Theorem 11.3.1. Fundamental Theorem of Algebra.
The field of complex numbers is algebraically closed; that is, every polynomial in \({\mathbb C}[x]\) has a root in \({\mathbb C}\text{.}\)
Section 11.3 The Fundamental Theorem of Algebra
It seems fitting that the last theorem that we will state and prove is the Fundamental Theorem of Algebra. This theorem was first proven by Gauss in his doctoral thesis. Prior to Gauss’s proof, mathematicians suspected that there might exist polynomials over the real and complex numbers having no solutions. The Fundamental Theorem of Algebra states that every polynomial over the complex numbers factors into distinct linear factors.
Proof.
Suppose that \(E\) is a proper finite field extension of the complex numbers. Since any finite extension of a field of characteristic zero is a simple extension, there exists an \(\alpha \in E\) such that \(E = {\mathbb C}( \alpha )\) with \(\alpha\) the root of an irreducible polynomial \(f(x)\) in \({\mathbb C}[x]\text{.}\) The splitting field \(L\) of \(f(x)\) is a finite normal separable extension of \({\mathbb C}\) that contains \(E\text{.}\) We must show that it is impossible for \(L\) to be a proper extension of \({\mathbb C}\text{.}\)
Suppose that \(L\) is a proper extension of \({\mathbb C}\text{.}\) Since \(L\) is the splitting field of \(f(x) = (x^2 + 1)\) over \({\mathbb R}\text{,}\) \(L\) is a finite normal separable extension of \({\mathbb R}\text{.}\) Let \(K\) be the fixed field of a Sylow \(2\)-subgroup \(G\) of \(G(L/{\mathbb R})\text{.}\) Then \(L \supset K \supset {\mathbb R}\) and \(|G( L / K )| =[L:K]\text{.}\) Since \([L : {\mathbb R}] = [L:K][K:{\mathbb R}]\text{,}\) we know that \([K:{\mathbb R}]\) must be odd. Consequently, \(K = {\mathbb R}(\beta)\) with \(\beta\) having a minimal polynomial \(f(x)\) of odd degree. Therefore, \(K = {\mathbb R}\text{.}\)
We now know that \(G(L/{\mathbb R})\) must be a \(2\)-group. It follows that \(G(L / {\mathbb C})\) is a \(2\)-group. We have assumed that \(L \neq {\mathbb C}\text{;}\) therefore, \(|G(L / {\mathbb C})| \geq 2\text{.}\) By the first Sylow Theorem and the Fundamental Theorem of Galois Theory, there exists a subgroup \(G\) of \(G(L/{\mathbb C})\) of index \(2\) and a field \(E\) fixed elementwise by \(G\text{.}\) Then \([E:{\mathbb C}] = 2\) and there exists an element \(\gamma \in E\) with minimal polynomial \(x^2 + b x + c\) in \({\mathbb C}[x]\text{.}\) This polynomial has roots \(( - b \pm \sqrt{b^2 - 4c}\, ) / 2\) that are in \({\mathbb C}\text{,}\) since \(b^2 - 4 c\) is in \({\mathbb C}\text{.}\) This is impossible; hence, \(L = {\mathbb C}\text{.}\)
Although our proof was strictly algebraic, we were forced to rely on results from calculus. It is necessary to assume the completeness axiom from analysis to show that every polynomial of odd degree has a real root and that every positive real number has a square root. It seems that there is no possible way to avoid this difficulty and formulate a purely algebraic argument. It is somewhat amazing that there are several elegant proofs of the Fundamental Theorem of Algebra that use complex analysis. It is also interesting to note that we can obtain a proof of such an important theorem from two very different fields of mathematics.
