Let $R$ be a commutative ring with unit element, let $I$ be an ideal in $R$. Let $A = R[[X]]$, the ring of formal power series with coefficients in $R$. Let $I_A$ be the ideal generated by elements of $I$ in $A$. Let $\varphi : A \rightarrow (R/I)[[X]], \sum_{n=0}^{\infty} a_nX^n \mapsto \sum_{n=0}^{\infty} (a_n + I)X^n$ be the canonical reduction homomorphism.
Now I was wondering: is it always true, that $I_A = \ker(\varphi)$? We can directly see that $I_A \subseteq \ker(\varphi)$, since if we have a power series with coefficients in $I$, these coefficients will all be send to $0$. For "$\supseteq$" we can take $P\in \ker(\varphi)$, so $P(X) = \sum_{n=0}^\infty p_n X^n$ gets send to $0$ by $\varphi$, so $p_n + I = 0 + I$, which means that all the $p_n$ are already elements of $I$. But now I am confused: We know that every $p_n X^n$ is an element of $I_A$, and because ideals are closed under addition, it should follow that $P \in I_A$, or does that only work in the case of finite sums? I was able to prove the equality in the case of $R$ being noetherian without such infinite sums, since we can then write the $p_n$ as finite sums over $I$-generating elements $i_1, ... , i_k$, so $p_n = \sum_{m=1}^k r_{n,m} i_m$, this way we can split $P$ into:
$$P(X) = \sum_{n=0}^\infty p_n X^n = \sum_{n=0}^\infty \sum_{m=1}^k r_{n,m} i_m X^n = \sum_{m=1}^k \left( i_m\sum_{n=0}^\infty r_{n,m} X^n \right)$$
Now we can see that $i_m\sum_{n=0}^\infty r_{n,m} X^n \in I_A$, but this time we have a finite sum of elements which are in $I_A$, so $P \in I_A$. Now is my question: Does this hold in general, even for non-noetherian rings, or is there a counterexample? If it's true, how can I prove it?