Theorem. Every non-constant polynomial $P \in \mathbb{C}[z]$ must have a zero.
Proof. Using the stereographic projection from the point $\infty=(0, \, 0, \, 1)$, we can extend the polynomial function $P \colon \mathbb{C} \to \mathbb{C}$ to a function $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ such that $f(\infty)=\infty$. A straightforward use of the change of coordinates $z=1/w$ shows that $f$ is smooth on the whole of $\mathbb{S}^2$, including the point $\infty$.
Let us remark that $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ has a finite number of critical points (and so a finite number of critical values), corresponding to the zeros of the derivative $P’$ and possibly to the point $\infty.$
If $0\in \mathbb{C}$ corresponds to a critical value for $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ then, by definition, it has a preimage via $P \colon \mathbb{C}^2 \to \mathbb{C}^2$, and there is nothing to prove.
We may therefore assume that $0\in \mathbb{C}$ corresponds to a regular value for $f \colon \mathbb{S}^2 \to \mathbb{S}^2$. The locus of these regular values is the complement of finitely many points on the sphere, so it is connected.
From the Implicit Function Theorem for differentiable manifolds, it follows that $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ is a local diffeomorphism when restricted to the locus of regular values. So the number of points which are preimages of a fixed one is locally constant on such locus, hence constant (being the locus connected).
Since $\deg(P) \geq 1$, there are surely (infinitely many) points $y \in \mathbb{S}^2$ such that $f^{-1}(y) $ is non-empty. Thus, the same must be true for the point corresponding to $0 \in \mathbb{C}.$
In other words, $0$ is in the image of $P$. $\square$
References.
[1] J. W. Milnor: Topology from the Differentiable Viewpoint, Princeton University Press 1965.
Nessun commento:
Posta un commento