# Fundamental theorem of Algebra using fundamental groups.

I tried to prove the fundamental theorem of algebra using fundamental groups. I am not able to understand the proof. Can anyone help me?

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How is the proof? From what book? What have you done, where are you stuck,...? –  DonAntonio Jan 29 at 16:15
I am guessing the proof by winding number: If $p$ is a polynomial with leading term $z^n$ then $p(re^{i\theta})$ winds $n$ times around the origin (when $\theta$ varies over $[0,2\pi]$) when $r$ is large, but $0$ times when $r=0$. Which yields a contradiction if $p$ has no zero. Am I on the right track? –  Harald Hanche-Olsen Jan 29 at 16:19
Here's the proof jeremykun.com/2012/01/22/… –  Mathematician Jan 29 at 16:23
Nice, @WaqasAliAzhar: where are you stuck? BTW, the proof in the book is a little more detailed (and long) and, perhaps, a little easier to grasp than the one in that site (which is a shortened copy of the one from the book), but you must know several things from algebraic topology, so again: where are you stuck? –  DonAntonio Jan 29 at 16:35

So the statement of the Fundamental theorem of Algebra is as follows: $$\text{Let}~ f(x) = x^n + c_1x^{n-1} + \cdots + c_{n-1}x + c_n \text{be a polynomial with } n > 0 \text{ and each }c_i \in \Bbb{C}\\ \text{ There are } n \text{ complex numbers (counted including multiplicities) } x_i \text{ such that } f(x_i) = 0.$$

We'll first prove the easier statement that there exists at least one root, which is the one in your book, but we can do better than that.

Assume to the contrary that $f$ has no roots in $\Bbb{C}$. We can therefore constuct the following function $g_r(s): I \to S^1$ for each $r \in \Bbb{R}$: $$g_r(s) = \frac{f(re^{2\pi is})/f(r)}{|f(re^{2\pi is})/f(r)|}$$

Notice how $g_r(0) = g_r(1) = 1$ and it is continuous, so it forms a closed loop in $S^1$, whose fundamental group we know to be isomorphic to the infinite group generated by a single generator. We also know, however that $g_r(s) = g(r,s)$ is continuous in $r$, so as $r$ varies we get a homotopy between $g_r(s)$ and $g_{r'}(s)$ for two $r, r'$. But $g_0(s) = \frac{f(0)/f(0)}{|f(0)/f(0)|} = 1$, and is therfore constant, and a constant loop, but any $g_r$ homotopes to $g_0$, implying $[g_r] = 0$.

Fix some $r_0$ that is large (at least larger than both $\sum |c_i|$ and $1$). For all $x$ on the circle of radius $r_0$ (i.e., $|x| = r_0$) we have the following: $$|x^n| = r_0^n = r_0\cdot r_0^{n-1} > (\sum |c_i|)|x^{n-1}| \geq |c_1x^{n-1} + \cdots + c_{n-1}x + c_n|$$

Because $|x^n| > t|c_1x^{n-1} + \cdots + c_{n-1}x + c_n|$ for $0 \leq t \leq 1$, the polynomial $$f_t(x) = x^n + t(c_1x^{n-1} + \cdots + c_{n-1}x + c_n)$$ has no roots on the circle of radius $r_0$. if, however, you repace $f$ with $f_t$ in our formula for $g_r$, you get, as $t$ goes from $0$ to $1$ a homotopy between $\omega_n$, the loop that winds $n$ times around the origin and the 0 loop, because if $$g_{r_0,t} = \frac{f_t(re^{2\pi is})/f_t(r)}{|f_t(re^{2\pi is})/f_t(r)|}$$

then $$g_{r_0,0}(s) = \frac{r^ne^{2\pi nis}/r^n}{|r^n/r^n|} = e^{2\pi nis} = \omega_n$$ but $g_{r_0,1}$ is homotopic to 0 as we have already seen. This implies $n=0$, contrary to assumption.

The stronger proof simply follows from using $f'(x) = f(x)/(x-c)$ given the root $c$ found in the prequel until the degree of the polynomial is 0.

I've tried to flesh out all of the details in order to guide through the proof as much as possible. If there's anything I can clarify let me know.

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