Here are a few more consequences of the theorem, which are either about the field of real numbers or about the relationship between the field of real numbers and the field of complex numbers:

The field of complex numbers is the algebraic closure of the field of real numbers.

Every polynomial in one variable $x$ with real coefficients is the product of a constant, polynomials of the form $x + a$ with $a$ real, and polynomials of the form $x 2 + a x + b$ with $a$ and $b$ real and $a 2 - 4 b < 0$ (which is the same thing as saying that the polynomial $x 2 + a x + b$ has no real roots).

Every rational function in one variable $x$ , with real coefficients, can be written as the sum of a polynomial function with rational functions of the form $a / (x - b) n$ (where $n$ is a natural number, and $a$ and $b$ are real numbers), and rational functions of the form $(a x + b) / (x 2 + c x + d) n$ (where $n$ is a natural number, and $a$ , $b$ , $c$ , and $d$ are real numbers such that $c 2 - 4 d < 0$ ). A corollary of this is that every rational function in one variable and real coefficients has an elementary primitive.

Every algebraic extension of the real field is isomorphic either to the real field or to the complex field.

[edit] References

A.-L. Cauchy, Cours d'Analyse de l'École Royale Polytechnique, 1^ère partie: Analyse Algébrique, 1992, Éditions Jacques Gabay, ISBN 2-87647-053-5

B. Fine and G. Rosenberger, The Fundamental Theorem of Algebra, 1997, Springer-Verlag, ISBN 0-387-94657-8

C. F. Gauss, “New Proof of the Theorem That Every Algebraic Rational Integral Function In One Variable can be Resolved into Real Factors of the First or the Second Degree”, 1799

C. Gilain, “Sur l'histoire du théorème fondamental de l'algèbre: théorie des équations et calcul intégral”, Archive for History of Exact Sciences, 42 (1991), 91–13

H. Kneser, “Der Fundamentalsatz der Algebra und der Intuitionismus”, Mathematische Zeitschrift, 46 (1940), 287–302, see also: M. Kneser: “Ergänzung zu einer Arbeit von Hellmuth Kneser über den Fundamentalsatz der Algebra”, Mathematische Zeitschrift, 177 (1981) 285–287

E. Netto and R. Le Vavasseur, “Les fonctions rationnelles §80–88: Le théorème fondamental”, in Encyclopédie des Sciences Mathématiques Pures et Appliquées, tome I, vol. 2, 1992, Éditions Jacques Gabay, ISBN 2-87647-101-9

R. Remmert, “The Fundamental Theorem of Algebra”, in Numbers, 1991, Springer-Verlag, ISBN 0-387-97497-0

D. E. Smith, “A Source Book in Mathematics”, 1959, Dover Publications, ISBN 0-486-64690-4

F. Smithies, “A forgotten paper on the fundamental theorem of algebra”, Notes & Records of the Royal Society, 54 (2000), 333–341

M. Spivak, Calculus, 1994, Publish or Perish, ISBN 0-914098-89-6

B. L. van der Waerden, Algebra I, 1991, Springer-Verlag, ISBN 0-387-97424-5

[edit] External links

Fundamental Theorem of Algebra - a collection of proofs
Eric W. Weisstein, Fundamental Theorem of Algebra at MathWorld.
Fundamental Theorem of Algebra Module by John H. Mathews

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Categories: Abstract algebra | Complex analysis | Field theory | Mathematical theorems

Fundamental theorem of algebra

From Wikipedia, the free encyclopedia

Contents

[edit] History

[edit] Proofs

[edit] Complex-analytic proofs

[edit] Topological proofs

[edit] Algebraic proofs

[edit] Corollaries

[edit] References

[edit] External links

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