# Lost about a simple proof of the Fundamental Theorem of Algebra

The following is a proof of the Fundamental Theorem of Algebra divided into three steps:

1) Let $$C$$ be the finite set of critical points , i.e. $$p′(z)=0$$ for all $$z∈C$$$$C$$ is finite by elementary algebra.

2) Remove $$p(C)$$ from the codomain and call the resulting open set $$B$$ and remove from the domain its inverse image $$p^{1-1}(p(C))$$, and call the resulting open set $$A$$. Note that the inverse image is again finite.

3) Now you get an open map from $$A$$ to $$B$$, which is also closed, because any polynomial is proper (inverse images of compact sets are compact). But $$B$$ is connected and so $$p$$ is surjective.

$$Q1$$: Why is $$B$$ open? If one was to presuppose that the codomain was open from the beginning then I would see how the claim follows, yet it isn't clear to my why the codomain is open.

$$Q2$$: Why is $$A$$ open? If $$p^{-1}(p(C))$$ is finite then, again, I would see how the claim follows, but I don't see why the latter is true.

$$Q3$$: I do not understand why the mapping is open nor closed.

$$Q4$$: I do not see why $$B$$ is connected nor why that implies $$p$$ is surjective.

I have read baby Rudin before, yet I feel like I'm missing something since there are many steps of the proof that I don't get.

Many ideas pop in my head (to apply the concept of continuity defined in terms of open sets to polynomials, for example) but I can't seem to use such ideas to fill in the gaps of my understanding.

I would appreciate any help.

• $p'(z)=0$ is a polynomial equation. Sep 16, 2019 at 17:22

Q1: $$B$$ is arrived at by removing $$p(C)$$ from the codomain of the polynomial, not the image of the polynomial. The image is what we are trying to determine, but the codomain is just $$\Bbb C$$. Personally, I think wherever you got this from is at fault for saying "codomain" instead of just $$\Bbb C$$. And of course $$\Bbb C$$ less a finite set of points will definitely be open.

Q2: Same thing, in reverse. The domain of a polynomial is $$\Bbb C$$.

Q3: Consider $$z_0 \in U \subset A$$, with $$U$$ open. For a small enough neighborhood of $$z_0$$, since $$p'(z_0) \ne 0$$, you can approximate $$p(z) \approx p(z_0) + (z-z_0)p'(z_0)$$ well enough that all points sufficiently close to $$p(z_0)$$ must have inverse images. I.e., p(U) must contain a neighborhood of $$p(z_0)$$. Since this holds for all $$z_0 \in U, p(U)$$ must be open.

I haven't spotted where they were going with their comments about polynomials being proper. Maybe if I'd ever studied "proper" maps, I would. But there are other ways to prove that $$p$$ is closed when restricted to $$A \to B$$. Since $$\Bbb C$$ is first-countable, one way is this:

Let $$D \subset A$$ be closed, and suppose that $$d$$ is an accumulation point of $$p(D)$$. Then there must be a sequence $$(z_i)_i \subset D$$ such that $$p(z_i) \to d$$. Now, no subsequence $$(z_{i_j})_j$$ of $$(z_i)$$ can diverge to $$\infty$$, because then $$p(z_{i_j}) \to \infty$$ (since $$p$$ is non-constant polynomial), but we know that $$p(z_{i_j}) \to d$$. Therefore $$(z_i)$$ must be bounded. And therefore it must have some convergent subsequence $$z_{i_j} \to w$$ in $$\Bbb C$$. By continuity $$p(w) = d \notin p(C)$$, and therefore $$w \in A$$. And since $$w$$ is the limit of a sequence of points in the closed set $$D, w \in D$$. Therefore $$d = p(w) \in p(D)$$. Since $$p(D)$$ contains all of its accumulation points, it is closed.

Q4: Again, $$B = \Bbb C \setminus p(C)$$, and you cannot disconnect $$\Bbb C$$ by removing only a finite set of points. Now $$A$$, considered as a space itself, is both open and closed in the subspace topology (in any topological space, the entire space is always both open and closed in its own topology). And $$p$$ when considered as a map $$A \to B$$ is both open and closed, which means it carries open sets to open sets and closed sets to closed sets. Therefore $$p(A)$$ is both open and closed. It is also non-empty. Since $$B$$ is connected, the only non-empty subset that is both open and closed is itself. Therefore $$p(A) = B$$.

• @Leo - I've added an argument for why $p : A\to B$ is a closed map, though it is different from the one aluded to in the proof. Sep 17, 2019 at 23:26
• An interesting consequence of this proof is that any topologically-complete first-countable topological field that cannot be disconnected by the removal of a finite number of points must be algebraicly complete as well. And I have little doubt that "first-countable" can be removed from the conditions. Sep 17, 2019 at 23:42