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Consider a ring $R$ and the ring of formal power series $R[[X]]$ over $R$. Note that $R[X]$ naturally embeds into $R[[X]]$. Now let $I$ be a finitely generated ideal of $R[[X]]$, say, $I=\langle f_1,\dots,f_n\rangle$.

What are the least necessary assumptions on $R$ so that we may assume $f_1,\dots,f_n\in R[X]$?

This is certainly possible if $R=K$ is a field. Then $K[[X]]$ is a discrete valuation ring, i.e. a local PID with maximal ideal $\mathfrak =\langle X\rangle$. In this case all ideals are of the form $\langle X^k\rangle$ which are polynomials. This is also possible for $R$ a complete local ring. In this case the Weierstraß preparation theorem holds which allows us to write any $f\in R[[X]]$ (uniquely) as the product of a unit and a so-called distinguished polynomial $F$. Hence we may replace $f$ by $F$ and are done.

I am not sure whether something similar is possible for, say, $R=\mathbb Z$.

Is there are general theorem on this situation? Both, having a coefficients in a field or working over a complete local ring are quite specific scenarios (and exclude on of the most basic rings: $\mathbb Z$). Moreover, is there a natural generalization to not necessarily finitely generated ideals (the given cases extend right away)?

Thanks in advance!

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    $\begingroup$ It might help to see $R[[x]]$ is the inverse limit of $R[x]/\langle x^n\rangle.$ Not sure how, though. $\endgroup$ Commented Aug 25, 2021 at 21:15
  • $\begingroup$ @ThomasAndrews Any progress on this? I thought about this but it doesn't seem to lead anywhere helpful. $\endgroup$
    – mrtaurho
    Commented Sep 1, 2021 at 18:08
  • $\begingroup$ Given the counterexample given below, it seems unlikely to work. $\endgroup$ Commented Sep 1, 2021 at 18:12
  • $\begingroup$ @ThomasAndrews Thank you anyways! It was an interesting (albeit not successful) way of approaching the question nonetheless. $\endgroup$
    – mrtaurho
    Commented Sep 1, 2021 at 21:40

1 Answer 1

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$\def\ZZ{\mathbb{Z}}$For $R=\ZZ$, there is no result like this. Let $\ZZ_p$ be the $p$-adic integers and let $\theta$ in $\ZZ_p$ with $\theta \equiv p \bmod p^2$ and $\theta$ transcendental over $\mathbb{Q}$. Notice that, if $f(x) \in \ZZ[[x]]$, it will make sense to evaluate $f(\theta)$ in $\ZZ_p$, since the sum $\sum f_n \theta^n$ will be $p$-adically convergent.

Now, we can construct a power series $g(x)$ in $\ZZ[[x]]$ of the form $p+g_1 x + g_2 x^2 + \cdots$ with $g(\theta)=0$. Indeed, if we have inductively chosen $g_1$, $g_2$, ..., $g_{n-1}$ in $\ZZ$ such that $p+g_1 \theta + g_2 \theta^2 + \cdots + g_{n-1} \theta^{n-1}$ is zero modulo $p^{n}$, then we can choose $g_n$ in $\ZZ$ such that $p+g_1 \theta + g_2 \theta^2 + \cdots + g_{n-1} \theta^{n-1}+g_n\theta^n$ is zero modulo $p^{n+1}$. Note also, since the constant term of $g$ is $p$, that $g$ is not the zero power series.

Then I claim that the ideal $\langle g(x) \rangle$ does not contain any nonzero element of $\ZZ[x]$ and thus cannot be generated by elements of $\ZZ[x]$. Indeed, suppose that $h(x) = g(x) u(x)$ with $h \in \ZZ[x]$ and $u \in \ZZ[[x]]$. Then $h(\theta) = g(\theta) u(\theta) = 0$. But $\theta$ is transcendental over $\mathbb{Q}$, so we deduce that $h$ is the zero polynomial.

I would expect you can generalize "complete local ring" to "complete semi-local ring", but I don't know how you'd go further than that.

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    $\begingroup$ This is a very nice counterexample for $R=\mathbb Z$! (+1) Thank you! I'll need some further time studying the construction in more detail, though. $\endgroup$
    – mrtaurho
    Commented Aug 31, 2021 at 19:03
  • $\begingroup$ Could you maybe elaborate on the inductive construction of $g$? Either I'm missing something or there is a typo somewhere. I'm not sure if you can define $g_n$ like this (at least I'm having some problems with $g_1$ and $g_2$ in particular). $\endgroup$
    – mrtaurho
    Commented Sep 1, 2021 at 15:32
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    $\begingroup$ There was an off by one in the indexing, which I have fixed, does that help? $\endgroup$ Commented Sep 1, 2021 at 17:43
  • $\begingroup$ Ah, yeah. That makes more sense now :) Thank you! Another question: do you know how to extend the theorem to complete semi-local rings (either by proof or by reference)? $\endgroup$
    – mrtaurho
    Commented Sep 1, 2021 at 18:07

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