Fundamental theorem of algebra

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with its imaginary part equal to zero.
Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed.
The theorem is also stated as follows: every non-zero, single-variable, degree "n" polynomial with complex coefficients has, counted with multiplicity, exactly "n" complex roots. The equivalence of the two statements can be proven through the use of successive polynomial division.
Despite its name, there is no purely algebraic proof of the theorem, since any proof must use some form of the analytic completeness of the real numbers, which is not an algebraic concept. Additionally, it is not fundamental for modern algebra; its name was given at a time when algebra was synonymous with theory of equations.
History.
Peter Roth, in his book "Arithmetica Philosophica" (published in 1608, at Nürnberg, by Johann Lantzenberger), 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 formula_1 although incomplete, has four solutions (counting multiplicities): 1 (twice), formula_2 and formula_3
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 degrees are either 1 or 2. However, in 1702 Leibniz erroneously said that no polynomial of the type (with real and distinct from 0) can be written in such a way. Later, Nikolaus Bernoulli made the same assertion concerning the polynomial , but he got a letter from Euler in 1742 in which it was shown that this polynomial is equal to
with formula_5
Also, Euler pointed out that
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 not be proved until more than a century later and using 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, but neither of which was complete. 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, only filled by Alexander Ostrowski in 1920, as discussed in Smale (1981). The first rigorous proof was published by Argand in 1806 (and revisited in 1813); it was also 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 incomplete 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, which 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.
Without using countable choice, it is not possible to constructively prove the fundamental theorem of algebra for complex numbers based on the Dedekind real numbers (which are not constructively equivalent to the Cauchy real numbers without countable choice). However, Fred Richman proved a reformulated version of the theorem that does work.
Proofs.
All proofs below involve some mathematical analysis, or at least the topological concept of continuity of real or complex functions. Some also use differentiable or even analytic functions. This fact has led to the remark that the Fundamental Theorem of Algebra is neither fundamental, nor a theorem of algebra.
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 "p"("z") with complex coefficients, the polynomial
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") whose dominant coefficient is 1 behaves like "zn" when |"z"| is large enough. A more precise statement is: there is some positive real number "R" such that:
when |"z"| > "R".
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 Maximum modulus principle (applied to 1/"p"("z")) implies then that "p"("z"0) = 0. In other words, "z"0 is a zero of "p"("z").
A variation of this proof does not require the use of the maximum modulus principle (in fact, the same argument with minor changes also gives a proof of the maximum modulus principle for holomorphic functions). If we assume by contradiction that "a" := "p"("z"0) ≠ 0, then, expanding "p"("z") in powers of "z" − "z"0 we can write
Here, the "cj" are simply the coefficients of the polynomial "z" → "p"("z" + "z"0), and we let "k" be the index of the first coefficient following the constant term that is non-zero. But now we see that for "z" sufficiently close to "z"0 this has behavior asymptotically similar to the simpler polynomial formula_10,
in the sense that (as is easy to check) the function
is bounded by some positive constant "M" in some neighborhood of "z"0. Therefore, if we define formula_12 and let formula_13, then for any sufficiently small positive number "r" (so that the bound "M" mentioned above holds), using the triangle inequality we see that
When "r" is sufficiently close to 0 this upper bound for |"p"("z")| is strictly smaller than |"a"|, in contradiction to the definition of "z"0. (Geometrically, we have found an explicit direction θ0 such that if one approaches "z"0 from that direction one can obtain values "p"("z") smaller in absolute value than |"p"("z"0)|.)
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 the 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
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
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" → +∞. But the number is also equal to "N" − "n" and so "N" = "n".
Still another complex-analytic proof can be given by combining linear algebra with the Cauchy theorem. To establish that every complex polynomial of degree "n" > 0 has a zero, it suffices to show that every complex square matrix of size "n" > 0 has a (complex) eigenvalue. The proof of the latter statement is by contradiction.
Let "A" be a complex square matrix of size "n" > 0 and let "In" be the unit matrix of the same size. Assume "A" has no eigenvalues. Consider the resolvent function
which is a meromorphic function on the complex plane with values in the vector space of matrices. The eigenvalues of "A" are precisely the poles of "R"("z"). Since, by assumption, "A" has no eigenvalues, the function "R"("z") is an entire function and Cauchy theorem implies that
On the other hand, "R"("z") expanded as a geometric series gives:
This formula is valid outside the closed disc of radius formula_20 (the operator norm of "A"). Let formula_21 Then
(in which only the summand "k" = 0 has a nonzero integral). This is a contradiction, and so "A" has an eigenvalue.
