# Matching power series to infinity

As pointed out by Makoto, on this question 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?

## 1 Answer

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?

• This is what I thought must happen. Glad to see a quick answer. (This, of course, answers all the other more complicated questions as well.) – Pace Nielsen Jan 22 '15 at 18:28
• Actually, note that this establishes an equivalence: the statement "if every coefficient of the powerseries is a multiple of $c$, then the series is a multiple of $c$" is equivalent (over ZF) to the axiom of countable choice. And, if we generalize the class of formal power series to index sets other than $\omega$ in the natural way, we get equivalences with other forms of choice, too. – Noah Schweber Jan 22 '15 at 19:40
• The construction I'm familiar with (namely, the Malcev-Neumann series) depends on supports which are well-ordered. So while you can get more general equivalences, I believe there may be some natural (algebraic) limitations. – Pace Nielsen Jan 22 '15 at 20:15
• I was thinking in a slightly different direction: off the top of my head, I think we have a reasonable ring of power series for any index set $I$ which is the underlying set of a monoid $(I, *)$ such that for each $x$ in $I$, there are only finitely many pairs $y_1, y_2$ with $y_1*y_2=x$. But maybe there's another obstacle? – Noah Schweber Jan 22 '15 at 20:20
• That will also work. There is no obstacle in that construction (other than finding the appropriate monoid $I$ corresponding to the choice axiom you want). – Pace Nielsen Jan 22 '15 at 20:38