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This is my first question; I hope it worthy of this awesome forum and its members.

Let $R$ be a commutative ring, perhaps with unit, perhaps not. As usual let $R[x]$ denote the ring of polynomials over $R$, and let $I$ be an ideal in $R[x]$. Let $D(I)$ be the subset of $I$ consisting of all $f(x) \in I$ of minimal degree. It is well known from elementary algebra that in the event that $R$ is a field, $D(I)$ consists of essentially one element $d(x)$ (up to multiplication by a unit of $R$), and that $I$ is the principal ideal generated by $d(x)$: $I = (d(x)) = Rd(x)$. In the case of general commutative $R$, does $D(I)$ generate $I$ in the sense that any $f(x) \in I$ may be written $f(x) = \sum_{i = 1}^{m} f_{i}(x)d_i(x)$ with $d_{i}(x)\in D(I)$ and $f_{i}(x) \in R[x]$? Does the existence of $1 \in R$ make any difference?

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I am afraid that your question is somewhat below the awesome level of this awesome forum. Anyway, this is wrong already for $R=\mathbb{Z}:$ the ideal $(2,x)$ of $\mathbb{Z}[x]$ is not generated by the degree 0 polynomials contained in it. (I hope this wasn't a homework question, by the way!) – Victor Protsak Aug 22 '10 at 22:14
It looks as though you're trying to rediscover the proof of Hilbert's basis theorem though.'s_basis_theorem – Simon Wadsley Aug 23 '10 at 7:50
First of all, thanks to Victor Prostak for providing a negative answer to my to my question. Upon re-thinking it, looking at the structure of $(2, x)$ in $Z[x]$, I remembered the ascending chain condition from a graduate course in abstract algebra I took quite awhile ago, and from several books on ring theory I read in its wake. I soon discovered the wiki page on the Hilbert basis theorem Simon Wadsely mentioned; and yes, that is the kind of thing I was reaching for, though I haven't fully digested it yet. Finally, I assure all it was not homework; I pursued the question on my own. – drbobmeister Aug 24 '10 at 0:51
there is no reason why this should be closed. – Martin Brandenburg Aug 25 '10 at 11:34

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