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.

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

1 Answer 1


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 $p(A) = B$.

  • $\begingroup$ @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. $\endgroup$ Sep 17, 2019 at 23:26
  • $\begingroup$ 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. $\endgroup$ Sep 17, 2019 at 23:42

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