caledonia Posted May 25, 2015 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 ?
mathematic Posted May 25, 2015 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.
caledonia Posted May 27, 2015 Author 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 ?
mathematic Posted May 27, 2015 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.
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