Separable Extensions, The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra

The field of complex numbers, denoted C, is algebraically closed. Every polynomial with complex coefficients has a complex root, and if we extract roots one by one, the entire polynomial splits. This is the fundamental theorem of algebra.

You've probably seen the proof based on analytic functions, but here is another, based on separable fields and galois theory.

The intermediate value theorem provides a positive square root for every positive real number, and a root to any odd degree polynomial in the reals, as x moves from -∞ to +∞. Therefore every irreducible polynomial in the reals has even degree.

The existance of real square roots implies a complex square root for z = a+bi. Let r be the radial distance from z to the origin, i.e. sqrt(a²+b²). Define y as follows and verify that y² = 2z.

y = sqrt(r+a) + sqrt(r-a)i

Remember that r ≥ a, so y is well defined. Divide y by the square root of 2 and find a square root for z. Thus there is no extension of C with dimension 2.

To show C is closed it is sufficient to show there are no finite dimensional extensions of C.

If E is such an extension it is also a finite separable extension over R. Separable, because the characteristic of R is 0. Take the normal closure to get a galois extension F.

Since C/R has dimension 2, the galois group G, for F over R, has even order. The first sylow theorem provides a subgroup H whose order is a power of 2, while the index of H is odd.

Let L be the extension of R that is fixed by H. This takes place inside a galois extension, so the dimension of L/R is the index of H in G, which is odd. Since L/R is finite and separable, apply the primitive element theorem and write L = R(u). The irreducible polynomial associated with u has odd degree, yet an odd degree polynomial has a root in R. The polynomial has degree 1, and L = R. Thus G = H, and |G| is a power of 2.

If G = Z₂ then F = C. For larger G, F properly contains C, and F/C is galois. Let G be the galois group of F/C. Let H be a subgroup of G with index 2. The field fixed by H is a quadratic extension of C, which is impossible. Therefore C is closed.

One Extension of the Reals

Let F be a proper extension of R. Let u be an element in F-R and let p(x) have root u. We know that p is irreducible in R, yet it splits in C. Let v be a root of p(x) taken from the extension C/R. Build an isomorphism from R(u) onto R(v). Since v is outside of R, R(v) = C. Thus we have an isomorphism from R(u) onto all of C. Now R(u) is closed, and F is algebraic over R(u), hence F = R(u), is isomorphic to C. The only extension of the reals is the complex numbers.

As a corollary, the only irreducible polynomials over the reals are quadratic. Since C is separable, the two roots are distinct. You know them better as conjugates in the complex plane.