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?

nothomework; I pursued the question on my own. – drbobmeister Aug 24 '10 at 0:51