Answer:
To find a splitting field for f, we first factor f completely over Q:
f = (2x^2 − 4x + 1)(x^4 + 1) = 2(x - 1/2)^2(x^4 + 1)
The roots of the quadratic factor are x = 1/2 ± 1/2√2, which are real, and the roots of the quartic factor are x = ±i. Therefore, a splitting field for f is Q(i, √2).
(a) Q(√2): The roots of the quadratic factor are real, so they are already in Q(√2). The roots of the quartic factor are ±i, which are not in Q(√2). Therefore, the Galois group of f over Q(√2) is isomorphic to Z/2Z.
(b) Q(2): The roots of the quadratic factor are not in Q(2). However, both i and √2 are in Q(2), so Q(2) contains a splitting field for f. Therefore, the Galois group of f over Q(2) is trivial, since there is only one automorphism of Q(2) that fixes Q.
(c) Q(i): The roots of the quadratic factor are not in Q(i). However, √2 is in Q(i), so Q(i) contains a splitting field for f. The Galois group of f over Q(i) is isomorphic to Z/2Z × Z/2Z, since there are two complex roots and two real roots, and the complex roots are conjugates of each other.
(d) Q(ζ): The roots of the quadratic factor are not in Q(ζ). However, both i and √2 are in Q(ζ), since ζ^4 = i and ζ^8 = √2. Therefore, Q(ζ) contains a splitting field for f. The Galois group of f over Q(ζ) is isomorphic to Z/2Z × Z/2Z, since there are two complex roots and two real roots, and the complex roots are conjugates of each other.
(e) R: The roots of the quadratic factor are real, so they are already in R. The roots of the quartic factor are ±i, which are not in R. Therefore, the Galois group of f over R is isomorphic to Z/2Z.