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There are many proofs of the fundamental theorem of algebra.

Which are the most beautiful proofs?

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closed as primarily opinion-based by N. F. Taussig, user133281, Daniel Fischer Dec 24 '15 at 19:46

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I kind of like the one using Louiseville's theorem. $\endgroup$ – Gregory Grant Dec 24 '15 at 15:51
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    $\begingroup$ This is awfully broad (there being many proofs) and a matter of taste... $\endgroup$ – vonbrand Dec 24 '15 at 16:01
  • $\begingroup$ Milnor's book topology from the differentiable viewpoint, there is a marvellous proof of the fundamental theorem of algebra in the first chapter using stereographic projection, compactness and regular values I strongly suggest to read it because is very fruitfully helpful $\endgroup$ – Salvatore Dec 24 '15 at 16:05
  • $\begingroup$ How do you like the proofs listed in en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Proofs ? $\endgroup$ – lhf Dec 24 '15 at 16:08
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    $\begingroup$ See also mathoverflow.net/questions/10535/… $\endgroup$ – Wojowu Dec 24 '15 at 16:47
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Let $P\in\mathbb{C}[X]$ of degree greater than $1$. Assume by contradiction that $P$ has no roots in $\mathbb{C}$ and let define: $$f:=\frac{1}{P}.$$ $f$ is holomorphic on $\mathbb{C}$ and is bounded since: $$\lim_{z\to\pm\infty}|f(z)|=0.$$ Therefore, using Liouville's theorem, $f$ is constant and so is $P$, a contradiction.

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  • $\begingroup$ Yes on one hand extremely elegant. On the other hand, you need Liouville's theorem which is non-trivial to prove. So if you had to prove it from scratch, I doubt going through Liouville would be the fastest route. $\endgroup$ – Gregory Grant Dec 24 '15 at 16:06
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    $\begingroup$ Louiville is non-trivial to prove and not intuitive. I like the homotopy approach because there is a strong topological intuition going on, even if the topology part is actually quite hard to formalize. $\endgroup$ – Thomas Andrews Dec 24 '15 at 16:09
  • $\begingroup$ From what I know there is no straightforward proof of d'Alembert's theorem, all the proofs I am aware of involve some machinery. However, I agree with both of you, Liouville's theorem need some non-intuitive work to be proved. $\endgroup$ – C. Falcon Dec 24 '15 at 16:09
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How about a proof in two parts:

$\bullet$ If $R$ is a real-closed field, then $R[X]/(X^2+1)$ is algebraically closed.

$\bullet$ $\mathbb R$ is a real-closed field.

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In my opinion, the most beautiful proofs are the purely algebraic ones. We need a little analysis, but only enough to show that every odd degree polynomial over $\mathbb{R}$ has a root, which follows from the Intermediate Value Theorem (we can't really do better than this, since we must use completeness of $\mathbb{R}$ in some way for the theorem to hold).

Suppose that $\mathbb{R}\subset F$ is a finite extension of fields. If we can show that it has degree $\leq 2$, then the FTA follows.

Let $G$ be the Galois group $\operatorname{Gal}(F/\mathbb{R})$, and let $H$ be a Sylow $2$-subgroup of $G$. $H$ has odd index, so the fixed field $F^H$ of $H$ has odd degree. But a nontrivial extension of $\mathbb{R}$ of odd degree would give us an odd-degree polynomial with no root, so we have $F^H = \mathbb{R}\implies H=G$, so $G$ is a $2$-group.

If $|G|=1$ or $|G|=2$, we are done. If $|G|\geq 4$, then $G$ has a subgroup of index $2$ containing a subgroup of index $4$ (by general facts about $p$-groups). By Galois, these correspond to a degree $2$ extension of $\mathbb{C}$, which is impossible by the quadratic formula.

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Homotopy approach

Assume that a $p$ exists with $\deg p>0$ and $p(z)\neq 0$ for all $z\in\mathbb C$.

The homotopy approach is to note that a polynomial $p$ sends circles around zero with "large enough" radius to a path that goes around the origin $n$ times, counterclockwise, where $n=\deg p$. But the circles of radius tiny radius go to paths that circle zero not at all, and these paths are trivially homotopic.

Then we just note that there is no homotopy in the punctured plane $\mathbb C\setminus 0$.

