Let $A$ be a commutative ring, $A[[X]]$ be the ring of formal power series over $A$. Let $I$ be an ideal of $A[[X]]$, do we have $$\bigcap^{\infty}_{k=0} (I + (X^k)) = I?$$
Thanks you in advance of any help.
I found myself even having trouble dealing with the case where $I$ is principal. If $f(X), f_0(X)\in A[[X]]$, and $$X^k | (f(X) - f_0(X)g_k(X))$$ for all $k$, must there be a $g(X)$ such that $f(X) = f_0(X)g(X)$?
If $A$ is noetherian, the claim is an immediate consequence of the Krull intersection theorem. I have managed to come up with a nice non-noetherian counterexample.
Let us take any field $k$ and let $$G:=\{\underset{n\in\mathbb{N}}{\sum}g_ne_n\in\underset{n\in \mathbb{N}}{\prod}\mathbb{N}e_n\mid \forall n\in \mathbb{N},a_n\le a_0\},$$ an additive monoid, where $\mathbb{N}$ contains $0$. We introduce $2$ different orders $\preceq, \le$ on $G$. Set $g=\sum g_ne_n \preceq h=\sum h_ne_n \iff \forall n\in \mathbb{N}, g_n\le h_n$. $\le$ is the lexicographic order on $G$ where $\sum 0e_n$ is minimal. $\le$ is a total order.
Set $M:=\underset{g\in G}{\prod}{k\lambda_g}$ and its $k$-subspace $A:=\{\underset{g\in H}{\sum}a_g\lambda_g\in M\mid H\subset G, a_g\in k\}$ where we can take any $H$ and $a_g$ satisfying: (1) $\forall g\in G, \#\{h\in H\mid h\preceq g\}\lt \infty$ and (2) $H’:=\{h\in H\mid a_h\neq 0\}$ either is empty or has a (the) minimal element with respect to $\le$. Condition (1) is clearly closed under addition. To see that (2) is closed under addition too, assume for $\sum a_g\lambda_g, \sum b_g\lambda_g\in A$,$\sum a_g\lambda_g+ \sum b_g\lambda_g=\sum(a_g+b_g)\lambda_g\neq \sum 0\lambda_g$. Let $H’’=\{h=\sum h_ne_n\in G\mid a_h+b_h\neq 0\}$ and take $h=\sum h_n e_n\in H’’$. Then, for $H’’’=\{h’\in H’’\mid h\preceq \sum h_0e_n\}, \#H’’’\lt \infty.$ The minimal element of $H’’’$ with respect to $\le$ is also the minimal element of $H$ w.r.t. $\le$.
We can define multiplication on $A$ by $$(\sum a_g\lambda_g)(\sum b_g\lambda_g)=\underset{g\in G}{\sum}(\underset{h+h’=g}{\sum}a_hb_{h’})\lambda_g$$ by the condition (1). In fact, condition (1) and (2) is preserved under this multiplication so that $A$ is a ring. For $a\in A\setminus\{0\}$, write $v(a)\in G$ for the minimal element in (2). Set $v(0)=+\infty$ by convention. For this $v$, we have: (i) $\forall a_1, \dots, a_n\in A, v(a_1+\dots+a_n)\ge \min_i\{v(a_i)\}.$ (ii)$\forall a_1, a_2\in A,v(a_1a_2)=v(a_1)+v(a_2).$ (iii) If $v(a)=0, a$ is invertible in $A$. To see (iii), if $a=a_0\lambda_{0}+b$, $0\prec v(b)$, then $a_0^{-1}(1-(b/a_0)+(b/a_0)^2-\dots)$ is well-defined and is the inverse to $a$. Be careful, $A$ is not a valuation ring.
Let us take $\mu=\sum_{n=0}^\infty e_n, \mu_i=(\sum_{n=0}^\infty e_n) -e_i$ for $i\ge 1$. Set $x=\lambda_\mu, x_i= \lambda_{\mu_i}$. Set $I=(x-x_iX^i:i=1,2,\dots)\subset A[[X]]$. It is easily seen that for any $1\le k \lt \infty, x\in I+(X^k)$. We would like to show $x\notin I$ by contradiction. Suppose for $f_i(X)= \sum a_{ij}X^j \in A[[X]], 1\le i\le n, \sum_{i=1}^n f_i(X)(x-x_iX^i)=x.$ Looking at the coefficients of $X^0$, $ x=(a_{10}+\dots +a_{n0})x, \sum_n 0e_n=v(1)=v(a_{10}+\dots +a_{n0})\ge \min_i \{v(a_{i0})\} \ge \sum_n 0e_n$ so that there exists at least one $i$ such that $v(a_{i0})=\sum 0e_n$, i.e., $a_{i0}^{-1}\in A$, and consequently $f_i$ is invertible in $A[[X]]$. Multiplying by $f_i^{-1}(X)$, we get $$\sum_{0\le j \le n,j\neq i} f_i^{-1}(X)f_j(X)(x-x_jX^j)+x-x_iX^i=f_i^{-1}(X)x.$$ Looking at the coefficients of $X^i$,$$\sum_{0\le j \le n,j\neq i}(b_jx+c_jx_j)-dx=x_i.$$ Define $v_n$ by $v(y)=\sum_nv_n(y)e_n$ . Then, seeing $b_jx, c_jx_j$ and $dx$ as elements of $M$, they are infinite sums of terms of the form $l_g\lambda_g$, $0\neq l_g\in k, v_i(\lambda_g)\ge 1.$ On the other hand, since $x_i=\lambda_{\mu_i}$, the left hand side must have a term of the form $l_{\mu_i}\lambda_{\mu_i}.$ But $v_i(\lambda_{\mu_i})=0,$ contradiction.