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

c=\pm\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}

d=\pm\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}

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

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Bishal Deb

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

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