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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 $n \geq 1$ that each monic polynomial of degree $n$ in ${\mathbf R}[X]$ is a product of linear and quadratic polynomials. This is clear for $n = 1$ and 2, so from now on let $n \geq 3$ 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 $\mathbf C \rightarrow \mathbf C$. Of course a nonconstant one-variable real polynomial is a proper mapping $\mathbf R \rightarrow \mathbf R$, 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 $P_d$ be the space of monic polynomials of degree $d$: $$x^d + a_{d-1}x^{d-1} + \cdots + a_1x + a_0.$$ By induction, every polynomial in $P_1, \dots, P_{n-1}$ is a product of linear and quadratic polynomials. We will show every polynomial in $P_n$ is a product of a polynomial in some $P_k$ and $P_{n-k}$ where $1 \leq k \leq n-1$ and therefore is a product of linear and quadratic polynomials.

For $n \geq 3$ and $1 \leq k \leq n-1$, define the multiplication map $$\mu_k \colon P_k \times P_{n-k} \rightarrow P_n \ \ \text{ by } \ \ \mu_k(g,h) = gh.$$ Let $Z_k$ be the image of $\mu_k$ in $P_n$ and $$Z = \bigcup_{k=1}^n bigcup_{k=1}^{n-1} Z_k.$$ These are the monic polynomials of degree $n$ which are composite. We want to show $Z = P_n$. To achieve this we will look at topological properties of $\mu_k$.

We can identify $P_d$ with ${\mathbf R}^d$ by associating to the polynomial displayed way up above the vector $(a_{d-1},\dots,a_1,a_0)$. This makes $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ a continuous mapping. The key point is that $\mu_k$ is a proper mapping: its inverse images of compact sets are compact. To explain why $\mu_k$ is proper, we will use an idea of Pushkar' to "compactify" $\mu_k$ 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 $\mu_k$ is proper is more complicated and I will not be translating it!)

Let $Q_d$ be the nonzero real polynomials of degree $\leq d$ considered up to scaling. There is a bijection
$Q_d \rightarrow {\mathbf P}^d({\mathbf R})$ associating to a class of polynomials $[a_dx^d + \cdots + a_1x + a_0]$ in $Q_d$ the point $[a_d,\dots,a_1,a_0]$. In this way we make $Q_d$ a compact Hausdorff space. The monic polynomials $P_d$, of degree $d$, embed into $Q_d$ in a natural way and are identified in ${\mathbf P}^d({\mathbf R})$ with a standard copy of ${\mathbf R}^d$.

Define $\widehat{\mu}_k \colon Q_k \times Q_{n-k} \rightarrow Q_n$ by $\widehat{\mu}_k([g],[h]) = [gh]$. This is well-defined and restricts on the embedded subsets of monic polynomials to the mapping $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$. In natural homogeneous coordinates, $\widehat{\mu}_k$ is a polynomial mapping so it is continuous. Since projective spaces are compact and Hausdorff, $\widehat{\mu}_k$ is a proper map. Then, since $\widehat{\mu}_k^{-1}(P_n) = P_k \times P_{n-k}$, restricting $\widehat{\mu}_k$ to $P_k \times P_{n-k}$ shows $\mu_k$ is proper.

Since proper mappings are closed mappings, each $Z_k$ is a closed subset of $P_n$, so $Z = Z_1 \cup \cdots \cup Z_{n-1}$ is closed in $P_n$. Topologically, $P_n \cong {\mathbf R}^n$ is connected, so if we could show $Z$ is also open in $P_n$ then we immediately get $Z = P_n$ (since $Z \not= \emptyset$), 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 $P_n$ that are near $f$ are also in $Z$. The inverse function theorem is a natural tool to use in this context: supposing $f = \mu_k(g,h)$, is the Jacobian determinant of $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ nonzero at $(g,h)$? If so, then $\mu_k$ has a continuous local inverse defined in a neighborhood of $f$.

To analyze $\mu_k$ near $(g,h)$, we write all (nearby) points in $P_k \times P_{n-k}$ as $(g+u,h+v)$ where $\deg u \leq k-1$ and $\deg v \leq n-k-1$ (allowing $u = 0$ or $v = 0$ too). Then $$\mu_k(g+u,h+v) = (g+u)(h+v) = gh + gv + hu + uv = f + (gv + hu) + uv.$$ 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 $\deg u < \deg g$ or $u = 0$ and $\deg v < \deg h$ 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 $\deg u < \deg g$ and $\deg v < \deg h$. Therefore the Jacobian of $\mu_k$ at $(g,h)$ is invertible if $g$ and $h$ are relatively prime and not otherwise.

