Section 50, Number 8 50.9 Example Let 2 be the real cube root of 2, as usual. Now x - 2 does not split in Q(/2), forQV2) < R and only one zero of x³ – 2 is real. Thus x³ – 2 factors in (Q(/2))[x] intoV2 and an irreducible quadratic factor. The splitting field E of x³ – 2a linear factor xover Q is therefore of degree 2 over Q(V2). Then[E: Q] = [E : Q(2)][Q(/2) : Q] = (2)(3) = 6.We have shown that the splitting field over Q of x-2 is of degree 6 over Q.We can verify by cubing thatandare the other zeros of x'-2 in C. Thus the splitting field E of x-2 over Q isQ/2, i/3). (This is not the same field as Q(2, i, V3), which is of degree 12 over Q.)Further study of this interesting example is left to the exercises (see Exercises 7, 8, 9,16, 21, and 23).| 8. Refer to Example 50.9 for Exercise 8.Exercise 8: What is the order of G(Q(VZ, iv3)/Q)?

Name: Section 50, Number 8 50.9 Example Let 2 be the real cube root of 2, as
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Description: Section 50, Number 8 50.9 Example Let 2 be the real cube root of 2, as usual. Now x - 2 does not split in Q(/2), forQV2) < R and only one zero of x³ – 2 is real. Thus x³ – 2 factors in (Q(/2))[x] intoV2 and an irreducible quadratic factor. The splitt