اقتباس : المشاركة الأصلية كتبت بواسطة أسامه رشوان [ مشاهدة المشاركة ]

هذا جزء من سؤالك الثالث وسامحني لأن الترجمة بالعربي في هذ التخصص تفرق في المعنى لذا السؤال بالإنجليزية وإجابته 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

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