 Solution

ID: past-exam-of-the-number-theory-course-of-the-university-of-oxford/2011/a/1/b/iii/solution

F_{7}

x^{3} + x^{2} + x + 1 = (x + 1) (x^{2} + 1) .

7

are

1, 2, 4

, so

- 1

is not a square. Adjoining a root of

x^{2} + 1

produces

F_{49}

and already gives both roots. Its Frobenius automorphism has order

2

and interchanges them, whence

Gal (f) ≅ C_{2} .

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