Let's say we got $C_f := k[x,y]/(f)$ where $f$ an irreducible poly in $k[x,y]$, and $k$ alg. closed. Say we want factor the ideal $(x-a)$ explicitly as a product of prime ideals (I think this was building up to the fact that we can only do that when our noetherian domain of dimension 1 is a Dedekind domain).
We write $f(a,y) := c \Pi_{i=1}^s (y-b_i)^{e_i}$ (with $b_i \neq b_j$ whenever $i \neq j$ and $c \in k$) so that $f(x,y) = (x-a)g(x,y) + c \Pi_{i=1}^s (y-b_i)^{e_i}$ with $g(x,y) \in k[x,y]$ and $M_i := (x-a , y-b_i)$ would be the maximal ideals of $C_f$ that contain $(x-a)$.
A couple of nights ago I understood that $M_i^{e_i}$ is generated by $\{(x-a)^n (y-b_i)^m \}$ such that $n + m = e_i$ and $n> 0, m \geq 0$. So $M_i^{e_i} \subset (x-a , (y-b)^{e_i})$ - which I think is because $(y-b_i)^{e_i - z} \in C_f$ for $0 < z < e_i$, so you can always get one of those generators I mentioned above by picking one of those latter polynomials.
So they claim $\Pi_{i = 1}^s (y - b_i)^{e_i} \in (x-a)C_f$ since $f(x,y) = (x-a) g(x,y) + c \Pi_{i = 1}^s (y - b_i)^{e_i} = 0$. I would think this is because you could write $\Pi_{i=1}^s (y - b_i)^{e_i} = -c^{-1}(x-a)g(x,y)$, but $g(x,y) \in k[x,y]$, not $C_f$, so it stands to reason you could get something different, like $[g(x,y)]$ in $C_f$ that might mess something up. It's just that $\neg \square (g(x,y) \in C_f)$ and it is confusing me.
So, in particular, how do they justify $\Pi_{i = 1}^s (y - b_i)^{e_i} \in (x-a)C_f$ ?
EDIT: They also write that $\Pi_{i=1}^s M_i^{e_i} \subset (x-a_i , \Pi_{i=1}^s (y-b_i)^{e_i})$, just before the line I am asking about. I thought I would include this to maybe clear up any remaining confusion upon reading this.