Of course, this requires a lot of machinery to prove rigorously, but I've always liked it because it makes intuitive sense, imaging trying to stretch a rubber band that does not "go around zero" to go around zero a total of at least once, without passing through zero...

This homotopy approach is much like a multi-dimensional "intermediate value theorem." If $f:\mathbb R^n\to\mathbb R^n$ sends a sphere to something that wraps around the origin in some non-trivial way (that is, a map that is not homotopically constant in $\mathbb R^n\setminus\{0\}$) then it must send some point on the interior of that sphere to the origin.

Constructive proofs exist

There exist entirely constructive proofs of FTA. In reverse mathematics, FTA is a result in $RCA_0$, which, roughly, models constructive number theory.

You can Google "constructive proof of fundamental theorem of algebra" to find specific examples.

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Consider $p: \mathbb R^2 \to \mathbb R^2$ a non-constant complex polynomial. Its Jacobian determinant for each $z \in \mathbb R^2$ is given by $\|p'(z)\|$, thus the determinant is zero at finitely many points, that is, the zeros of the polynomial $p'(z)$. Then $p(\mathbb R^2)$ is open (which is an exercise in analysis). On the other hand, $$\lim _{z \to \infty} p(z) = \infty$$

Then $p(\mathbb R^2)$ is closed. As $\mathbb R^2$ is connected we have that $p(\mathbb R^2) = \mathbb R^2$, that is , $p$ is surjective. That concludes the proof.

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  • $\begingroup$ Isn't the jacobian determinant $|p'(z)|^2$? $\endgroup$ – Thomas Andrews Dec 24 '15 at 16:13
  • $\begingroup$ Also, "the zeros of the polynomial" means "the zeroes of the polynomial $p'$," presumably? Very easy to read that as referring to $p$. $\endgroup$ – Thomas Andrews Dec 24 '15 at 16:18
  • $\begingroup$ Topologically, $p$ is continuous, open, and proper; the image of a proper map between locally compact Hausdorff spaces must be closed. $\endgroup$ – Dustan Levenstein Dec 24 '15 at 16:49
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This blog post has a probabilistic proof based on Brownian motion.

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Via Galois Theory:

WLOG, let $p(x)\in\mathbb{R}[x]$ and let $F$ be the splitting field of $p$, embedded in some algebraic closure of $\mathbb{R}$. Since $F(i)$ is the splitting field of the square free part of $p(x)(x^2-1)$, it is Galois over $\mathbb{R}$.

Let $G$ be the Galois group of $F(i)$ over $\mathbb{R}$. If there was an odd number dividing $|G|$, by Galois correspondence there is an odd degree subextension. This contradicts the fact that the intermediate value theorem tells us that every odd degree polynomial over $\mathbb{R}$ has a root in $\mathbb{R}$, so it must be that $G$ has order $2^n$ for some $n$.

Let $G'$ be the Galois group of $F(i)$ over $\mathbb{C}$. This is the splitting field of some polynomial dividing $p(x)$ so it is Galois. By the above, $|G'|=2^k$, where we wish to show $k=0$. By Cauchy's Theorem for $p$-groups, if $k>0$ then there exists a subgroup of $G'$, and therefore a field extension of $\mathbb{C}$ that has order $2$, but that contradicts the quadratic equation, as the quadratic equation tells us that there are no degree $2$ irreducible polynomials over $\mathbb{C}$. Therefore $|G'|=1$. However, $2|G'|\geq |G|$ so $2\geq |G|$.

It follows by the Fundamental Theorem of Galois Theory that $[F(i):\mathbb{R}]\leq 2$. Since $$2=[F(i):\mathbb{R}]=[F(i):F][F:\mathbb{R}]$$ we have that $[F:\mathbb{R}]=2$ precisely when $F=\mathbb{C}$ and that $[F(i):F]=2$ precisely when $F=\mathbb{R}$

However, $F$ was an arbitrary finite degree field extension, so we have that an arbitrary finite degree field extension of $\mathbb{R}$ is $\mathbb{C}$ precisely when it is non trivial, and we are done.

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See the book The Fundamental Theorem of Algebra by Fine and Rosenberger, which contains eight proofs.

See also The Fundamental Theorem of Algebra: A Visual Approach by Velleman.

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