Here is a translation into English of a second "real" proof from the journal Ilya mentioned in his answer. This proof is due to Petya Pushkar', it is found at
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mp&paperid=7&option_lang=eng
and it is based on the notion of the degree of a map. Recall that for a smooth proper mapping of oriented manifolds, its degree is defined by picking a regular value and adding up the signs of the determinants of the differential of the mapping at the points in the inverse image. That the degree is well-defined is rather complicated to prove, but it explains the following topological fact.
Fact: Let $M^n$ and $N^n$ be smooth connected oriented manifolds and
$f colon M^n rightarrow N^n$ be a smooth proper mapping of degree
not equal to zero. Then $f$ is surjective.
To prove the Fundamental Theorem of Algebra in a "real" version, we will focus on polynomials of even degree (any of odd degree have a real root). We will show that
any real polynomial of degree $2n$ can be factored into a product of $n$ polynomials
of the second degree.
We identify each monic polynomial $x^d + a_{d-1}x^{d-1} + cdots + a_1x + a_0$ with the point $(a_{d-1},dots,a_0)$ in $mathbf R^d$. We will be particularly interested in the space of monic quadratic polynomials $x^2 + ax + b$, which are identified with the plane $mathbf R^2$. Consider the multiplication mapping
$$
u colon (mathbf R^2)^n rightarrow mathbf R^{2n} text{ where} (f_1,f_2,dots,f_n) mapsto f_1f_2cdots f_n.
$$
Proving the Fundamental Theorem of Algebra amounts to showing that $u$ is surjective.
First we show $u$ is proper.
For any $d geq 1$, identify the nonzero polynomials of degree at most $d$, considered up to scaling by nonzero real numbers, with $mathbf P^d(mathbf R)$ by $[a_dx^d + a_{d-1}x^{d-1} +cdots + a_0] mapsto
[a_d,a_{d-1},dots,a_0]$. (The polynomials of exact degree $d$, after being scaled to be monic, are a copy of $mathbf R^d$ in $mathbf P^d(mathbf R)$.) Consider the multiplication mapping
$$
widehat{u} colon (mathbf P^2(mathbf R))^n rightarrow mathbf P^{2n}(mathbf R) text{ where} ([f_1],[f_2],dots,[f_n]) mapsto [f_1f_2cdots f_n].
$$
The mapping $widehat{u}$ is proper since it is defined on a compact manifold and is continuous.
The mapping $widehat{u}$ is a natural "compactification" of the mapping $u$. The space
$(mathbf P^2(mathbf R))^n$ can be written as the union of $(mathbf R^{2})^n$ and an "infinitely distant part" $B_1$ ($n$-tuples of polynomials of degree at most 2 where at least one polynomial has degree less than 2), while $mathbf P^{2n}(mathbf R)$ can be written as the union of $mathbf R^{2n}$ and an "infinitely distant part" $B_2$ (polynomials of degree less than $2n$). From this point of view, $widehat{u}$ on $(mathbf R^2)^n$ agrees with $u$ and, clearly, $widehat{u}^{-1}(B_2) = B_1$. Therefore the map $u$ is proper.
Next we show the degree of $u$ is equal to $n!$.
Orient the space of monic polynomials of degree 2 (we denote this space as $mathbf R^2$) arbitrarily and give $(mathbf R^2)^n$ the product orientation (as a product of oriented manifolds). As an exercise, show
the polynomial $p(x) = prod_{i=1}^n (x^2+i)$ is a regular value of the mapping $u$. (Hint:
This polynomial is a product of distinct monic irreducibles. Now use the description of the regular values of the multiplication mappings $mu_k$ in Pukhlikov's proof of the Fundamental Theorem of Algebra, which is written in a separate answer on this page.)
The polynomial $p(x)$ has $n!$ inverse images under $u$: all ordered $n$-tuples
with coordinates $x^2+i$ for $i = 1,dots,n$. Let's prove that these points
all contribute the same sign to the degree.
The mapping $u$ is invariant under permutations of its arguments, and
any such permutation preserves orientation (exercise). Therefore the
sign of the determinant of the differential at all the inverse images is the same, which shows $u$ has degree $n!$. By the topological fact at the start, $u$ is surjective, so all monic real polynomials of degree $2n$ are a product of monic quadratic real polynomials.
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