We conclude that if $f \in Z$ can be written somehow as a product of nonconstant relatively prime polynomials then a neighborhood of $f$ in $P_n$ is inside $Z$. Every $f \in Z$ 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 $P_n$: all $(x+a)^n$ for real $a$ if $n$ is odd and all $(x^2+bx+c)^{n/2}$ for real $b$ and $c$ if $n$ is even. (Note when $n$ is even that $(x+a)^n = (x^2 + 2ax + a^2)^{n/2}$.) We have shown $Z - Y$ is open in $P_n$. This is weaker than our hope of showing $Z$ is open in $P_n$. But we're in good shape, as long as we change our focus from $P_n$ to $P_n - Y$. If $n = 2$ then $Y = P_2$ and $P_2 - Y$ is empty. For the first time we will use the fact that $n \geq 3$.

Identifying $P_n$ with ${\mathbf R}^n$ using polynomial coefficients, $Y$ is either an algebraic curve ($n$ odd) or algebraic surface ($n$ even) sitting in ${\mathbf R}^n$. For $n \geq 3$, the complement of an algebraic curve or algebraic surface in ${\mathbf R}^n$ for $n \geq 3$ 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 $P_n$, $Z \cap (P_n - Y) = Z - Y$ is closed in $P_n - Y$. The inverse function theorem tells us that $Z - Y$ is open in $P_n$, so it is open in $P_n - Y$. Therefore $Z - Y$ is a nonempty open and closed subset of $P_n - Y$. Since $P_n - Y$ is connected and $Z - Y$ is not empty, $Z - Y = P_n - Y$. Since $Y \subset Z$, we get $Z = P_n$ and this completes Pukhlikov's "real" proof of the Fundamental Theorem of Algebra.

Mы доказывали, доказывали и наконец доказали. Ура! :)

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 $n \geq 1$ that each monic polynomial of degree $n$ in ${\mathbf R}[X]$ is a product of linear and quadratic polynomials. This is clear for $n = 1$ and 2, so from now on let $n \geq 3$ 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 $\mathbf C \rightarrow \mathbf C$. Of course a nonconstant one-variable real polynomial is a proper mapping $\mathbf R \rightarrow \mathbf R$, 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 $P_d$ be the space of monic polynomials of degree $d$: $$x^d + a_{d-1}x^{d-1} + \cdots + a_1x + a_0.$$ By induction, every polynomial in $P_1, \dots, P_{n-1}$ is a product of linear and quadratic polynomials. We will show every polynomial in $P_n$ is a product of a polynomial in some $P_k$ and $P_{n-k}$ where $1 \leq k \leq n-1$ and therefore is a product of linear and quadratic polynomials.

For $n \geq 3$ and $1 \leq k \leq n-1$, define the multiplication map $$\mu_k \colon P_k \times P_{n-k} \rightarrow P_n \ \ \text{ by } \ \ \mu_k(g,h) = gh.$$ Let $Z_k$ be the image of $\mu_k$ in $P_n$ and $$Z = \bigcup_{k=1}^n Z_k.$$ These are the monic polynomials of degree $n$ which are composite. We want to show $Z = P_n$. To achieve this we will look at topological properties of $\mu_k$.

We can identify $P_d$ with ${\mathbf R}^d$ by associating to the polynomial displayed way up above the vector $(a_{d-1},\dots,a_1,a_0)$. This makes $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ a continuous mapping. The key point is that $\mu_k$ is a proper mapping: its inverse images of compact sets are compact. To explain why $\mu_k$ is proper, we will use an idea of Pushkar' to "compactify" $\mu_k$ 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 $\mu_k$ is proper is more complicated and I will not be translating it!)

