Question

6. (16 points) Let CE C be a primitive n-th root of unity. Let X = 6 + 1/5. (a) (4 points) Show that Q(5) R = Q(1). (b) (4 points) Let f be the minimal polynomial of over Q. Show that Q(x) is a splitting field of f over Q. (c) (4 points) Show that Gal(Q(^)/Q) – (Z/nZ)* / (-1). (d) (4 points) Find the minimal polynomial of 2 cos(27/9) over Q.

Answer 1

Solution : Let, 0 9=5+111. a) It is immediate that monterat Q() SQ (I)R. Conversely Consider any element REQ (5) DR, there are ao -- amta, such that, x= dota, Zt - tams.then (because = 5") => M=Xtī -(ao ta, 37-tam sm) ... + (asta, s t - tams So that, x= 20 + (5+ 7) + - +49 (1945) -do + 1 (1+ (%) + -- + am ( JM + 5") But, by Symmetric polynomials) Every 3m + 5 is a polynomial of 3+5=2. Thus, a cold). This finishes the proof.

b) we consider the Galois Extension Q(2) cQ (3) The minimal polynomial & Go Eoloro is g(0) =*-3x+1 So that, (2) cQ (5) is degree 2. Any degree 2 Extension is normal; . By fundamental theorom of Galois theory, Q CQ (2) is Galois . In particular a(a) is splitting field of f(x). c) As mentioned in (b) n« Gal. (003)/0) (zhoz)x Gallec3/9cm)) t} C atila do that, d) Let 5=e Note that he facos = 4 cos 27/9 - 3008 29

-473-32 58 B-62+1=0 Thus, if g (W = 8x² - 6x +1 then the minimal Polynomial of dis a factor of it. But only possible rational roots of g(x) = 823-6*+1 are F1, I llei and these are not roots of g(x)=883884 (by inspection). Thus, g(x) = 82² -6Xt1 is voeducible It is the minimal polynomial. Hence,