6.3 Splitting Fields

1.
a. x3 + 1 = (x + 1)(x2x + 1), and the roots of x2x + 1 are img and img so img But img so it follows that img Since img is a root of the polynomial x2 + 3, and since x3 + 3 is irreducible over img (no root), then it is the minimal polynomial of img Hence img
c. f = (x2 − 7)(x2 + 1) so img. Thus,

img

Moreover, because x2 − 2 is irreducible over and because x2 + 1 has no root in Hence by the multiplication ...

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