This proof, due to D'Alembert, relies on the following three facts:

• Every polynomial with real coefficients which is of odd order has a real root. (This is a corollary of the intermediate value theorem.
• Every second order polynomial with complex coefficients has two complex roots.
• For every polynomial $p$ with real coefficients, there exists a field $E$ in which the polynomial may be factored into linear terms. (For more information, see the entry ``splitting field''.)

Note that it suffices to prove that every polynomial with real coefficients has a complex root. Given a polynomial with complex coefficients, one can construct a polynomial with real coefficients by multiplying the polynomial by its complex conjugate. Any root of the resulting polynomial will either be a root of the original polynomial or the complex conjugate of a root.

The proof proceeds by induction. Write the degree of the polynomial as $2^n (2m+1)$ . If $n = 0$ , then we know that it must have a real root. Next, assume that we already have shown that the fundamental theorem of algebra holds whenver $n < N$ . We shall show that any polynomial of degree $2^N (2m+1)$ has a complex root if a certain other polynomial of order $2^{N-1} (2m' + 1)$ has a root. By our hypothesis, the other polynomial does have a root, hence so does the original polynomial. Hence, by induction on $n$ , every polynomial with real coefficients has a complex root.

Let $p$ be a polynomial of order $d = 2^N (2m+1)$ with real coefficients. Let its factorization over the extension field $E$ be $$p(x) = (x - r_1) (x - r_2) \cdots (x - r_d)$$ Next construct the $d(d-1)/2 = 1$ polynomials $$q_k (x) = \prod_{i < j} (x - r_i - r_j - k r_i r_j)$$ where $k$ is an integer between $1$ and $d(d-1)/2 = 1$ . Upon expanding the product and collecting terms, the coefficient of each power of $x$ is a symmetric function of the roots $r_i$ . Hence it can be expressed in terms of the coefficients of $p$ , so the coefficients of $q_k$ will all be real.

Note that the order of each $q_k$ is $d(d-1)/2 = 2^{N-1} (2m+1) (2^N (2m+1) - 1)$ . Hence, by the induction hypothesis, each $q_k$ must have a complex root. By construction, each root of $q_k$ can be expressed as $r_i + r_j + k r_i r_j$ for some choice of integers $i$ and $j$ . By the pigeonhole principle, there must exist integers $i, j, k_1, k_2$ such that both $$u = r_i + r_j + k_1 r_i r_j$$ and $$v = r_i + r_j + k_2 r_i r_j$$ are complex. But then $r_i$ and $r_j$ must be complex as well. because they are roots of the polynomial $$x^2 + bx + c$$ where $$b = -{k_2 u + k_1 v \over (k_1 + k_2)}$$ and $$c = {u - v \over k_1 - k_2}$$

Note: D'Alembert was an avid supporter of the famous French philosophical encyclopaedia. Therefore it is a fitting tribute to have his proof appear in the web pages of this encyclopaedia.

Reference: Jean le Rond D'Alembert, "Recherches sue le calcul integral" Histoire de l'Academie Royale des Sciences et Belles Lettres), annee MDCCXLVI, Berlin 1748, 182-224

R. Argand, "Reflesions sur la nouvelle theorie d'analyse." Annales de mathematiques 5, 197-209, 1814