Referring to Example 50.9, show that G(Q(³√2, i√3)/Q(i√3)) ≈ (Z₃, +).

Data from in Example 50.9

Transcribed Image Text:

- Let 2 be the real cube root of 2, as usual. Now x³ 2 does not split in Q(V2), for Q(³2) < R and only one zero of x³ - 2 is real. Thus x³ - 2 factors in (Q(2)) [x] into a linear factor x 2 and an irreducible quadratic factor. The splitting field E of x³ - 2 over Q is therefore of degree 2 over Q(2). Then [E : Q] = [E : Q(√2)][Q(√2) : Q] = (2)(3) = : 6. We have shown that the splitting field over Q of x³ - 2 is of degree 6 over Q. We can verify by cubing that √√2 √2-1 are the other zeros of x³ 2 in C. Thus the splitting field E of x³ 2 over Qis Q(√2, i√3). (This is not the same field as Q(√2, i, √√3), which is of degree 12 over Q.) ·1+i√3 2 and -1-i√3 2