I was trying to find a suitable proof of the Fundamental Theorem of Algebra at an undergraduate level which avoided abstract linear algebra, as I have not yet begun it. However, I came across this
http://www.math.washington.edu/~aloveles/ArchivedMaterials/Math300/FundamentalTheoremOfAlgebra.pdf
which is seemingly comprehensible and straightforward, and logical but I was wondering if it is rigourous or even valid?
Best Answer
I think the one of the more accessible proof of Fundamental Theorem of Algebra is the following argument (due to Lagrange). I learned it from Pierre Samuel's Algebraic Theory of Numbers. It uses induction on the highest power of $2$ dividing the degree of $f$.
Let $f(x)$ be a polynomial with coefficients in $\mathbb{C}$. We need to show that $f(x)$ has exactly $d=\deg(f)$ roots. It suffices to show that every polynomial has some root in $\mathbb{C}$. Because then we can factor $f(x)=(x-a_1) f_1(x)$ (where $a_1$ is a root of $f(x)$), and then apply the same argument to $f_1(x)$ to get $f_1(x)=(x-a_2)f_2(x)$. Doing this $d$ times (formally: we use induction) would give $f(x)=(x-a_1)\cdots (x-a_{d})$.
Okay, so let's show that $f(x)$ has some complex root. We might as well assume $f$ is monic, i.e. $f(x)=x^{d}+\mathsf{lower \ order \ terms}$. By replacing $f(x)$ by $f(x)\overline{f(\overline{x})}$ where $\overline{z}$ is the complex conjugation, it suffices to assume that $f(x)$ has real coefficients. This is because $f(x)$ and $f(x)\overline{f(\overline{x})}$ have same roots, but the latter polynomial has real coefficients! Indeed, if $f(x)=a_d x^{d}+\cdots + a_1 x + a_0$, then $\overline{f(\overline{x})} = \overline{a_d} x^{d}+\cdots + \overline{a_1} x + \overline{a_0}$, so their product will have real coefficients.
We write $d=\deg(f)=2^{n} m$ where $m$ is odd, and proceed by induction on $n$. When $n=0$, the degree $\deg(f)$ is odd. By intermediate value theorem, it is easy to see that $f$ has a real root. Indeed, $\displaystyle\lim_{x\to-\infty} f(x) =-\infty$ and $\displaystyle\lim_{x\to\infty} f(x)=\infty$. So $f$ must cross the real axis (thereby achieving a zero) at some point. Now we assume the induction hypothesis for the case $n-1$. A theorem from field theory tells us that $f$ has some splitting field, meaning that we can find some bigger field $K$ (containing $\mathbb{C}$) over which $f$ has $d=2^{n}m$ roots, say $x_1, x_2, …, x_d\in K$ (in reality, $K=\mathbb{C}$, and that's exactly what we are trying to prove!). Fix $c\in\mathbb{R}$ and consider the following polynomial: $$ G_{c}(x) = \prod_{i<j} (x-(x_i+x_j+cx_ix_j)) $$ The coefficients of $G_{c}(x)$ are symmetric polynomials in $\{x_i\}$ with real coefficients. By Fundamental Theorem of Symmetric Polynomials, any such symmetric polynomial can be written as a some polynomial (with real coefficients) in $\sum_{i} x_i$, $\sum_{i<j} x_i x_j$, $\sum_{i<j<k} x_i x_j x_k$, …, $x_1x_2\cdots x_d$. But these elementary symmetric functions are precisely the coefficients of $f$, which are real. To see this, expand $f(x)=(x-x_1)\cdots(x-x_d)$. So the conclusion is that $G_{c}(x)$ has real coefficients. Now, $\deg(G_{c})={d\choose 2}=d(d-1)/2=2^{n-1} (2^{n}m-1)$. Since $2^{n}m-1$ is odd, we can apply induction hypothesis to deduce that $G_{c}$ has a root $z_{c}\in\mathbb{C}$. So $z_{c}=x_i+x_j+cx_ix_j$ for some pair of index $(i, j)$ with $i<j$. But number of such indices is finite, while there are infinitely many choices for $c\in\mathbb{R}$. Hence, we can find distinct $c$ and $d$ in $\mathbb{R}$ such that $$ z_{c}=x_i+x_j+cx_ix_j \ \ \ \ \ z_{d}=x_i+x_j+dx_ix_j $$ Since $c\neq d$, by using appropriate linear combination of these equations, it is easy to solve for $x_i+x_j$ and $x_ix_j$. In other words, find complex numbers $w_1, w_2$ such that $x_i+x_j=w_1$ and $x_ix_j = w_2$. These two equations can be combined to give a quadratic equation in $x_i$. The good old quadratic formula applies and shows $x_i\in\mathbb{C}$. So we have found a root of $f$ in $\mathbb{C}$! Whew!