Suppose that p(z) is a nonconstant polynomial with no roots. The complex plane with additional point ∞ is homeomorphic to the 2-sphere. At each z in the plane, let the vector at z be 1/p(z), which is defined since p(z) is nonzero everywhere. As |z| goes to infinity, p(z) goes to 0, since p(z) is not constant, and hence the vector field is 0 at ∞. This gives a vector field on the 2-sphere with only one critical point, whereas Morse Theory tells us there must be at least two critical points.
I think this is correct. What perplexes me is, what prevents defining a nonzero constant vector field likewise, which would have no critical points?
Thanks for your help.
Morse function says that for any Morse function on $S^2$, it has at least two critical point. However, you constructed a vector field instead of a Morse function. The right machinery to use here should be Poincare-Hopf index theorem, which says the index of a vector field is equal to the Euler characteristic of the space. In this case, $\chi(S^2)=2$.
Consider your vector field given by $1/p(z)$, which has only one isolated zero at infinity. If you invert that, i.e. taking $z\mapsto 1/z$, your vector field becomes $1/p(1/z)$ with an isolated zero at the origin.
Let $p(x)=a_0+a_1x+\cdots+a_nx^n$ with nonzero $a_n$. Then $$\frac{1}{p(1/z)}=\frac{z^n}{a_0z^n+a_1z^{n-1}+\cdots+a_n}$$ which is a degree-$n$ function. So the vector field has index $n$.
According to Poincare-Hopf, we have $n=2$. But we know that the Fundamental Theorem of Algebra is true for $n=2$ because we have the quardratic formula.