Let $A$ be a commutative ring with identity and $A[x]$ the polynomial ring over $A$.
Given $I$ as an ideal of $A[x]$, we have the following informations:
- For integers $0\leq d\leq n$, the set of degree $d$ coefficients in degree $n$ polynomial $$I_d^n = \{a_d\in A: a_0+...a_dx^d+...a_nx^n\in I\}$$ is an ideal in $A$, and $I_s^m\subset I_d^n$ if $m\leq n$, $s\leq d$ and $m-s = n-d$.
This would imply that the set $I^d_d$ of leading coefficients of polynomials in $I$ with degree $d$ is an ideal. Further more the set of leading coefficients in $I$ is an ideal as the filtered union $I^L = \bigcup_{d\geq 0} I^d_d.$
If condition 1. is true for a subset $S\subset A[x]$, is $S$ is an ideal? I am struggling to find a counterexample.
Note that the above sets $I^n_d$ and $I^L$ are extensively used in the proof of Hilbert's Basis Theorem.
The counterexample is obvious, but I would like to see if we can add more fundamental properties other than condition 1. to get a stronger condition than ideals. I will re-edit the question, ask it later, and attach the link below.
Here is the new question post. The above thought will be used in a similar way.
Consider $A = \mathbb{Z}$, $S := \mathbb{Z}[x] - \{2(x+1) \}$.
$S^n_d = \mathbb{Z}$ forany $n,d \geq 0$.
Maybe you should post conditions on $S$ instead of $S^n_d$.