Question

G. Shorter Questions Relating to Automorphisms and Galois Groups Let F be a field, and K a finite extension of F. Suppose a, b E K. Prove parts 1-3: 1 If an automorphism h of K fixes Fand a, then h fixes F(a). 3 Aside from the identity function, there are no a-fixing automorphisms of a(). [HINT: Note that aV2 contains only real numbers.] 4 Explain why the conclusion of part 3 does not contradict Theorem 1.

Answer 1

1. F(a) is a finite extension of F, being a subfield of K . Hence each element of F(a) is a polynomial in a over F and hence any automorphism fixing F and a fixes all polynomial expressions of a also. Thus F(a) is fixed under such automorphisms. 2. Each ge F(a, b)* fixes F, a and b. Hence g є F(a)' and g є F(b). Conversely, g є F(a)in F(b)* g E F(a, b) 3. Consider K-Q(3) where B-21/3 . K contains only real numbers. Hence any automorphism fixing Qand 3 fixes K also. 4. In (1) K is a finite extension of F, but in(3) K is an extension of Q of infinite order. Hence there is no contradiction g fixes F, a and b . Hence 1/3