30 Nov Fundamental Theorem of Algebra using Galois theory
In this article we shall prove that is algebraically closed. Here we consider as a splitting field of the polynomial .
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 has a square root in . For let
We choose the sign of and such that has the same sign as b. Then we can see that .
Now we show that every polynomial splits in . This is equivalent to showing that the splitting field of over is . Let be the splitting field of over for a fixed . As has characteristic zero, is a separable polynomial and hence is a Galois extension. Let .
Let be a 2-Sylow subgroup of . Let , the subfield of E fixed by . Then degree of the extension is which is odd. Hence, for any it's minimal polynomial over has odd degree. But as any odd degree polynomial has a root in , . Thus, and hence is a 2-group.
If it has a subgroup of index . Then degree of over is 2 and hence is generated by the squareroot of an element in . But as has all its square roots we get that . Hence,
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