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Going through my Commutative Algebra notes I found out that I don't know the answer to this question.

Let $A$ be a commutative ring with unit and let $f(T):=\sum_{k=0}^\infty a_k T^k \in A[[T]]$ be a power series. It is a standard fact from Commutative Algebra that $1/f$ exists in $A[[T]]$ iff $a_0\in U(A)$, where $U(A)$ denotes the group of units of $A$.

Is there some characterization of all $f\in U(A[[T]])$ with the property that $1/f\in A[T]$?

Even though I have posed the problem for general $A$, I'd be happy to know the answer even for $A=\mathbb{C}$ only.

Some remarks:

  1. Such $f$ form a commutative submonoid of $U(A[[T]])$.
  2. If $f=\sum_{k=0}^n a_k T^k$ is in fact polynomial, then it is again a standard fact from Commutative Algebra that $1/f\in A[T]$ iff $a_0\in A^\times$ and $a_1,\dots,a_n\in\operatorname{nil}A$, so this special case is settled. Furthermore one might be inclined to conjecture for general $f=\sum_{k=0}^\infty a_k T^k$ that one needs $a_1, a_2, \ldots \in \operatorname{nil}A$, but this is clearly false as the simplest example $$ \frac{1}{1-T} = \sum_{k=0}^\infty T^k $$ shows.
  3. Writing $f=1/p$ for $p\in A[T]$ and $$ f(1/T) = \sum_{k=0}^\infty a_k T^{-k} = \frac{1}{p(1/T)} = \frac{T^m}{q(T)} $$ for some $q\in A[T]$ and $m=\deg p=\deg q$, this becomes a very special case of the problem of determining when a power series is a rational function in disguise.
  4. If $A=\kappa$ is an algebraically closed field, then $p(T)$ always splits into linear factors $p(T)=(T-\alpha_1)\ldots(T-\alpha_m)$, where wlog. we have assumed $a_0=1$, so $f$ is a product of a bunch of geometric series $$ f(T) = \sum_{k=0}^\infty a_k T^k = (-1)^m \bigg( \prod_{j=1}^m \frac{1}{\alpha_j} \bigg) \bigg(\prod_{j=1}^m \sum_{k_j=0}^\infty \frac{1}{\alpha_j^{k_j}} T^{k_j} \bigg) $$ In particular, $$ (-1)^m \bigg( \prod_{j=1}^m \frac{1}{\alpha_j} \bigg) = 1 $$ and $$ a_n = \sum_{k_1+\dots+k_m=n} \frac{1}{\alpha_1^{k_1} \ldots \alpha_m^{k_m}}, $$ but I don't see how this helps (if at all).

Is there at least an a priori way to determine $m$ in the case $A=\kappa$?

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    $\begingroup$ I don't know what kind of answer you're looking for beyond the answer you get by multiplying by the polynomial inverse, which is roughly "the coefficients satisfy a linear recurrence relation, with a special set of initial conditions." Over an integral domain you can get a closed form in terms of the roots of the polynomial inverse using partial fraction decomposition over the algebraic closure of the field of fractions, but this isn't particularly helpful in recognizing $f$. It's even more unclear what to do in general. $\endgroup$ Oct 6, 2018 at 5:20
  • $\begingroup$ @QiaochuYuan: I thought there might be something more to it, since it is a "middle" case between the inverses of power series and inverses of polynomials, which are well known. I guess I was hoping for too much. $\endgroup$
    – M.G.
    Oct 6, 2018 at 13:32

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