We discussed 3 proofs of this theorem in this post, each with a different flavor. Today I’m going to give another easy proof, yet with very different ideas.
We start with the following, which is well-known.
Lemma. Every rational function $f$ on $\mathbb{C}$ can be extended to a meromorphic function on the Riemann sphere $S^2$.
Proof. $f(1/z)$ is also rational, so is meromorphic around $0$.
Remark. The converse of this is also true. If $\varphi:S^2\to S^2$ is holomorphic, we can find some rational function $f$ such that $\varphi-f$ has no poles, i.e. it is holomorphic. Since $S^2$ is compact, $\operatorname{im}(\varphi-f)$ is disjoint from some neighborhood of $\infty$. By Liouville’s theorem, $\varphi-f$ is constant.
Theorem.(FTA) Every non-constant rational function $f\in\mathbb{C}(X)$ has a zero. Of course, it might be at $\infty$, but not if $f$ is a polynomial.
Proof.(Geometric) Let $Y\subset S^2$ be the set of regular values of $f$, including all points outside of the image. Put $X=f^{-1}(Y)$.
By the inverse function theorem, $f\vert_X$ is a local diffeomorphism. Moreover, since $f$ is rational, the fiber over every point is finite. We claim that the fiber size
\[y\mapsto\vert f^{-1}(y)\vert\]is locally constant on $Y$. Indeed, let $U_x$ be the open neighborhood of each $x\in f^{-1}(y)$. Then $\left(\bigcap_xf(U_x)\right)\setminus f\left(S^2\setminus\bigcup_xU_x\right)$ is an open neighborhood of $y$ where the fiber size is constant. (Note that this works even if $f^{-1}(y)=\varnothing$.)
Since $f’$ is nonzero rational, $\vert S^2\setminus Y\vert<\infty$. Therefore $Y$ is connected, so the fiber size is a constant $b$. If $b=0$, $f$ is constant. Thus $b>0$, which means that $Y\subset\operatorname{im}(f)$. Then
\[S^2\setminus\operatorname{im}(f)\subset Y\subset\operatorname{im}(f),\]thus $\operatorname{im}(f)=S^2$. In particular, $0\in\operatorname{im}(f)$.
By directly looking at the property of $Y$, we actually get an even simpler proof.
Proof.(Topological) As $f\vert_X$ is a local diffeomorphism, $U:=\operatorname{im}(f\vert_X)$ is open in $S^2$. Observe that $\operatorname{im}(f)$ is closed and is a union of $U$ with a finite set. We conclude that $\vert\partial U\vert<\infty$.
Then $U$ is clopen in $S^2\setminus\partial U$, a connected space. Therefore $\operatorname{im}(f)\supset\overline{U}=S^2$.
Post date: 2025/03/06
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