When is product of polynomial and power series a polynomial?

Question

I am currently pondering over the following question:

For a field $K$, denote by $K[X]$ and $K[[X]]$ the ring of multivariate polynomials and the ring of multivariate power series, respectively. Given $p \in K[X]$ and $s \in K[[X]]$, under what conditions is the product $p \cdot s$ again a polynomial, i.e., an element in $K[X]$?

Clearly, if $p = s^{-1}$, then $p \cdot s = 1 \in K[X]$. More generally, if $p$ and $s$ can be factored as $p = qt^{-1}$ and $s = rt$, with $q,r \in K[X]$ and $t \in K[[X]]$ invertible (s.t. $t^{-1}$ is a polynomial), then $p \cdot s = q r \in K[X]$.

Now, my suspicion would be that $p\cdot s$ is a polynomial if and only if we are in the situation illustrated above, where some $t$ cancels with $t^{-1}$. However, so far, I was not able to prove this (nor to find a counterexample).

Maybe someone can help me by either finding a proof, providing a different characterisation, or a counterexample. Any help is highly appreciated!

Answer 1

Let $q$ be the greatest common divisor of $p$ and $ps$ in the ring of polynomials, let $r = (ps)/q$, and let $u= p/q$. Then a factorization of the type you want exists if $u$ is not in the ideal of $K[X]$ generated by all the variables, since then $u^{-1}$ is a power series $t$. Hence $ps = qr = t u q r = t p r$ and dividing both sides by $p$ we get $s=tr$ as desired.

So it suffices to check that $u$ is not in the ideal of $K[X]$ generated by all the variables. If this is the case then $u$ must have some irreducible factor $\pi$ which is not in that ideal. Now by the definition of gcd, $r$ is not a multiple of $\pi$ in the ring of polynomials, but it is a multiple of $\pi$ in the ring of power series. Then the completion of $K[X]/(\pi, r)$ is $K[[X]]/(\pi)$ which is absurd as if $X$ is $n$ variables then $K[X]$ has dimension $n$ so $K[X]/(\pi)$ has dimension $n-1$ and thus $K[X]/(\pi,r)$ has dimension $n-2$ but $K[[X]]/(\pi)$ has dimension $n-1$.