proof of fundamental theorem of algebra (due to d’Alembert)

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 . We shall show that any polynomial of degree $2^N(2m+1)$ has a complex root if a certain other polynomial of order $2^{N-1}(2m^{\prime }+1)$ 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

$$p(x)=(x-r_1)(x-r_2)\mathrm{\cdots }(x-r_d)$$

Next construct the $d(d-1)/2=1$ polynomials

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

$$b=-\frac{k_{2}u+k_{1}v}{(k_1+k_2)}$$

and

$$c=\frac{u-v}{k_1-k_2}$$

Note.  D’Alembert was an avid supporter (in fact, the co-editor) of the famous French philosophical encyclopaedia. Therefore it is a fitting tribute to have his proof appear in the web pages of this encyclopaedia.