Fundamental Theorem of Algebra/Proof 4

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Theorem

Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.


Proof

Let $f$ be a polynomial with coefficients in $\R$, and let $f$ have degree $n$.

By Fundamental Theorem of Arithmetic, there exists $k \in \N$ such that $n = 2^k \cdot m$ with $m$ an odd number.

We proceed by induction on $k$.

Note that $f$ has roots $\alpha_1, \ldots, \alpha_n$ in some splitting field.

It suffices to show that one of these roots is a complex number.


Base case

$k = 0$

$f$ is a polynomial with real coefficients and odd degree. Hence, $f$ has a root in $\R \subset \C$ by Rolle's Theorem.


Induction hypothesis

For some integer $k > 0$, every polynomial with real coefficients of degree $2^{k - 1} \cdot m$, with $m$ an odd integer, has a root in $\C$.

For each nonzero $\lambda \in \Z$, construct the polynomial

$\ds \map {g_\lambda} x = \prod_{i \mathop < j} \paren {x - \paren {\alpha_i + \alpha_j} - \lambda \alpha_i \alpha_j} $

By Viète's Formulas, each coefficient of $g_\lambda$ is a symmetric polynomial in $\alpha_i + \alpha_j + \lambda \alpha_i \alpha_j$.

Thus, the coefficients of $g_\lambda$ are invariant under permutations of the roots of $g_\lambda$.

But a permutation of the roots of $g_\lambda$ is equivalent to permuting the indices $\closedint 1 n$ of the roots of $f$.

Thus, the coefficients of $g_\lambda$ are invariant under permutations of the roots of $f$.

By definition, each coefficient of $g_\lambda$ is a symmetric polynomial in $\set {\alpha_1, \ldots, \alpha_n}$.

By Fundamental Theorem of Symmetric Polynomials, each coefficient of $g_\lambda$ is a polynomial in the elementary symmetric polynomials in $\set {\alpha_1, \ldots, \alpha_n}$.

But by Viète's Formulas, these are exactly the coefficients of $f$.

Thus, each coefficient of $g_\lambda$ is a polynomial in the coefficients of $f$, which are real numbers.

Since Real Numbers form Field, the coefficients of $g_\lambda$ are real numbers.


Further, $g_\lambda$ has one linear factor for each unordered pair of roots of $f$.

Thus, the degree of $g_\lambda$ is

\(\ds \binom n 2\) \(=\) \(\ds \frac {n \cdot \paren {n - 1} } 2\) Binomial Coefficient with Two
\(\ds \) \(=\) \(\ds \frac {2^k \cdot m \cdot \paren {2^k \cdot m - 1} } 2\) expression of $n$
\(\ds \) \(=\) \(\ds 2^{k - 1} \cdot m \cdot \paren {2^k \cdot m - 1}\)

Since $m$ and $2^k \cdot m - 1$ are odd, the degree of $g_\lambda$ is of the form $2^{k-1} \cdot m'$, with $m' = m \cdot \paren {2^k \cdot m - 1}$ an odd integer.

Combined with the fact that $g_\lambda$ has real coefficients, the induction hypothesis implies that $g_\lambda$ has a root in $\C$.

Thus, we have infinitely many polynomials $g_\lambda$ with a complex root of the form $\alpha_i + \alpha_j + \lambda \alpha_i \alpha_j$ for some $i, j\in \closedint 1 n$.

Since there are infinitely many polynomials and only finitely many pairs $\tuple {i, j}$, it follows from the Infinite Pigeonhole Principle that there exist distinct $\lambda, \mu \in \Z$ and a particular pair of roots $\alpha_x, \alpha_y$ such that:

$\alpha_x + \alpha_y + \lambda \alpha_x \alpha_y \in \C$

and

$\alpha_x + \alpha_y + \mu \alpha_x \alpha_y \in \C$

Since Complex Numbers form Field, the difference of these numbers $\paren {\lambda - \mu} \alpha_x \alpha_y \in \C$.

Since $\lambda$ and $\mu$ are distinct, $\lambda - \mu \ne 0$, and so $\frac {\paren {\lambda - \mu} \alpha_x \alpha_y}{\lambda - \mu} = \alpha_x \alpha_y \in \C$.

Since $\alpha_x \alpha_y \in \C$, it follows that $\lambda \alpha_x \alpha_y \in \C$, and so $\alpha_x + \alpha_y + \lambda \alpha_x \alpha_y - \lambda \alpha_x \alpha_y = \alpha_x + \alpha_y \in \C$.

Thus, $\map h x = x^2 - \paren {\alpha_x + \alpha_y} x + \alpha_x \alpha_y$ is a quadratic equation with complex coefficients.

By Viète's Formulas, the roots of $\map h x$ are $\alpha_x$ and $\alpha_y$.

By Solution to Quadratic Equation, the roots of $\map h x$ are:

$\dfrac {- \alpha_x - \alpha_y \pm \sqrt{\paren {\alpha_x + \alpha_y}^2 - 4 \alpha_x \alpha_y} } 2$

which are complex numbers by Square Root of Complex Number in Cartesian Form and Complex Numbers form Field.

Thus, $\alpha_x$ and $\alpha_y$ are complex numbers.

Since $\alpha_x$ is a root of $f$, this means $f$ has a complex root. Thus, any polynomial with real coefficients has a root in $\C$.

Now let $\ds \map f x = \sum_{k \mathop = 0}^n c_k x^k$ be a polynomial with complex coefficients, and let $\map {\overline{f} } x$ denote $\ds \sum_{k \mathop = 0}^n \overline{c_k} x^k$, where an overline denotes a complex conjugate.

Observe:

\(\ds \overline {\map f x \map {\overline f} x}\) \(=\) \(\ds \map {\overline f} x \overline {\map {\overline f} x}\) Conjugation of Polynomials Preserves Multiplication
\(\ds \) \(=\) \(\ds \overline {\map f x} \map f x\)

Thus, $\map h x = \map f x \map {\overline f} x$ equals its conjugate, and so has real coefficients by Complex Number equals Conjugate iff Wholly Real.

Since $\map h x$ has real coefficients, it has a complex root, $\alpha$.

By Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs, $\overline{\alpha}$ is also a root of $h$.

Since $\map h \alpha = \map f \alpha \map {\overline f} \alpha = 0$, $\alpha$ is a root of either $f$ or $\overline f$.

If $\alpha$ is a root of $f$, then we are done.

If $\alpha$ is a root of $\overline f$, then $\map {\overline f} \alpha = 0$, and so

\(\ds \sum_{k \mathop = 0}^n \overline {c_k} \alpha^k\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^n c_k \overline \alpha^k\) \(=\) \(\ds 0\) taking conjugates, applying Product of Complex Conjugates
\(\ds \leadsto \ \ \) \(\ds \map f {\overline \alpha}\) \(=\) \(\ds 0\)

as desired.

$\blacksquare$