00:01
Hello students, given that characteristic of f is equals to 0 and suppose that f belongs to f of t is irreducible of prime degree and k by f is the splitting field of f, then f is solvable by radicals if and only if k equals to f of r i comma r j for any two roots r i comma r j of f belongs to k.
00:23
This is not true in general.
00:25
That's why we take an counter example.
00:28
So, let us consider a polynomial.
00:34
Let us consider a polynomial f x equals to x to the power 5 minus 2 which is belongs to q x.
00:51
This polynomial, this is irreducible by einstein criteria with the prime number p equals to 2.
01:32
Hence, the splitting field of f, the splitting field of f over q is given by k is equals to q of alpha comma omega where alpha is the root of f and omega is the primitive fifth root of unity.
02:20
So, the galois group of f, the galois group of f over q is isomorphic to the is isomorphic to the symmetric group s 5 which is not a solvable group which is not a solvable group...