I also want to say that there is not such thing as a "purely algebraic proof". Because the construction of the real numbers, and thereby the complex numbers, is topological! To develop the algebraic properties of the real and complex numbers we must know the topology of the reals. However, when we say "an algebraic proof" we mean to say that we can prove the theorem just using the algebraic properties of the reals and complex numbers alone without using any other additional non-algebraic facts.

**Analytic Proof:**The standard analytic proof of the fundamental theorem of algebra uses Liouville's theorem from complex analysis. But that proof is too boring because it is too commonly used. Instead we will give a stronger proof using Rouche's theorem. This is a better analytic proof because it gives a bound on the size of zeros of the polynomial. Liouville's theorem simply asserts a zero exist, it does not put a bound on them. Rouche's theorem will allow us to bound the zeros. Hence this is a better proof.

First let us recall what Rouche's theorem says. We will state it a little less general to keep the discussion simpler. But let f and g be analytic functions on an open set containing the closed disk $z\leq r$ for some r>0. Suppose that f dominates g on the boundary i.e. $|g(z)|<|f(z)|$ on $|z|=r$. Then f has as many zeros as (f-g) inside the disk |z|< r (counting multiplicities).

With this result we can prove the fundamental theorem. WLOG, assume that $p(z)=z^n+c_{n-1}z^{n-1}+...+c_1z+c_0$. If $z\geq 1$ then $z^{n-1} \geq z^k$ for all $0\leq k < n$. Therefore, $|c_{n-1}z^{n-1}+...+c_1z+c_0| \leq C|z|^{n-1}$. Where $C = \Sigma_k |c_k|$. If $r>\max(1,C)$ then on $|z|=r$ we see that $C|z|^{n-1} = Cr^{n-1} < r^n =|z|^n$. Thus, $z^n$ dominates $c_{n-1}z^{n-1}+...+c_1z+c_0$ on $|z|=r$. By Rouche's theorem we immediately see that $p(z)$ has as many zeros as $z^n$ inside $|z|=r$, which is n. Not only did we prove a zero exists, but we have a bound. We know all these zeros must lie inside this disk |z|< r

**Algebraic Proof:**We need some lemmas before we jump into the proof. The first is that there is no proper odd degree extension of $\mathbb{R}$. If F is a field extension with $[F:\mathbb{R}]$ odd then $F=\mathbb{R}$. The reason is very simple, pick any $\alpha \in F - \mathbb{R}$ (assuming this set is non-empty). Then $[\mathbb{R}(\alpha):\mathbb{R}]$ divides $[F:\mathbb{R}]$ therefore it must be odd. Which means $\alpha$ is a root of a odd degree irreducible polynomial over $\mathbb{R}$. However, all odd degree polynomials over $\mathbb{R}$ have a real zero. This is a standard basic fact about odd polynomials which follows from the intermediate value theorem.

The other lemma that we need is that there is no quadratic field over $\mathbb{C}$. This is because if $[F:\mathbb{C}]=2$ then $F=\mathbb{C}(\alpha)$. Thus, $\alpha$ is a root of a quadratic polynomial over $\mathbb{C}$. But all quadratic polynomials have roots in $\mathbb{C}$ - because the quadratic formula is always applicable in this case.

If K/F is a field extenion the "normal closure" of this extension is the extension N/K where N is defined to be the splitting field of $\{ \min (F,a): a\in K \}$. This in the way is the smallest extension over K that makes N/F normal. If K/F is normal then N=K. It is not hard to show that if $K=F(a_1,...,a_n)$ i.e. $K/F$ is finite, then $N$ is the splitting field over $\{ \min (F,a_k): 1\leq k \leq n \}$, so that N/F is finite also.

Let us begin with the proof. To say that $\mathbb{C}$ always has a zero over any non-constant complex polynomial is the same as saying that $\mathbb{C}$ is algebraically closed. If $A$ is the algebraic closure over $\mathbb{C}$ and $a\in A$ then $[\mathbb{C}(a):\mathbb{C}]<\infty$. Therefore, if we want to prove that $A=\mathbb{C}$ i.e. it is algebraically closed it suffices to prove that $\mathbb{C}$ has no proper finite field extensions.

