\f\fPROOF: We already know that every finite extension is an iterated extension. We will now show that any extension F(a, b) is equal to F(c) for some c. Using this result several times in succession yields our theorem. (At each stage, we reduce by 1 the number of elements that must be adjoined to F in order to get the desired extension.) Well, given F(a, b), let A(x) be the minimum polynomial of a over F, and let B(x) be the minimum polynomial of b over F. Let K denote any extension of F which contains all the roots al, . . . , an of A(x) as well as all the roots b1,...,bmofB(x) Let al be a and let b1 be b. Let t be any nonzero element of F such that a. -- a b'-b for everyisl andjal i Cross multiplying and setting c = a + tb, it follows that c % at. + (bj, that is, {54 c-tbjaa, forallislandjal Define h(x) by letting h(x) = A(c - tx); then h(b) = A(c - tb) = 0 x-w-J while for every j =?'= 1. h(b,) = A(c - tbj) si 0 H-l-NI-FJ a any at "nm, b is me only common mat of h(x) and B(x). we will wow that be F(c), hence also a = c - tb E F(c), and there- fore mg, mg F(c). But ce F(a, b), so Mr) g F(a, b). Thus F(a, b) = So. u . A my prove that b 6 He). Let p(x) be the minimum tam - , ' umadggrceofpu) is 1, then p(x) is x - b, so His" 2 deg p(x) 2 2 and get a an" Int nitil uttar); mum be multiples of p(x) l k ' lair: fi'a'i'iil u um polynomial of b . urnui nr 'lzltnoh ch 22, they must havC ' "'5'Wlt'l Hawai-{willi . ' '4 that b IS the" only , ;, an): iuldqulluillli?'