Here is the proof by Pukhlikov (1997) at
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mp&paperid=6&option_lang=eng
which Ilya mentioned as being only in Russian so far. What I present below is not a literal translation (as if anyone on this site cares...).
The argument will use only real variables: there is no use of complex numbers anywhere.
The goal is to show for every ngeq1 that each monic polynomial of degree n in mathbfR[X] is a product of linear and quadratic polynomials.
This is clear for n=1 and 2, so from now on let ngeq3 and assume by induction
that nonconstant polynomials of degree less than n admit
factorizations into a product of linear and quadratic polynomials.
First, some context: we're going to make use of proper mappings. A complex-variable proof on this page listed by Gian depends on the fact that a nonconstant one-variable complex polynomial is a proper mapping mathbfCrightarrowmathbfC. Of course a nonconstant one-variable real polynomial is a proper mapping mathbfRrightarrowmathbfR, but that is not the kind of proper mapping we will use. Instead, we will use the fact (to be explained below) that multiplication of real one-variable polynomials of a fixed degree is a proper mapping on spaces of polynomials. I suppose if you find yourself teaching a course where you want to give the students an interesting but not well-known application of the concept of a proper mapping, you could direct them to this argument.
Now let's get into the proof. It suffices to focus on monic polynomials and their monic factorizations.
For any positive integer d, let Pd be the space of monic polynomials
of degree d:
xd+ad−1xd−1+cdots+a1x+a0.
By induction, every polynomial in P1,dots,Pn−1 is a product of linear and quadratic polynomials. We will show every polynomial in Pn is a product of
a polynomial in some Pk and Pn−k where 1leqkleqn−1 and therefore
is a product of linear and quadratic polynomials.
For ngeq3 and 1leqkleqn−1, define the multiplication map
mukcolonPktimesPn−krightarrowPntextbymuk(g,h)=gh. Let Zk be the image of muk in Pn and
Z=bigcupn−1k=1Zk.
These are the monic polynomials of degree n which are composite. We want
to show Z=Pn. To achieve this we will look at topological properties of muk.
We can identify Pd with mathbfRd by associating to the polynomial displayed way up above the vector (ad−1,dots,a1,a0). This makes mukcolonPktimesPn−krightarrowPn a continuous mapping. The key point
is that muk is a proper mapping: its inverse images of compact sets are
compact. To explain why muk is proper, we will use an idea of Pushkar' to "compactify" muk
to a mapping on projective spaces. (In the journal where Pukhlikov's paper appeared, the paper by Pushkar' with his nice idea comes immediately afterwards. Puklikov's own approach to proving muk is proper is more complicated and I will not be translating it!)
Let Qd be the nonzero real polynomials of degree leqd considered
up to scaling. There is a bijection
QdrightarrowmathbfPd(mathbfR) associating to a class of polynomials
[adxd+cdots+a1x+a0] in Qd the point [ad,dots,a1,a0].
In this way we make Qd a compact Hausdorff space.
The monic polynomials Pd, of degree d, embed into
Qd in a natural way and are identified in mathbfPd(mathbfR)
with a standard copy of mathbfRd.
Define widehatmukcolonQktimesQn−krightarrowQn
by widehatmuk([g],[h])=[gh].
This is well-defined and restricts on the embedded subsets of monic polynomials to the
mapping mukcolonPktimesPn−krightarrowPn. In natural homogeneous coordinates, widehatmuk is a polynomial mapping so
it is continuous. Since projective spaces are compact and Hausdorff,
widehatmuk is a proper map. Then, since
widehatmu−1k(Pn)=PktimesPn−k,
restricting widehatmuk to PktimesPn−k shows muk is proper.
Since proper mappings are closed mappings,
each Zk is a closed subset of Pn, so Z=Z1cupcdotscupZn−1
is closed in Pn. Topologically, PncongmathbfRn is connected,
so if we could show Z is also open in Pn then we immediately
get Z=Pn (since Znot=emptyset), which was our goal. Alas, it will not be easy to show Z is open directly, but a modification of this
idea will work.
We want to show that if a polynomial f is in Z then all polynomials in Pn that are near
f are also in Z. The inverse function theorem is a natural tool to use in
this context: supposing f=muk(g,h), is the Jacobian determinant of
mukcolonPktimesPn−krightarrowPn nonzero at (g,h)?
If so, then muk has a continuous local inverse defined in a neighborhood of f.
To analyze muk near (g,h), we
write all (nearby) points in PktimesPn−k as
(g+u,h+v) where deguleqk−1 and degvleqn−k−1 (allowing u=0 or v=0 too). Then
muk(g+u,h+v)=(g+u)(h+v)=gh+gv+hu+uv=f+(gv+hu)+uv.
As functions of the coefficients of u and v, the coefficients of gv+hu are all linear
and the coefficients of uv are all higher degree polynomials.
If g and h are relatively prime then
every polynomial of degree less than n is uniquely of the form
gv+hu where degu<degg or u=0 and degv<degh or v=0,
while if g and h are not relatively prime then
we can write gv+hu=0 for some nonzero polynomials
u and v where degu<degg and degv<degh.
Therefore the Jacobian of muk at (g,h) is invertible if g and h are relatively prime
and not otherwise.
We conclude that if finZ can be written somehow as a product
of nonconstant relatively prime polynomials then a neighborhood of f in Pn is inside Z.
Every finZ is a product of linear and quadratic polynomials, so
f can't be written as a product of nonconstant relatively prime
polynomials precisely when it is a power of a linear or quadratic polynomial. Let Y be all these "degenerate" polynomials in Pn:
all (x+a)n for real a if n is odd and all (x2+bx+c)n/2 for real b and c if n is even. (Note when n is even that (x+a)n=(x2+2ax+a2)n/2.) We have shown Z−Y is open in Pn. This is weaker than our hope of
showing Z is open in Pn. But we're in good shape, as long as
we change our focus from Pn to Pn−Y. If n=2 then Y=P2 and P2−Y is empty.
For the first time we will use the fact that ngeq3.
Identifying Pn with mathbfRn using polynomial coefficients,
Y is either an algebraic curve (n odd) or algebraic surface (n even) sitting in
mathbfRn. For ngeq3, the complement of an algebraic curve
or algebraic surface in mathbfRn for ngeq3 is path connected, and thus
connected.
The set Z−Y is nonempty since (x−1)(x−2)cdots(x−n) is in it. Since Z is closed in Pn, Zcap(Pn−Y)=Z−Y is closed in Pn−Y.
The inverse function theorem tells us that
Z−Y is open in Pn, so it is open in Pn−Y.
Therefore Z−Y is a nonempty open and closed subset of Pn−Y.
Since Pn−Y is connected and Z−Y is not empty, Z−Y=Pn−Y.
Since YsubsetZ, we get Z=Pn and this completes Pukhlikov's "real" proof
of the Fundamental Theorem of Algebra.
Mы доказывали, доказывали и наконец доказали. Ура! :)