Let $Q_d$ be the nonzero real polynomials of degree $\leq d$ considered up to scaling. There is a bijection
$Q_d \rightarrow {\mathbf P}^d({\mathbf R})$ associating to a class of polynomials $[a_dx^d + \cdots + a_1x + a_0]$ in $Q_d$ the point $[a_d,\dots,a_1,a_0]$. In this way we make $Q_d$ a compact Hausdorff space. The monic polynomials $P_d$, of degree $d$, embed into $Q_d$ in a natural way and are identified in ${\mathbf P}^d({\mathbf R})$ with a standard copy of ${\mathbf R}^d$.

Define $\widehat{\mu}_k \colon Q_k \times Q_{n-k} \rightarrow Q_n$ by $\widehat{\mu}_k([g],[h]) = [gh]$. This is well-defined and restricts on the embedded subsets of monic polynomials to the mapping $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$. In natural homogeneous coordinates, $\widehat{\mu}_k$ is a polynomial mapping so it is continuous. Since projective spaces are compact and Hausdorff, $\widehat{\mu}_k$ is a proper map. Then, since $\widehat{\mu}_k^{-1}(P_n) = P_k \times P_{n-k}$, restricting $\widehat{\mu}_k$ to $P_k \times P_{n-k}$ shows $\mu_k$ is proper.

Since proper mappings are closed mappings, each $Z_k$ is a closed subset of $P_n$, so $Z = Z_1 \cup \cdots \cup Z_{n-1}$ is closed in $P_n$. Topologically, $P_n \cong {\mathbf R}^n$ is connected, so if we could show $Z$ is also open in $P_n$ then we immediately get $Z = P_n$ (since $Z \not= \emptyset$), 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 $P_n$ that are near $f$ are also in $Z$. The inverse function theorem is a natural tool to use in this context: supposing $f = \mu_k(g,h)$, is the Jacobian determinant of $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ nonzero at $(g,h)$? If so, then $\mu_k$ has a continuous local inverse defined in a neighborhood of $f$.

To analyze $\mu_k$ near $(g,h)$, we write all (nearby) points in $P_k \times P_{n-k}$ as $(g+u,h+v)$ where $\deg u \leq k-1$ and $\deg v \leq n-k-1$ (allowing $u = 0$ or $v = 0$ too). Then $$\mu_k(g+u,h+v) = (g+u)(h+v) = gh + gv + hu + uv = f + (gv + hu) + uv.$$ 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 $\deg u < \deg g$ or $u = 0$ and $\deg v < \deg h$ 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 $\deg u < \deg g$ and $\deg v < \deg h$. Therefore the Jacobian of $\mu_k$ at $(g,h)$ is invertible if $g$ and $h$ are relatively prime and not otherwise.

We conclude that if $f \in Z$ can be written somehow as a product of nonconstant relatively prime polynomials then a neighborhood of $f$ in $P_n$ is inside $Z$. Every $f \in Z$ 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 $P_n$: all $(x+a)^n$ for real $a$ if $n$ is odd and all $(x^2+bx+c)^{n/2}$ for real $b$ and $c$ if $n$ is even. (Note when $n$ is even that $(x+a)^n = (x^2 + 2ax + a^2)^{n/2}$.) We have shown $Z - Y$ is open in $P_n$. This is weaker than our hope of showing $Z$ is open in $P_n$. But we're in good shape, as long as we change our focus from $P_n$ to $P_n - Y$. If $n = 2$ then $Y = P_2$ and $P_2 - Y$ is empty. For the first time we will use the fact that $n \geq 3$.

Identifying $P_n$ with ${\mathbf R}^n$ using polynomial coefficients, $Y$ is either an algebraic curve ($n$ odd) or algebraic surface ($n$ even) sitting in ${\mathbf R}^n$. For $n \geq 3$, the complement of an algebraic curve or algebraic surface in ${\mathbf R}^n$ for $n \geq 3$ 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 $P_n$, $Z \cap (P_n - Y) = Z - Y$ is closed in $P_n - Y$. The inverse function theorem tells us that $Z - Y$ is open in $P_n$, so it is open in $P_n - Y$. Therefore $Z - Y$ is a nonempty open and closed subset of $P_n - Y$. Since $P_n - Y$ is connected and $Z - Y$ is not empty, $Z - Y = P_n - Y$. Since $Y \subset Z$, we get $Z = P_n$ and this completes Pukhlikov's "real" proof of the Fundamental Theorem of Algebra.

Mы доказывали, доказывали и наконец доказали. Ура! :)