TITLE: The proofs of the fundamental Theorem of Algebra
QUESTION [17 upvotes]: There are many proofs of the fundamental theorem of algebra.
Which are the most beautiful proofs?

REPLY [3 votes]: Via Galois Theory:
WLOG, let $p(x)\in\mathbb{R}[x]$ and let $F$ be the splitting field of $p$, embedded in some algebraic closure of $\mathbb{R}$. Since $F(i)$ is the splitting field of the square free part of $p(x)(x^2-1)$, it is Galois over $\mathbb{R}$. 
Let $G$ be the Galois group of $F(i)$ over $\mathbb{R}$. If there was an odd number dividing $|G|$, by Galois correspondence there is an odd degree subextension. This contradicts the fact that the intermediate value theorem tells us that every odd degree polynomial over $\mathbb{R}$ has a root in $\mathbb{R}$, so it must be that $G$ has order $2^n$ for some $n$.
Let $G'$ be the Galois group of $F(i)$ over $\mathbb{C}$. This is the splitting field of some polynomial dividing $p(x)$ so it is Galois. By the above, $|G'|=2^k$, where we wish to show $k=0$. By Cauchy's Theorem for $p$-groups, if $k>0$ then there exists a subgroup of $G'$, and therefore a field extension of $\mathbb{C}$ that has order $2$, but that contradicts the quadratic equation, as the quadratic equation tells us that there are no degree $2$ irreducible polynomials over $\mathbb{C}$. Therefore $|G'|=1$. However, $2|G'|\geq |G|$ so $2\geq |G|$.
It follows by the Fundamental Theorem of Galois Theory that $[F(i):\mathbb{R}]\leq 2$. Since $$2=[F(i):\mathbb{R}]=[F(i):F][F:\mathbb{R}]$$ we have that $[F:\mathbb{R}]=2$ precisely when $F=\mathbb{C}$ and that $[F(i):F]=2$ precisely when $F=\mathbb{R}$
However, $F$ was an arbitrary finite degree field extension, so we have that an arbitrary finite degree field extension of $\mathbb{R}$ is $\mathbb{C}$ precisely when it is non trivial, and we are done.