caledonia Posted May 25, 2015 Share Posted May 25, 2015 I have seen asserted that for polynomials f and h, f divides h implies that fbar divides hbar, where "bar" indicates reduction modulo p, a prime number say. But when I test this with sample polynomials, it does not seem to be true. For example modulo 3 and f = 2x + 1, h = 6x2 + 7x + 2, then fbar = 2x + 1, hbar = x + 2. What is wrong here please ? Link to comment Share on other sites More sharing options...
mathematic Posted May 25, 2015 Share Posted May 25, 2015 The quotient of h over f is 3x+2 = 2(modulo 3). 2fbar=4x+2=x+2(modulo 3), which is hbar. Link to comment Share on other sites More sharing options...
caledonia Posted May 27, 2015 Author Share Posted May 27, 2015 Thank you. But how about this case (mod 3) : f = 11x + 2, g = 5x + 7, then fg = 55x2 + 87x + 14, fbar = 2x + 2, gbar = 2x + 1, (fg)bar = x2 + 2. How can fbar (or gbar) divide (fg)bar ? Link to comment Share on other sites More sharing options...
mathematic Posted May 27, 2015 Share Posted May 27, 2015 fbar x gbar = 4x2+6x+2=x2+2 (mod 3) I don't know why you had a problem with this. Link to comment Share on other sites More sharing options...
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