Finally, Rouché's theorem gives perhaps the shortest proof of the theorem.
Topological proofs.
Suppose the minimum of |"p"("z")| on the whole complex plane is achieved at "z"0; it was seen at the proof which uses Liouville's theorem that such a number must exist. We can write "p"("z") as a polynomial in "z" − "z"0: there is some natural number "k" and there are some complex numbers "ck", "c""k" + 1, ..., "cn" such that "ck" ≠ 0 and:
If "p"("z"0) is nonzero, it follows that if "a" is a "k"th root of −"p"("z"0)/"ck" and if "t" is positive and sufficiently small, then |"p"("z"0 + "ta")| < |"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 the polynomial "p"("z") has no roots, and consequently is never equal to 0. Think of the polynomial as a map from the complex plane into the complex plane. It maps any circle |"z"| = "R" into a closed loop, a curve "P"("R"). We will consider what happens to the winding number of "P"("R") at the extremes when "R" is very large and when "R" = 0. When "R" is a sufficiently large number, then the leading term "zn" of "p"("z") dominates all other terms combined; in other words,
When "z" traverses the circle formula_25 once counter-clockwise formula_26 then formula_27 winds "n" times counter-clockwise formula_28 around the origin (0,0), and "P"("R") likewise. At the other extreme, with |"z"| = 0, the curve "P"(0) is merely the single point "p"(0), which must be nonzero because "p"("z") is never zero. Thus "p"(0) must be distinct from the origin (0,0), which denotes 0 in the complex plane. The winding number of "P"(0) around the origin (0,0) is thus 0. Now changing "R" continuously will deform the loop continuously. At some "R" the winding number must change. But that can only happen if the curve "P"("R") includes the origin (0,0) for some "R". But then for some "z" on that circle |"z"| = "R" we have "p"("z") = 0, contradicting our original assumption. Therefore, "p"("z") has at least one zero.
Algebraic proofs.
These proofs of the Fundamental Theorem of Algebra must make use of the following two facts about real numbers that are not algebraic but require only a small amount of analysis (more precisely, the intermediate value theorem in both cases):
The second fact, together with the quadratic formula, implies the theorem for real quadratic polynomials. In other words, algebraic proofs of the fundamental theorem actually show that if "R" is any real-closed field, then its extension "C" = "R"() is algebraically closed.
By Induction.
As mentioned above, it suffices to check the statement "every non-constant polynomial "p"("z") with real coefficients has a complex root". This statement can be proved by induction on the greatest non-negative integer "k" such that 2"k" divides the degree "n" of "p"("z"). Let "a" be the coefficient of "zn" in "p"("z") and let "F" be a splitting field of "p"("z") over "C"; in other words, the field "F" contains "C" and there are elements "z"1, "z"2, ..., "zn" in "F" such that
If "k" = 0, then "n" is odd, and therefore "p"("z") has a real root. Now, suppose that "n" = 2"km" (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:
Then the coefficients of "qt"("z") are symmetric polynomials in the "zi" 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)"nan". So "qt"("z") has in fact "real" coefficients. Furthermore, the degree of "qt"("z") is "n"("n" − 1)/2 = 2"k"−1"m"("n" − 1), and "m"("n" − 1) is an odd number. So, using the induction hypothesis, "qt" has at least one complex root; in other words, "zi" + "zj" + "tzizj" 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 "zi" + "zj" + "tzizj" and "zi" + "zj" + "szizj" are complex (for the same "i" and "j"). So, both "zi" + "zj" and "zizj" 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 "zi" and "zj" are complex numbers, since they are roots of the quadratic polynomial "z"2 −  ("zi" + "zj")"z" + "zizj".
Joseph Shipman showed in 2007 that the assumption that odd degree polynomials have roots is stronger than necessary; any field in which polynomials of prime degree have roots is algebraically closed (so "odd" can be replaced by "odd prime" and this holds for fields of all characteristics). For axiomatization of algebraically closed fields, this is the best possible, as there are counterexamples if a single prime is excluded. However, these counterexamples rely on −1 having a square root. If we take a field where −1 has no square root, and every polynomial of degree "n" ∈ "I" has a root, where "I" is any fixed infinite set of odd numbers, then every polynomial "f"("x") of odd degree has a root (since has a root, where "k" is chosen so that ). Mohsen Aliabadi generalized Shipman's result in 2013, providing an independent proof that a sufficient condition for an arbitrary field (of any characteristic) to be algebraically closed is that it has a root for every polynomial of prime degree.
