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**Subject Mathematics Abstract Algebra**

See Question.pdf

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Z[x] is not PID because if we consider an ideal <2,x> which is principal then <2,x> = <h(x)> for some h(x) belongs to Z[x] now we consider p(x) belongs to Z[x] such that p(x).h(x) = 2. Since multiplication of two polynomial is constant therefore deg (h(x)) = 0 . and then h(x) = +1 or -1 . therefore <h(x)> = Z[x] = <2,x>

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