Question #282132

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


1
Expert's answer
2021-12-23T17:33:07-0500

Solution:

I={x24x+3,x3+3x2x3}={(x1)(x3),(x+3)(x21)}={(x1)(x3),(x+3)(x1)(x+1)}={(x1)(x3)f(x)+(x+3)(x1)(x+1)q(x):f(x),q(x)Z[x]}={(x1)[(x3)f(x)+(x+3)(x+1)q(x)]:f(x),q(x)Z[x]}={(x1)g(x):g(x)=(x3)f(x)+(x+3)(x+1)q(x):f(x),q(x)Z[x]}I=<(x1)>=<p(x)>, where p(x)=(x1)Z[x]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]


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