From Galois Theory.
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. Since the normal closure of "K" over R still has a finite degree over C (or R), we may assume without loss of generality that "K" is a normal extension of R (hence it is a Galois extension, as every algebraic extension of a field of characteristic 0 is separable). Let "G" be the Galois group of this extension, and let "H" be a Sylow 2-subgroup 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 by way of contradiction that ["K":C] > 1, we conclude that the 2-group Gal("K"/C) contains a subgroup of index 2, so there exists a subextension "M" of C of degree 2. However, C has no extension of degree 2, because every quadratic complex polynomial has a complex root, as mentioned above. This shows that ["K":C] = 1, and therefore "K" = C, which completes the proof.
Geometric proofs.
There exists still another way to approach the fundamental theorem of algebra, due to J. M. Almira and A. Romero: by Riemannian geometric arguments. The main idea here is to prove that the existence of a non-constant polynomial "p"("z") without zeros implies the existence of a flat Riemannian metric over the sphere S2. This leads to a contradiction since the sphere is not flat.
A Riemannian surface ("M", "g") is said to be flat if its Gaussian curvature, which we denote by "Kg", is identically null. Now, Gauss–Bonnet theorem, when applied to the sphere S2, claims that
which proves that the sphere is not flat.
Let us now assume that "n" > 0 and
for each complex number "z". Let us define
Obviously, "p*"("z") ≠ 0 for all "z" in C. Consider the polynomial "f"("z") = "p"("z")"p*"("z"). Then "f"("z") ≠ 0 for each "z" in C. Furthermore,
We can use this functional equation to prove that "g", given by
for "w" in C, and
for "w" ∈ S2\{0}, is a well defined Riemannian metric over the sphere S2 (which we identify with the extended complex plane C ∪ {∞}).
Now, a simple computation shows that
since the real part of an analytic function is harmonic. This proves that "Kg" = 0.
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 the relationship between the field of real numbers and the field of complex numbers:
Bounds on the zeros of a polynomial.
While the fundamental theorem of algebra states a general existence result, it is of some interest, both from the theoretical and from the practical point of view, to have information on the location of the zeros of a given polynomial. The simpler result in this direction is a bound on the modulus: all zeros ζ of a monic polynomial formula_38 satisfy an inequality |ζ| ≤ "R"∞, where
Notice that, as stated, this is not yet an existence result but rather an example of what is called an a priori bound: it says that "if there are solutions" then they lie inside the closed disk of center the origin and radius "R"∞. However, once coupled with the fundamental theorem of algebra it says that the disk contains in fact at least one solution. More generally, a bound can be given directly in terms of any p-norm of the "n"-vector of coefficients formula_40 that is |ζ| ≤ "Rp", where "Rp" is precisely the "q"-norm of the 2-vector formula_41 "q" being the conjugate exponent of "p", formula_42 for any 1 ≤ "p" ≤ ∞. Thus, the modulus of any solution is also bounded by
for 1 < "p" < ∞, and in particular
(where we define "an" to mean 1, which is reasonable since 1 is indeed the "n"-th coefficient of our polynomial). The case of a generic polynomial of degree "n",
is of course reduced to the case of a monic, dividing all coefficients by "an" ≠ 0. Also, in case that 0 is not a root, i.e. "a"0 ≠ 0, bounds from below on the roots ζ follow immediately as bounds from above on formula_47, that is, the roots of
Finally, the distance formula_49 from the roots ζ to any point formula_50 can be estimated from below and above, seeing formula_51 as zeros of the polynomial formula_52, whose coefficients are the Taylor expansion of "P"("z") at formula_53
Let ζ be a root of the polynomial
in order to prove the inequality |ζ| ≤ "Rp" we can assume, of course, |ζ| > 1. Writing the equation as
and using the Hölder's inequality we find
Now, if "p" = 1, this is
thus
In the case 1 < "p" ≤ ∞, taking into account the summation formula for a geometric progression, we have
thus
and simplifying,
Therefore
holds, for all 1 ≤ "p" ≤ ∞.