Give a Galois-theoretic proof of the fundamental theorem of algebra. What input from real analysis is needed?
Solution
We prove that \(\CC\) has no nontrivial finite algebraic extension.
The analytic input is:
- every real polynomial of odd degree has a real root, by the intermediate value theorem;
- every positive real number has a real square root.
The second fact implies that every complex number has a complex square root.
1. The real field has no nontrivial finite extension of odd degree.
Let \(E/\RR\) be finite of odd degree. Since the extension is separable, the primitive element theorem gives \(E=\RR(\alpha)\). The minimal polynomial of \(\alpha\) has odd degree, hence has a real root. Being irreducible over \(\RR\), it must therefore be linear. Thus \(E=\RR\).
2. The complex field has no quadratic extension.
In characteristic different from \(2\), every quadratic extension has the form \(\CC(\sqrt a)\) for some \(a\in\CC^\times\). But every complex number has a square root in \(\CC\), so such an extension is trivial.
3. Let \(L/\RR\) be a finite Galois extension containing \(\CC\). Then \(\Gal(L/\CC)\) is trivial.
Put \[ G=\Gal(L/\RR), \qquad H=\Gal(L/\CC). \] Since \([\CC:\RR]=2\), the subgroup \(H\) has index \(2\) in \(G\).
By 1, \(|G|\) has no odd prime divisor: if an odd prime \(p\) divided \(|G|\), Cauchy’s theorem would give a subgroup of order \(p\), whose fixed field would yield a nontrivial odd-degree finite extension at the corresponding step. Hence \(G\) is a \(2\)-group, and so is \(H\).
2. Therefore \(H=1\), so \(L=\CC\).
4. Every nonconstant polynomial in \(\CC[x]\) has a root in \(\CC\).
Let \(f\in\CC[x]\) be nonconstant and let \(L\) be the Galois closure over \(\RR\) of a splitting field of \(f\). Then \(L/\RR\) is finite Galois and contains \(\CC\). By 3, \(L=\CC\). Hence all roots of \(f\) already lie in \(\CC\).
Thus \(\CC\) is algebraically closed.