(New page: Proof: We claim that the ideal <x,2> subset Z[x] is not principal. Suppose, to the contrary, that <x,2> = <p(x)> for some p(x) in Z[x]. p(x) cannot be a constant polynomial, because this...) |
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We claim that the ideal <x,2> subset Z[x] is not principal. Suppose, to the contrary, that <x,2> = | We claim that the ideal <x,2> subset Z[x] is not principal. Suppose, to the contrary, that <x,2> = | ||
| − | <p(x)> for some p(x) in Z[x]. p(x) cannot be a constant polynomial, because this constant would have to be even, and then | + | <p(x)> for some p(x) in Z[x]. p(x) cannot be a constant polynomial, because this constant would have to be even, |
| − | x is not in <p(x)>. Furthermore, p(x) has a nonzero constant term, since otherwise 2 is not in <p(x)>. | + | and then x is not in <p(x)>. Furthermore, p(x) has a nonzero constant term, since otherwise 2 is |
| − | Since x is in <p(x)>, there exists a polynomial q(x) in Z[x] such that x = q(x)p(x), and since | + | not in <p(x)>. Since x is in <p(x)>, there exists a polynomial q(x) in Z[x] such that x = q(x)p(x), and since |
| − | constant term of q(x)p(x) is zero, q(x) has degree at least 1. But this means x = q(x)p(x) has | + | the constant term of q(x)p(x) is zero, q(x) has degree at least 1. But this means x = q(x)p(x) has |
degree at least 2, a contradiction. | degree at least 2, a contradiction. | ||
MATH IS SO FUN. | MATH IS SO FUN. | ||
Revision as of 16:34, 5 November 2008
Proof:
We claim that the ideal <x,2> subset Z[x] is not principal. Suppose, to the contrary, that <x,2> = <p(x)> for some p(x) in Z[x]. p(x) cannot be a constant polynomial, because this constant would have to be even, and then x is not in <p(x)>. Furthermore, p(x) has a nonzero constant term, since otherwise 2 is not in <p(x)>. Since x is in <p(x)>, there exists a polynomial q(x) in Z[x] such that x = q(x)p(x), and since the constant term of q(x)p(x) is zero, q(x) has degree at least 1. But this means x = q(x)p(x) has degree at least 2, a contradiction.
MATH IS SO FUN.
