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1.6. Quadratic field extensions 19 Proof. It is clear that F ⊆ F ( √ a), so what has to be shown is that F ( √ a) is a field. But note that if x1,x2,y1,y2 ∈ F , then (x1 + y1 √ a) (x2 + y2 √ a) = (x1x2 + y1y2a) + (x1y2 + x2y1) √ a ∈ F ( √ a). Similar arithmetic shows that the sum and difference of any two elements of F ( √ a) belongs to F ( √ a). Also, assuming x2 + y2 √ a = 0, x1 + y1 √ a x2 + y2 √ a = x1 + y1 √ a x2 + y2 √ a · x2 − y2 √ a x2 − y2 √ a = (x1x2 − y1y2a) x2 2 + y2 2 + (−x1y2 + x2y1)√ x2 2 + y2 2 a ∈ F ( √ a), completing the proof that F ( √ a) is a field. Of course if √ a ∈ F , then F ( √ a) = F . When equality fails, we will say that F ( √ a) is a proper extension of F . Also note that we could iterate the above process. For example, suppose a ∈ F , a 0, and √ a / F . So we form the proper extension F ( √ a). Now suppose b ∈ F ( √ a), b 0, and √ b / F ( √ a). Therefore we can form the proper extension F ( √ a)( √ b) of F ( a), which is also an extension of F . This leads us to the following definition. 1.6.4. Definition. If F is a field and K is an extension of F , say that K is a simple quadratic extension of F if there is a positive number a in F such that √ a / F and K = F ( √ a). Say that a field F is a quadratic extension of Q if there are fields F0,...,Fn such that: (a) F0 = Q and Fn = F (b) for 1 ≤ k ≤ n, Fk is a simple quadratic extension of Fk−1 Say that F is a quadratic extension of Q of degree n if there are n + 1 fields F0,F1,...,Fn as just described and no such chain of simple quadratic extensions can be found with fewer than n + 1 fields. It is not difficult to show that the intersection of any collection of fields contained in R is also a field (Exercise 5). So if we are given two fields F and K, we can talk about the field generated by F and K as the smallest field containing F ∪ K, that is, the field {L : L is a field and L ⊇ F ∪ K}. If F and K are quadratic extensions of Q, then it is true but not so clear that the field they generate is also a quadratic extension of Q. 1.6.5. Proposition. If F and K are quadratic extensions of Q of degrees n and m, respectively, then the field generated by F and K is a quadratic extension of Q of degree at most n + m. More generally, if F1,...,Fn are quadratic extensions of Q, then the field generated by F1,...,Fn is a qua- dratic extension of Q.