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prove that Z[x] is not principle ideal domain????????

الجواب
Consider the set S of all polynomials in Z[x] whose constant term is even.

Check that S is an ideal. This is a pretty straightforward exercise in the definitions. (If you have theorems around, you could do it less directly: it's the inverse image of an ideal [the even integers] under a ring homomorphism [namely, the homomorphism Z[x] -> Z given by evaluating at 0], and any set of this form is necessarily an ideal.)

To see that S is not principal, suppose it were, and that p is a generator for S. Thus

S = {p(x) q(x): q(x) in Z[x]}.

If p(x) had a nonzero degree d, then every nonzero polynomial of the form p(x) q(x) would have degree at least d. Since S contains nonzero constant polynomials (e.g. 2) we conclude that p(x) cannot have nonzero degree. Thus p(x) is a constant k.

We cannot have k = 0 (because S is not {0}), nor k = 1 or -1 (because S is not Z[x]). If we had another k that did the job, we would conclude that the coefficients of every element of S had a number |k| > 1 as a common divisor. But x + 2 (for example) is a polynomial whose coefficients (1 and 2) have no common divisors of this form