Answer to Question #282132 in Abstract Algebra for Chinmoy kumar Bera

Question

cosider the ideal I={x^2-4x+3,x^3+3x^2-x-3} of the ring .find polynomial p such that I=<p>

Answer 1

Solution:

$I=\{x^2-4x+3,x^3+3x^2-x-3\} \\=\{(x-1)(x-3),(x+3)(x^2-1)\} \\=\{(x-1)(x-3),(x+3)(x-1)(x+1)\} \\=\{(x-1)(x-3)f(x)+(x+3)(x-1)(x+1)q(x):f(x),q(x)\in Z[x]\} \\=\{(x-1)[(x-3)f(x)+(x+3)(x+1)q(x)]:f(x),q(x)\in Z[x]\} \\=\{(x-1)g(x):g(x)=(x-3)f(x)+(x+3)(x+1)q(x):f(x),q(x)\in Z[x]\} \\I=<(x-1)>=<p(x)>,\ where\ p(x)=(x-1)\in Z[x]$