Fundamental Theorem of Algebra using Galois theory

In this article we shall prove that \mathbb{C} is algebraically closed. Here we consider \mathbb{C} as a splitting field of the polynomial x^2+1.

The proof uses very little analysis and most of it is Galois theory. The only facts from analysis which will be used in proof are :

  • Positive real numbers have square roots.
  • Every polynomial of odd degree with real coefficients has a real root.

Both of these facts are consequences of the intermediate value theorem.

We first show that every element of \mathbb{C} has a square root in \mathbb{C}. For a,b\in\mathbb{R} let



We choose the sign of c and d such that cd has the same sign as b. Then we can see that (c+\iota d)^2=a+\iota b.

Now we show that every polynomial f(x)\in \mathbb{R}[x] splits in \mathbb{C}. This is equivalent to showing that the splitting field of f(x)(x^2+1) over \mathbb{R} is \mathbb{C}. Let E be the splitting field of f(x)(x^2+1) over \mathbb{R} for a fixed f. As \mathbb{R} has characteristic zero, f(x)(x^2+1) is a separable polynomial and hence E/\mathbb{R} is a Galois extension. Let G=Gal(E/\mathbb{R}).

Let H be a 2-Sylow subgroup of G. Let M=E^H, the subfield of E fixed by H. Then degree of the extension M/\mathbb{R} is |G/H| which is odd. Hence, for any \alpha\in M it's minimal polynomial over \mathbb{R} has odd degree. But as any odd degree polynomial has a root in \mathbb{R}, \alpha\in\mathbb{R}. Thus, M=\mathbb{R} and hence G is a 2-group.

If G\neq (1) it has a subgroup N of index 2. Then degree of E^N over C is 2 and hence is generated by the squareroot of an element in \mathbb{C}. But as \mathbb{C} has all its square roots we get that G=(1). Hence, E=\mathbb{R}


Source : Field and Galois Theory Notes, James Milne

Print Friendly, PDF & Email
The following two tabs change content below.

Bishal Deb

Bishal Deb is an undergraduate student at Chennai Mathematical Institute. He is an editor of Gonit Sora.

Latest posts by Bishal Deb (see all)

READ:   AHSEC 2016 Question Papers
No Comments

Post A Comment