As pointed out by Makoto, on [this question](https://mathoverflow.net/questions/154469/can-we-prove-that-the-ring-of-formal-power-series-over-a-noetherian-ring-is-noet/194416#194416) about power series rings and the axiom of choice, an idea I had needed the axiom of dependent choice to work. However, the construction raises another interesting problem which I'm hopeful is easier to solve.


Let $A$ be a commutative ring with $1$. Let $I\in A[[x]]$ be a finitely generated ideal of the power series ring. If we don't assume the axiom of dependent choice, is it possible to have an element $f(x)\in A[[x]]\setminus I$ such that for every $n\in \mathbb{N}$ there exists $f\_n(x)\in I$ with $x^n|(f(x)-f\_n(x))$?


Intuitively, this says that we can build a linear combination, from the generators of $I$, which matches $f$ to any *finite* degree, but there is no combination that matches $f$ completely. If we write $I=(g\_1(x),\ldots, g\_k(x))$ and we let $I'$ be the ideal generated by the coefficients of all the $g$'s and $f$, I'm most interested in the case when the leading terms of $g\_1(x),\ldots, g\_k(x)$ generate $I'$. In this case, the linear combinations can be built up in compatible ways (degree by degree), but could it still not be possible to get all of $f$?


Edited to add: Here is an even simpler question. Let $a\in A$ and let $f(x)\in A[[x]]$ be such that every coefficient of $f(x)$ lives in the ideal generated by $a$. Can we say that $f$ lives in the ideal generated by $a$, without using a choice axiom?



Matching power series to infinity
I believe the answer to your simpler question is "no:" Fix an infinite sequence of sets $A\_i$ ($i\in\omega$) for which no choice function exists. Now let $A$ be the free commutative ring generated by the set $$(\bigcup A\_i)\cup\{d\_i: i\in\omega\}\cup\{c\},$$ modulo the relations $$ca=d\_i\quad\mbox{ for every }i\in\omega, a\in A\_i.$$ Now each coefficient of the power series $$\sum\_{i\in\omega}d\_ix^i$$ is a multiple of $c$, but the whole series does not live in the ideal generated by $c$.


I might be missing something?


