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?


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?

  • $\begingroup$ 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.) $\endgroup$ – Pace Nielsen Jan 22 '15 at 18:28
  • 1
    $\begingroup$ 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. $\endgroup$ – Noah Schweber Jan 22 '15 at 19:40
  • $\begingroup$ 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. $\endgroup$ – Pace Nielsen Jan 22 '15 at 20:15
  • 1
    $\begingroup$ 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? $\endgroup$ – Noah Schweber Jan 22 '15 at 20:20
  • $\begingroup$ 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). $\endgroup$ – Pace Nielsen Jan 22 '15 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.