A rootless polynomial

The fundamental theorem of algebra proved by Carl Friedrich Gauss in the last year of the 18th century states the non-existence of some objects, as many other important theorems of mathematics do. Namely, it states that there is no non-constant polynomial of the complex variable which vanishes at no point of the complex plane C. (Those who have never heard of complex numbers may wish to read Z. Marciniak's article about thee-dimensional multiplication first.)

Up to this day many different proofs of Gauss's fundamental theorem have been conceived, each of them of great beauty. Unfortunately, none of them can be presented here with all the details. Nevertheless, to satisfy the curious, here are the main ideas of two different proofs which use no algebraic methods at all.

The first proof uses two lemmas which immediately imply the fundamental theorem of algebra.

Lemma 1. If P:C -> C is a polynomial, then there is a point z0 in C at which the function |P| attains its infimum.

The proof stems from the observation that for points z lying beyond the closed circle K(0,R) for sufficiently large R the modulus of the polynomial is quite big, so the infima of the two sets

      { |P(z)|: z arbitrary in C} and { |P(z)|: z in K(0,R)}

are equal. The function |P| being continuous, it attains its infimum on the compact set K(0,R).

Lemma 2. Let P be a polynomial of positive degree. If w is such a complex number that |P(w)| is smaller than or equal to |P(z)| for all complex z, then P(w)=0.

The proof of this lemma is more difficult and technical, but only elementary knowledge about complex numbers is required.

The easiest way to proceed is to apply reductio ad absurdum. Multipliying the polynomial by a constant and moving its graph on the plane we can assume that w=0 and P(w)=a0>0. Assume moreover that the polynomial P contains no positive powers of z with exponent less than k. Hence

Writing

for

(with |z| sufficiently small) we get

Now it is not hard to prove that by including the remaining parts of the polynomial in the absolute value nothing is spoiled: there are finitely many of them and in the neighborhood of zero their significance is infinitely less than that of ak zk. We leave to the persistent reader the determination of the exact value of "delta".

The second proof uses the so-called

Liouville's theorem. Let f:C -> C be a bounded function. If f is differentiable at every point, i.e. for each z0 in C there is a finite limit

then f is identically equal to a constant.

The proof of this fact can be found in any book on complex function theory. It may be useful to observe, though, that Liouville's theorem becomes false in the world of real numbers (consider e.g. f(x)=1/(x2+1).

Let's assume now that P is a complex polynomial such that P(z) is nonzero for all z. Then it can be observed that the function given by f(z)=1/P(z) satisfies all the assumptions of Liouville.s theorem (to see that it is bounded note, as in the proof of Lemma 1, that the modulus of the polynomial cannot be small beyond some large circle). Hence 1/P is constant and consequently also P is constant, which completes the proof of the fundamental theorem of algebra.

P.S.