# A "minimal" proof of the fundamental theorem of algebra

When I was in graduate school, I came up with what I think is a nice proof of the fundamental theorem of algebra.  At the time, I wrote it up here somewhat formally; I thought it might make a nice blog post, since the formal write-up obscures the very simple underlying ideas.  The goal was to use the minimal amount of technology possible -- in the end I use just a little algebra and some elementary point-set topology, as well as the implicit function theorem.

Recall that we'd like to prove:

Theorem (The Fundamental Theorem of Algebra).  Every non-constant polynomial over the complex numbers has a root.

The general strategy is this: let $$M_n$$ be the space of monic polynomials of degree $$n$$ over $$\mathbb{C}$$, that is, $$\mathbb{C}^n$$.  Let $$D\subset M_n$$ be the discriminant locus, that is, the set of polynomials which have a double root.  Let $$R\subset M_n\setminus D$$ be the set of polynomials with a complex root but with no double root.  We'll show

1. $$M_n\setminus D$$ is connected (indeed, the complement of any algebraic variety in affine space is connected in the complex-analytic topology).
2. $$R\subset M_n\setminus D$$ is open (this will follow from the implicit function theorem), and
3. $$R\subset M_n\setminus D$$ is closed (this is a simple limit argument).

In fact, both (2) and (3) above are true over $$\mathbb{R}$$, which shows that the set of polynomials with roots over $$\mathbb{R}$$ is a union of connected components of the complement of the discriminant locus.  In any case, now we're done -- we've shown that $$R$$ is a connected component of the connected space $$M_n\setminus D$$.  As $$R$$ is clearly non-empty, it is thus all of $$M_n\setminus D$$, so every polynomial has a root as desired.

The cubic discriminant, created using grapher.  Observe that one can see the the singular locus (the twisted cubic), which can be thought of as the set of cubics with a triple root.

### The discriminant locus and its complement

Recall that we let $$M_n=\{x^n+a_{n-1}x^{n-1}+\cdots+a_0, a_i\in \mathbb{C}\}$$ be the set of monic degree $$n$$ polynomials over the complex numbers; this space is isomorphic to $$\mathbb{C}^n$$.  We let $$D\subset M_n$$ be the discriminant locus, namely the set of polynomials with a double root.

Claim 0. There exists a polynomial $$g_n$$ such that $$D=\{(a_{n-1}, \cdots, a_0)\mid g_n(a_{n-1},\cdots, a_0)=0\}$$.  In particular, $$D$$ is an algebraic subvariety of $$M_n$$.

This is obviously very well-known, but let me sketch two proofs.

Proof 1.  Let $$f(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_0$$ be the generic polynomial of degree $$n$$, and let $$\alpha_1, \cdots, \alpha_n$$ be roots of $$f$$ in a finite extension of $$\mathbb{C}(t)$$.  Then $$f$$ has a double root if and only if $$\prod_{i\not=j} (\alpha_i-\alpha_j)=0.$$  But this is a symmetric polynomial in the $$\alpha_i$$, and hence by the fundamental theorem on symmetric functions, may be written in terms of the $$a_i$$ (which are the elementary symmetric polynomials in the $$\alpha_j$$).  $$\blacksquare$$

Proof 2.  $$f$$ has a double root if and only if $$f$$ and $$f'$$ have a factor in common.  By the Chinese Remainder Theorem, this occurs if and only if the natural map $$\phi: k[x]/(f(x)\cdot f'(x))\to k[x]/f(x)\oplus k[x]/f'(x)$$ is not an isomorphism (each component of this map is just the natural quotient map).  Thus $$f$$ has a double root if and only if $$\det(\phi)=0.$$  This is evidently the desired polynomial in the $$a_i$$.  $$\blacksquare$$

Remark.  Choosing the usual basis of $$k[x]/(f\cdot f'), k[x]/f, k[x]/f'$$ (that is, the basis consisting of powers of $$x$$, one obtains the usual formula for the discriminant very directly from this second proof.

We now prove our first claim, namely that the set of polynomials with no double roots is connected.

Claim 1.  $$M_n\setminus D$$, the set of monic polynomials of degree $$n$$ with no double root, is connected.

Proof.  In fact we show that if $$Y\subset \mathbb{C}^n$$ is the zero set of any polynomial $$g$$, then $$\mathbb{C}^n\setminus Y$$ is connected.  The desired claim is a special case.

