Bourbaki

Theorem 1

Tag 0BZW, permanent. Cite it as [Bourbaki, Tag 0BZW].

General Topology, chapter VIII, § 1. Complex numbers, quaternions

(d'Alembert-Gauss). The field C\mathbf{C} of complex numbers is algebraically closed.

For the proof it is enough to establish that (i) every element ≥0\geq 0 in R\mathbf{R} has a square root, and (ii) every polynomial of odd degree with coefficients in R\mathbf{R} has at least one root in R\mathbf{R}. The first of these assertions has already been proved in Chapter IV, § 3, no. 3. As to the second, if f(X)=a0Xn+a1Xn−1+⋯+anf(X) = a_0 X^n + a_1 X^{n-1} + \cdots + a_n is a polynomial of odd degree nn (a0≠0a_0 \neq 0) with real coefficients, we may write f(x)=a0xng(x)f(x) = a_0 x^n g(x) for x≠0x \neq 0, where

g(x)=1+a1a0x+⋯+ana0xn g(x) = 1 + \frac{a_1}{a_0 x} + \cdots + \frac{a_n}{a_0 x^n}

tends to +1+1 as xx tends to +∞+\infty or −∞-\infty. Hence there is a number a>0a > 0 such that f(a)f(a) has the sign of a0a_0 and f(−a)f(-a) the sign of −a0-a_0;

and so by Bolzano’s theorem (Chapter IV, § 6, no. 1, Theorem 2), ff has at least one root in [—a,a][—a, a].

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.