# Replacing ideal generators in $R[[X]]$ by polynomials

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)?

• It might help to see $R[[x]]$ is the inverse limit of $R[x]/\langle x^n\rangle.$ Not sure how, though. Commented Aug 25, 2021 at 21:15
• Given the counterexample given below, it seems unlikely to work. Commented Sep 1, 2021 at 18:12
• @ThomasAndrews Thank you anyways! It was an interesting (albeit not successful) way of approaching the question nonetheless. Commented Sep 1, 2021 at 21:40

$$\def\ZZ{\mathbb{Z}}$$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.
• 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. Commented Aug 31, 2021 at 19:03
• 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). Commented Sep 1, 2021 at 15:32