nphw7 - p | m 2 p | m . . . ( ll in the blanks !) p | n...

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Numbers and Polynomials Homework 7 Prove that if p and q are distinct prime numbers then Q [ p ] T Q [ q ] = Q . Solution Of course, the indicated intersection contains all rational numbers. For the reverse containment, let t Q [ p ] T Q [ q ]: say t = a + b p = c + d q for rational numbers a, b, c, d ; we shall show b = d = 0 and a = c whence t (= a = c ) Q as required. Rearrangement yields b p - d q = c - a Q whence squaring places ( b 2 p + d 2 q - 2 bd pq and consequently) bd pq in Q . If bd 6 = 0 then pq Q : say pq = m/n with m, n coprime integers; now m 2 = pqn 2
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Unformatted text preview: p | m 2 p | m . . . ( ll in the blanks !) p | n contadicts m, n being coprime. This proves that either b or d ( or both ) is (are) zero. However, if b 6 = 0 = d then b p-d q Q forces p Q ; this runs contrary to established fact, as does b = 0 6 = d . Conclusion: b = d = 0 and we are nished. Question : Where was the hypothesis p 6 = q used? 1...
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