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Fundamental theorem of algebra by the fundamental-group obstruction

Statement

Every nonconstant complex polynomial has a complex root.

Facts & Assumptions

Given: A nonconstant complex polynomial $p$ .

[F1]

A nonzero polynomial has a degree and a nonzero leading coefficient, and it is monic exactly when its leading coefficient is $1$ (Formal complex polynomials, evaluation, degree, leading coefficient, and monic polynomials).

[L1]

If a complex polynomial has no zero, then every normalized circle loop obtained from it is nullhomotopic (A root-free complex polynomial gives nullhomotopic normalized circle loops).

[L2]

For a monic complex polynomial of positive degree $n$ , every radius satisfying the strict leading-term bound gives a normalized circle loop of degree $n$ (The normalized large-radius loop of a monic degree- $n$ polynomial has degree $n$ ).

[L3]

A based circle loop is nullhomotopic exactly when its degree is zero (A based circle loop is nullhomotopic exactly when its degree is zero).

Proof

technique · contradiction

1.1

Suppose $p$ has no root. Since $p$ is nonconstant, it is nonzero and has degree $n \geq 1$ and leading coefficient $c \neq 0$ . Dividing every coefficient by $c$ gives a monic polynomial $q = c^{- 1} p$ of the same degree and with the same zero set, so $q$ is also root-free.

givenF1assume-contraalgebra

2.1

By [L1], every normalized radius- $R$ loop of $q$ is nullhomotopic, and therefore has degree zero by [L3].

step 1.1L1L3

2.2

Write $q (z) = z^{n} + \sum_{j < n} a_{j} z^{j}$ , put $S = \sum_{j < n} ∣ a_{j} ∣$ , and take $R = \max {1, S} + 1$ . Then $R > \max {1, S}$ , so [L2] says that the normalized radius- $R$ loop has degree $n$ .

step 1.1L2choose

3.1

Steps 2.1 and 2.2 assign the same loop both degree $0$ and degree $n$ , impossible because $n \geq 1$ . Hence the root-free assumption is false and $p$ has a complex root.

step 2.1step 2.2algebradischarge-contradiction∎

Depends on

Used by

The fundamental-group and minimum-modulus proofs of the fundamental theorem of algebra Remark

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, Theorem 1.8 (standard reference, not scraped)
J. Peter May, A Concise Course in Algebraic Topology, Chapter 1, §7 (standard reference, not scraped)