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

This proof, due to d’Alembert, relies on the following three facts:

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. 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

 $p(x)=(x-r_{1})(x-r_{2})\cdots(x-r_{d})$

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

 $q_{k}(x)=\prod_{i

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 $r_{i}+r_{j}+kr_{i}r_{j}$ 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 (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.

## References

• 1 Jean le Rond D’Alembert: “Recherches sur le calcul intégral”.   Histoire de l’Acadḿie Royale des Sciences et Belles Lettres, année MDCCXLVI, 182–224. Berlin (1746).
• 2 R. Argand: “Réflexions sur la nouvelle théorie d’analyse”.  Annales de mathématiques 5, 197–209 (1814).
Title proof of fundamental theorem of algebra (due to d’Alembert) ProofOfFundamentalTheoremOfAlgebradueToDAlembert 2013-03-22 14:36:06 2013-03-22 14:36:06 rspuzio (6075) rspuzio (6075) 10 rspuzio (6075) Proof msc 30A99 msc 12D99