Let $$x, y\in \mathbb{C}^n\setminus Y$$ be any two points, and let $$L=\{tx+(1-t)y\mid t\in \mathbb{C}\}$$ be the complex line connecting them.  Then $$L\cap Y=\{g|_L=0\}$$ is finite.  But $$L\simeq \mathbb{C}$$ and the complement of finitely many points in $$\mathbb{C}$$ is path-connected.  Thus there is a path connecting $$x$$ to $$y$$ which avoids $$Y$$.  As $$x, y$$ were arbitrary, this shows that $$\mathbb{C}^n\setminus Y$$ is path-connected.  $$\blacksquare$$

### The space of polynomials with roots

We now show that the space $$R$$ of polynomials with at least one root but no double root is both open and closed in $$M_n\setminus D$$.

Claim 2. $$R$$ is closed.

Proof.  Let $$\{f_i\}$$ be a sequence of polynomials in $$R$$ and $$\alpha_i$$ a sequence of complex numbers such that $$f_i(\alpha_i)=0$$.  Suppose that $$f=\lim_{i\to \infty} f_i,$$ with $$f\in M_n\setminus D$$.  Then the $$f_i$$ are bounded, and hence the $$\alpha_i$$ are bounded; hence they have some limit point $$\alpha$$.  But by continuity, $$f(\alpha)$$=0, so $$f\in R$$ as desired.  $$\blacksquare$$

Claim 3. $$R$$ is open.

Proof.  This is an exercise with the implicit function theorem.  Let $$f\in R$$, and let $$\alpha\in \mathbb{C}$$ be a root of $$f$$.  Let $$S\subset M_n\times \mathbb{C}$$ be the set of pairs $$(g, \beta)$$ such that $$g(\beta)=0$$, so that $$(f, \alpha)\in S$$. That is, $$S=\text{ev}^{-1}(0)$$ where $$\text{ev}: M_n\times \mathbb{C}\to \mathbb{C}$$ is the evaluation map $$\text{ev}: (g, \beta)\mapsto g(\beta).$$

As $$f\in R$$ it has no double roots, so $$f'(\alpha)\not=0$$.  Thus $$\text{ev}$$ is a submersion at $$(f, \alpha)$$, and we may apply the implicit function theorem in a neighborhood of $$f$$. In particular, there is an open neighborhood $$U$$ of $$f$$ and a function $$r: U\to \mathbb{C}$$ so that for $$g\in U$$, $$g(r(g))=0$$.  In particular, this neighborhood of $$f$$ is in $$R$$; as $$f$$ was arbitrary, $$R$$ is open.  $$\blacksquare$$

In particular:  $$R$$ is a connected component of $$M_n\setminus D$$.  But by Claim 1, $$M_n\setminus D$$ is connected.  $$R$$ is clearly non-empty (it contains $$\prod_{i=1}^n (x-i),$$ for example), so $$R=M_n\setminus D$$.

### Voila

And now we're done.  Indeed, if $$f\in D$$, $$f$$ has a double root and hence has a root (by induction on the degree of $$f$$).  But now if $$f\not\in D$$, then $$f\in M_n\setminus D=R$$.  But every polynomial in $$R$$ has a root by definition.

Remark.  One nice aspect of this proof is that it is completely analogous to Hensel's lemma.  Hensel's lemma tells us that over a $$p$$-adic field, if a polynomial away from the discriminant locus has a root, then any small perturbation of it has a root as well.  This is exactly Claim 3 above.  And indeed, one may consider the proof of Hensel's lemma to be an infinitesimal version of the implicit function theorem.  Of course the $$p$$-adic topology is too coarse to admit an analogue of Claim 1 (and it must be, since $$\mathbb{Q}_p$$ is not algebraically closed).

Surely the proof isn't new -- but I haven't been able to find it anywhere.  Does anyone know a reference?  Or is this in fact a new proof of the Fundamental Theorem of Algebra?

### Notes

Keith Conrad has alerted me to two proofs (by Pukhlikov and Pushkar) which seem related to this proof.  He has a writeup of these proofs here.  (Pukhlikov's original paper is in Russian.)

Several readers have commented that my definition of the discriminant locus is unclear.  To be precise, one should take the vanishing locus of the polynomial $$g_n$$ as the definition; then Claim 0 should be interpreted as the statement that this locus is the same as the set of polynomials which have a double root over their splitting field.  In particular, to conclude that a polynomial $$f$$ in the discriminant locus has a root over $$\mathbb{C}$$, one observes that there is a polynomial $$h$$ defined over $$\mathbb{C}$$ dividing both $$f, f'$$.  But $$h$$ has a root by induction on degree, so $$f$$ has a root.