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.