The Fundamental Theorem of Algebra by Degree

The Degree Contradiction

The Fundamental Theorem of Algebra says that every nonconstant polynomial with complex coefficients has a complex zero. Here the proof does not locate a zero or count all the zeros. Instead, it shows that assuming there are no zeros leads to two incompatible calculations of the degree of one circle map. On a sufficiently large circle, the polynomial’s normalized values have positive degree. But if the polynomial has no zero inside the circle, those same values extend over the disk, which forces the boundary degree to be zero.

We use \(\mathbb{C}\) with its usual topology, \(S^1=\{z\in\mathbb{C}:|z|=1\}\) with the subspace topology, and the closed disk \(D_R=\{z\in\mathbb{C}:|z|\le R\}\). Its boundary circle is \(C_R=\{z\in\mathbb{C}:|z|=R\}\). A map \(g:C_R\to S^1\) is a normalized boundary map for a polynomial \(p\) when \(p\) has no zeros on \(C_R\), and \(g(z)=p(z)/|p(z)|\) there.

In particular, the map \(e^{2\pi i t}\mapsto e^{2\pi i n t}\) has degree \(n\): its lift is \(\theta(t)=nt\), whose endpoint difference is \(n\). Multiplying its values by a fixed element of \(S^1\) does not change the degree, since this only adds a constant to a lift.

Why Large Circles Have Positive Degree

Proof. Write \(Q(z)=\sum_{k=0}^{n-1}a_kz^k\) and \(C=\sum_{k=0}^{n-1}|a_k|\). If \(C=0\), then \(p(z)=a_nz^n\), which is nonzero on every circle of positive radius, and its normalized boundary map has degree \(n\) by the power-map calculation above. Suppose \(C>0\). Choose \(R\ge1\) with \(R>C/|a_n|\). For \(|z|=R\), the triangle inequality gives \[ |Q(z)|\le\sum_{k=0}^{n-1}|a_k|R^k\le CR^{n-1}<|a_n|R^n. \] For \(0\le s\le1\), define \(P_s(z)=a_nz^n+sQ(z)\). On \(C_R\), \[ |P_s(z)|\ge |a_n|R^n-s|Q(z)|\ge |a_n|R^n-|Q(z)|>0. \] Thus \(P_s(z)/|P_s(z)|\) gives a homotopy of maps from \(C_R\) to \(S^1\), from the normalized leading monomial at \(s=0\) to the normalized polynomial at \(s=1\). Homotopy invariance makes their degrees equal. The normalized leading monomial on \(z=R e^{2\pi i t}\) is \((a_n/|a_n|)e^{2\pi i n t}\), which has degree \(n\). This also proves that \(p\) has no zeros on \(C_R\), as required for its normalized boundary map to be defined. \(\square\)

The strict inequality is important: it prevents the leading term and the lower-order terms from cancelling on the chosen circle. The radius is allowed to depend on the polynomial; the argument does not claim that every circle has degree \(n\).

An Extension over the Disk Forces Degree Zero

Proof. Parametrize \(C_R\) by \(R e^{2\pi i t}\). For \(0\le s\le1\), define a circle map \(G_s:S^1\to S^1\) by \(G_s(e^{2\pi i t})=f(sR e^{2\pi i t})\). This is a continuous homotopy: the point \(sR e^{2\pi i t}\) always belongs to \(D_R\), and at \(s=1\) the map is the boundary restriction. At \(s=0\), \(G_0\) is the constant map with value \(f(0)\). A lift of a constant map can be chosen constant, so its degree is \(0\). Homotopy invariance gives \(\deg(f|_{C_R})=\deg(G_0)=0\). \(\square\)

The lemma requires an extension to the entire disk, not merely a map on its boundary. For a polynomial with no zeros on \(D_R\), the formula \(f(z)=p(z)/|p(z)|\) supplies exactly such an extension: the denominator is nonzero everywhere on the disk.

The Fundamental Theorem of Algebra

Proof. Let \(p(z)=a_nz^n+\cdots+a_0\), where \(a_n\ne0\) and \(n\ge1\). Choose \(R\) large enough for the proposition above. Its normalized boundary map \(g:C_R\to S^1\) is defined and has degree \(n\). Suppose, for contradiction, that \(p\) has no zero anywhere in \(\mathbb{C}\). Then \(p(z)/|p(z)|\) is defined and continuous on all of \(D_R\), and it extends \(g\). The extension lemma forces \(\deg(g)=0\), while the proposition gives \(\deg(g)=n\). Since \(n\ge1\), these equalities contradict one another. Therefore \(p\) has a complex zero. \(\square\)

A Common Mistake and the Role of the Assumption

Common pitfall. It is incorrect to claim that the normalized boundary map of every degree-\(n\) polynomial has degree \(n\) on every circle. The calculation for \(p(z)=z-2\) on the unit circle gives degree \(0\), not \(1\). The correction is to choose a circle on which the leading term strictly dominates the sum of the lower-order terms; the proposition proves that such circles exist with arbitrarily large radius.

There is a second important distinction. Nonvanishing on the boundary alone does not give an extension of the normalized map over the disk. For \(p(z)=z\), the boundary map on the unit circle is \(z\mapsto z\), which has degree \(1\). But the proposed extension \(z\mapsto z/|z|\) is undefined at \(z=0\), where \(p\) vanishes. In the theorem’s contradiction, the assumption is that the polynomial has no zero anywhere; that assumption makes normalization continuous on the whole disk and allows the extension lemma to apply.

The proof therefore uses two separate facts with different hypotheses: large-radius dominance calculates the boundary degree, while absence of zeros throughout the disk supplies the extension. Their degrees conflict only when both facts refer to the same circle and the same boundary map.