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In mathematics, the fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. Equivalently, the field of complex numbers is algebraically closed.
Sometimes, this theorem stated as: 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. In other words, for every complex polynomial p of degree n > 0 the equation p(z) = 0 has exactly n complex solutions, counting multiplicities. Although this at first appears to be a stronger statement, it is an easy consequence of the other form of the theorem, through the use of successive polynomial division by linear factors.
In spite of its name, there is no known purely algebraic proof of the theorem, and many mathematicians believe that such a proof does not exist.See §1.9 of R. Remmert\'s text The fundamental theorem of Algebra. Besides, it is not fundamental for modern algebra; its name was given at a time in which algebra was basically about solving polynomial equations with real or complex coefficients.
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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 real 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 x4 = 4x − 3, although incomplete, has four solutions (counting multiplicities): 1 (twice), −1 + i√2, and −1 − i√2.
As will be mentioned again below, it follows from the fundamental theorem of algebra that every non-constant polynomial with real coefficients 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 x4 + a4 (with a real and distinct from 0) can be written in such a way. Later, Nikolaus Bernoulli made the same assertion concerning the polynomial x4 − 4x3 + 2x2 + 4x + 4, but he got a letter from Euler in 1742See section Le rôle d\'Euler in C. Gilain\'s article Sur l\'histoire du théorème fondamental de l\'algèbre: théorie des équations et calcul intégral. in which he was told that his polynomial happened to be equal to
where α is the square root of 4 + 2√7, whereas
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 + bi 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 18th 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.Concerning Wood\'s proof, see the article A forgotten paper on the fundamental theorem of algebra, by Frank Smithies. 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 the middle of the 19th century, the problem of finding a constructive proof of the fundamental theorem of algebra. He presented his solution, that amounts in modern terms to a combination of the Durand-Kerner method with the homotopy continuation principle, in 1891. Another proof of this kind was obtained by Hellmuth Kneser in 1940 and simplified by his son Martin Kneser in 1981.
All proofs below involve some analysis, at the very least the concept of continuity of real or complex functions. Some also use differentiable or even analytic functions.
Some proofs of the theorem only prove that any non-constant polynomial with real coefficients has some complex root. This is enough to establish the theorem in the general case because, given a non-constant polynomial with complex coefficients, the polynomial
has only real coefficients and, if is a zero of , then either or its conjugate is a root of .
A large number of non-algebraic proofs of the theorem use the fact (sometimes called “growth lemma”) that an th degree polynomial function whose dominant coefficient is 1 behaves like when is large enough. A more precise statement is: there is some positive real number such that
when .
Find a closed disk of radius centered at the origin such that whenever ≥ . The minimum of on , which must exist since is compact, is therefore achieved at some point in the interior of , but not at any point of its boundary. The minimum modulus principle implies then that . In other words, is a zero of .
Another analytic proof can be obtained along this line of thought observing that, since outside , the minimum of on the whole complex plane is achieved at . If , then is a bounded holomorphic function in the entire complex plane since, for each complex number , ≤ . Applying Liouville\'s theorem, which states that a bounded entire function must be constant, this would imply that is constant and therefore that is constant. This gives a contradiction, and hence
Yet another analytic proof uses argument principle. Let be a positive real number large enough so that every root of has absolute value smaller than ; such a number must exist because every non-constant polynomial function of degree has at most zeros. For each , consider the number
where is the circle centered at with radius oriented counterclockwise; then the argument principle says that this number is the number of zeros of in the open ball centered at with radius , which, since , is the total number of zeros of . On the other hand, the integral of along divided by is equal to . But the difference between the two numbers is
The numerator of the rational expression being integrated has degree at most and the degree of the denominator is . Therefore, the number above tends to as tends to +∞. But the number is also equal to and so .
Let ∈ C be such that the minimum of on the whole complex plane is achieved at ; it was seen at the proof which uses Liouville\'s theorem that such a number must exist. We can write as a polynomial in : there is some natural number and there are some complex numbers , , …, such that ≠ and that
It follows that if is a th root of and if is positive and sufficiently small, then , which is impossible, since is the minimum of on .
For another topological proof by contradiction, suppose that has no zeros. Choose a large positive number such that, for , the leading term of dominates all other terms combined; in other words, such that ··· . As traverses the circle once counter-clockwise, , like , winds times counter-clockwise around . At the other extreme, with , the “curve” is simply the single (nonzero) point , whose winding number is clearly . If the loop followed by is continuously deformed between these extremes, the path of also deforms continuously. Since has no zeros, the path can never cross over as it deforms, and hence its winding number with respect to will never change. However, given that the winding number started as and ended as , this is absurd. Therefore, has at least one zero.
These proofs use two facts about real numbers that require only a small amount of analysis (more precisely, the intermediate value theorem):
The second fact, together with the quadratic formula, implies the theorem for real quadratic polynomials.
As mentioned above, it suffices to check the statement “every non-constant polynomial with real coefficients has a complex root”. This statement can be proved by induction on the greatest non-negative integer such that divides the degree of . Let be a splitting field of (seen as a polynomial with complex coefficients); in other words, the field contains C and there are elements , , …, in such that
If , then is odd, and therefore has a real root. Now, suppose that (with odd and ) and that the theorem is already proved when the degree of the polynomial has the form with odd. For a real number , define:
Then the coefficients of are symmetric polynomials in the \'s with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the elementary symmetric polynomials, that is, in , , …, . So has in fact real coefficients. Furthermore, the degree of is , and is an odd number. So, using the induction hypothesis, has at least one complex root; in other words, is complex for two distinct elements and from {…}. Since there are more real numbers than pairs , one can find distinct real numbers and such that and are complex (for the same and ). So, both and are complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree has a complex root by the quadratic formula. It follows that and are complex numbers, since they are roots of the quadratic polynomial .
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.A proof of the fact that this suffices can be seen here. 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.
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:
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