Question

1.13 Show that the elements 4 and 2(1+V-3) in Z[V-3] have no g.cd
Please explain steps taken and why

Answer 1

Let R = Z(squareroot B) = {a + b squareroot - 3/a, b epsilon Z} with norm N asN(a + b squareroot -3) = a^2 + 3b^2 Note: - R = z (squareroot B) has only units plusminus 1 suppose d = gcd (x, y) exists then we have d/x and d/y therefore x = da y = db for some a, b epsilon R for alpha = m + squareroot - 3 n epsilon R N(alpha) = m^2 + 3n^2 which is non negative integer Now N(x) = N(d) N(a) and N(y) = n(d) N(b) 4.4 = N(d) N(a) and 4.4 = N(d) N(b) implies N(d)/16 On the other hand, 2/x and 2/y therefore 2/d implies d = 2d' therefore N(d) = 4 N(d') implies 4/N(d) similarly 1 + squareroot -3 is also common divisor of x and y we have 4/N(d) Thus N(d) = 4, 8, 16 But a^2 + 3 b^2 = 8 has not solution is Z since a^2 + 3b^2 = 4 or 16