I'm studying for an exam in rings and fields and I took out this page of my notes to study it the other day but I must have left in in the library. grr I can seam to find the proof of this anywhere online of construct it myself. If anyone could help It would be greatly appreciated. I am almost sure this will be on the exam.
Hey thanks I just saw that I have to use the result of
w=exp(2*Pi*i/7) and A=w + w^6
Finding the minimal polynomial or w over Q.
I know this is the polynomial Q[x] of least degree which A is a root.
Im cant really get any further than that.
(pity cos the proof using Fermat primes is so much nicer )
Solution 2: Here we are assuming knowledge of Galois theory. Let . Since it means the Galois groups is cyclic. Let by the Galois group so that for . It is simple to check that generates the Galois group. Consider the polynomial . Note that leaves the polynomial fixed, therefore fixes and therefore the coefficients of lie in . This means . Since degree of cannot be over and it solves it means the degree of is a divisor of three i.e. it must be three. Thus, the minimal polynomial is . Note that this not a power of two therefore the number is not constructible.