(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> = | + | <nowiki>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)>. |
| − | <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.</nowiki> |
| − | 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 | + | Hey thanks, this helped a lot, whoever you are... |
| − | degree at least 2, a contradiction. | + | -Tim |
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| − | + | ||
Latest revision as of 21:11, 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.
Hey thanks, this helped a lot, whoever you are... -Tim
