Am Math Monthly 107 (2000), 842-843

## Fundamental Theorem of AlgebraYet Another Proof

Anindya Sen

Theorem. The Fundamental Theorem of Algebra. Let

 P(z) = a0zn + a1zn-1 + ... + akzn - k + ... + an

be a polynomial of degree n ≥ 1 with complex numbers ai as coefficients. Then P has a root, i.e., there is a φ C such that P(φ) = 0.

We prove the theorem by showing that Image(P) = C.

We assume the standard result that a complex polynomial P: CC is a proper map, i.e., P-1(A) is compact whenever AC is compact. (P is continuous, and |P(x)| → ∞ as |x| → ∞. Hence, if A C is closed and bounded, so is P-1(A). Hence, P is proper.)

Let f: UR2 be a differentiable map of an open set UR2 to R2. A point xU is said to be a regular point of f if Df(x): R2R2 is invertible. Otherwise, x is said to be a critical point of f. A point y R2 is said to be a critical value of f if it is the image of a critical point.

With this notation in mind, we first prove

Lemma 1. Let K be the set of critical values of P. Then K and P-1(K) are both finite subsets of C.

Proof: The critical points of P are the points at which P’(z) = 0. Since P’ is a polynomial of degree n — 1, there are at most n — 1 critical points. Since each critical value is the image of a critical point, K has at most n — 1 points. Now each critical value has at most n inverse images, hence, P’(K) has at most n(n — 1) points. (We use the fact that a complex polynomial of degree k has at most k roots. The proof of this result does not use the fundamental theorem of algebra.)

Lemma 2. Let X = C \ P-1(K) and Y = C \ K. Then P(X) = Y.

Proof. Lemma 1 ensures that both X and Y are open connected subsets of C. Also, observe that all points in X are regular points of P, i.e., DP(x) is invertible for all x X.

Since P: CC is proper and C is locally compact, it follows that Image(P) is closed in C. P(X) = Y ∩ Image(P). Hence, P(X) is closed in Y.

Let y P(X). Then, y = P(x) for some xX. Since x is a regular point, the inverse function theorem tells us that there are open neighbourhoods U, V of x, y respectively, such that P: U → V is bijective. Hence every point yP(X) has an open neighbourhood also contained in P(X). Hence, P(X) is open in Y. Since Y is connected it follows that P(X) = Y.

Now, by definition, K ⊂ Image(P) and Lemma 2 tells us that C \ K Image(P). Hence, Image(P) = C The proof of the theorem is complete.

The crucial idea in this proof is that the plane remains connected after removing finitely many points. All the other results used hold for polynomials from R to R as well.

Our approach to the fundamental theorem of algebra is similar to arguments used in [1] to investigate proper, smooth maps with non-negative Jacobian between connected, orientable manifolds.

### REFERENCE

1. A. Nijenhuis and R. W. Richardson, Jr., A theorem on maps with non-negative jacobians, Michigan Math. J. 9 (1962) 173—176.