Fundamental Theorem of Algebra

Every non constant polynomial over has a root in .

Proof

By contradiction.
Let with no root, (wlog monic)

Take st
On , there can be no roots.
Let for .
Then also has no roots in .
Define:

Well defined because is non zero on

is a Homotopy of loops.

For , get loop ie
Set
well defined because is assumed to have no roots

.

So (by homotopy invariance)

so

is a homotopy that gives as loops,
so which is a contradiction.