Given a ring $A$ and a finitely generated ideal $I\subseteq A$, then $I A[\![X]\!]=I[\![X]\!]$. Conversely, assuming that the equality holds, must $I$ be finitely generated?
If the set of generators of $I$ is countable, say $(c_n)_{n\in \mathbb N}$, the series $(c_nX^n)_{n\in \mathbb N}$ can be expressed as a finite sum of the form: $$\sum_{i=1}^m (c_{n_i}(a_{i,n}X^n)_{n\in \mathbb N})_i$$ where $n_i\in \mathbb N$ for all $i$ and $a_{i,n}\in A$ for all $i,n$. (A priori, in the sum above there should be a generic $b_i\in I$ instead of $c_{n_i}$; however every such $b_i$ is a finite sum of generators, and the number of terms $b_i$ is finite too). Thus $c_n=\sum_{i=1}^m (c_{n_i}a_{i,n})_i$, for all $n$, meaning that the set $\{c_{n_i}\}$, whose cardinality is $m$, generates $I$.
What if $I$ is not countably generated? My guess is that this result does not hold anymore, but I can't think of a counterexample. Do you have any suggestion? Thank you
I believe you don't really need countable generation. Indeed, take $ (t_n) $ to be a countable sequence of elements of $I$ that are not contained in any finitely generated subideal of $I$. Then $\sum t_nX^n \in I[[X]]$, but the coefficients of any element of $IA[[X]]$ are contained in a finitely generated subideal of $I$.