As a simple example: $$ \mathbb{Z}[n]/ \langle n^2 - an - b \rangle $$
We know this is simply all the linear expressions $\{cn+d \mid c,d \in \mathbb{Z}\}$
But is there an explicit way to describe the homomorphism $\mathbb{Z}[n] \rightarrow \mathbb{Z}[n]/ \langle n^2 - an - b \rangle$?
E.g., take a poly $\displaystyle P(n) = \sum_{j=0}^{\infty} \alpha_j n^j$
One can split this up in to even and odd powers and substitute in $n^2 = an + b$ to reduce it. This gives various constraints on $\alpha_k$ such as the obvious $\alpha_0 = d,\, \alpha_1 = c,\, \alpha_k = 0$ for $k>1$; there are more complex constrains though. I'm interested in the general representation of these complex constraints in the general case.
E.g., there are polynomials with an "infinite" number of non-zero coefficients that map to the a line(since that is all we are working with). E.g., what are all the polynomials that map to $3n + 4$?
I can't figure this out because when reducing $n^k = f(n)\cdot n^m,\; \deg(f(n)) = k-m$ one has ends up an infinite recursion: $$ P(n) = \sum_{j=0}^{\infty} \alpha_j n^j = \sum_{j=0}^{\infty} \sum_{i=0}^{k-1} \alpha_{jk + i} n^{jk + i} = \sum_{j=0}^{\infty} \sum_{i=0}^{k-1} \alpha_{jk + i} (f(n)\cdot n^m)^j n^i$$
And one is left trying unwind an infinite number of nested $n^k$ values. Of course at each stage we can a new set of constrains on the coefficients but one can't generalize the results this way it seems.
So, simply, I just want to explicitly construct/parameterize the equivalence classes of a homomorphism on a polynomial ring. In this case I'm ok with just using a simple ideal.
\langle n^2 - an - b \ranglefor better ideal angle brackets: $\langle n^2 - an - b \rangle$ – Sammy Black Jan 23 '24 at 17:19