Let F be a finite field extension of $\mathbb{C}$. Thus, clearly F is also a finite field extension of $\mathbb{R}$. Since the characteristic of $\mathbb{R}$ is zero it is a perfect field which means that $F/\mathbb{R}$ is a seperable extension. However, it is not necessarily normal. To make this extension normal we work with the normal closure instead. Let $N$ be the normal closure of the extension $F/\mathbb{R}$. We will argue that $N=\mathbb{C}$ to complete the proof.

Since $N/\mathbb{R}$ is a normal and separable extension it is a Galois extension. Let $G=\text{Gal}(N/\mathbb{R})$. Now $G = [N:\mathbb{R}]$ which is even because $[\mathbb{C}:\mathbb{R}]=2$ divides $[N:\mathbb{R}]$. Thus, $G$ has a Sylow 2-subgroup by the Sylow Theorems. Let P be a Sylow 2-subgroup.

Let $K=N^P$ i.e. let $K$ be the fixed field under the group $P$. Then $[G:P] = [K:\mathbb{R}]$ (by the degree correspondence of Galios theory). But $[G:P]$ is odd since $P$ is a Sylow 2-subgroup, which means $K$ is an odd extension over $\mathbb{R}$, but then it means $K=\mathbb{R}$ which forces $P=G$. Thus, we have shown that G is in fact a 2-group i.e. a group of order a power of $2$. Therefore, $G_1=\text{Gal}(N/\mathbb{C})$ is a 2-group also (unless it has order $2^0=1$). By Sylow's theorems again we know there is a maximal subgroup M of $G_1$. Hence, $[G_1:M]=2$, if E is the fixed field under M then $[E:\mathbb{C}]=2$. But this is impossible. Hence, $G_1$ must be a trivial group, hence $N = \mathbb{C}$.

**Geometric Proof:**Here we will assume standard results from algebraic topology. We will assume that the polynomial has the special form $p(z) = z^n + c_{n-1}z^{n-1}+...+c_1 z + c_0$ where $\Sigma_k |c_k| < 1$. The general case follows easily from this special case.

Let $S^1 = \{ z\in \mathbb{C} : |z| = 1\}$ be the unit circle. Then we know that $\pi_1 (S^1,1) = \mathbb{Z}$. Consider the continuous function $f : (S^1,1) \to (S^1,1)$ defined by $f(z) = z^n$. This map induces a group homomorphism $f_*:\pi_1(S^1,1)\to \pi_1(S^1,1)$. Since $\pi(S^1,1)$ is an infinite cyclic group with generator consisting of the loop $[ e^{2\pi i t} ]$ it follows that $f_* [ e^{2\pi i t} ] = e^{2\pi i n t}$. Henceforth, we can think of $f_*$ as acting on the group $\mathbb{Z}$ by multiplication by $n$. The important fact that we need about $f_*$ is that it is a monomorphism (injective).

Now consider the map $g: (S^1,1) \to (\mathbb{C}^{\times},1)$. If we let $i:(S^1,1) \to (\mathbb{C}^{\times},1)$ then $i_*$, the induced homomorphism between homotopy groups, is a monomorphism. This is because there is a retract from $\mathbb{C}^{\times}$ to $S^1$. Since $g = if$ it follows $g_* = i_* f_*$, but $f_*$ was a monomorphism, $g_*$ being a composition of monomorphism is a monomorphism.

Assume that $p(z)$ has no zero then we can define a map $p: B^2 \to \mathbb{C}^{\times}$. Let $h$ be the restriction of $p$ to $S^1$. The map $h$ is nulhomotopic because it extends to a continuous function of the ball $B^2$ which is contractible.

If we define $H:S^2\times I\to \mathbb{C}^{\times}$ by $H(z,t) = z^n + t(c_{n-1}z^{n-1}+...+c_1z+c_0)$ then $H$ is a homotopy between $h$ and $g$. Notice, to prove that this is a homotopy we use the fact that the sum of the coefficients $c_k$ is strictly less than 1.

Therefore, $g$ is nulhomotopic. But this is non-sense because $g_*$ is injective, if $g$ was nulhomotopic then $g_*$ would have to be the trivial homomorphism. We have arrived at a contradiction. Therefore, it must be that $p(z)$ has a root.

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I am curious to see more proofs of this theorem. What are some of your favorite proofs of the Fundamental Theorem?