Advanced Algebra I
Homework 14 due on Jan. 9, 2004
(1) If F is a finite extension of Q, then F has only a finite number of roots of unity.
(2) Which roots of unity are contained in Q(√
d), where d is a square-free integer?
(3) Let F/K be a finite extension over a finite field K. Show that the norm and trace maps are surjective.
(4) Examine that the property of being radical under ”extension”,”lifting”, and ”compositum”.
(5) Find a polynomial which is not solvable by radicals. And ex- plain why.
Part of answer to the problem 4:
Let K ⊂ E ⊂ F be field extensions. We are going to give an example that F/K is radical but E/K is not radical.
Let F0 be the splitting field of f (x) = x3− x − 1 over Q. It’s clear that the Galois group Gf is solvable. Hence F0 is contained in a radical extension F/Q.
Let u ∈ F0 be a root of f (x). We claim that E := Q(u)/Q is not radical. Suppose on the other hand that Q(u)/Q is radical, that is Q(u) = Q(v1, ..., vn). However, [Q(u) : Q] = 3, we can therefore assume that Q(u) = Q(v) for some v3 = d ∈ Q.
Thus u = av2 + bv + c. Since T (u) = T (v) = T (v2) = 0, we have T (c) = 0, hence c = 0. We can write u = av2 + bv. By direct computation,
u3− u − 1 = a(1 + 3abd)v2+ b(1 + 3abd)v + a3d2+ b3d + 1 = 0.
One has
a(1 + 3abd) = 0 b(1 + 3abd) = 0 a3d2+ b3d + 1 = 0
If 3abd + 1 = 0, then a2+ 9ab2− 3b4 = 0 has no non-zero solution in Q. If 3abd+1 6= 0, then a = b = 0, which is also impossible. Therefore, Q(u) can’t be a radical extension over Q.
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