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The observation in this answer and this comment suggest the following:

If $R$ is a domain and $f \in R[x]$ is monic and not a unit in $R[[x]]$, then if $f$ is irreducible in $R[x]$ it is also irreducible in $R[[x]]$.

Is this true, and if so how do we prove it? It doesn't seem obvious.

The statement when $R$ is a field seems trivial, since if $f$ is not a unit as a power series it is always reducible, since it has a zero constant term and so it can be factored as some power of $x$ times another polynomial.

Anakhand
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1 Answers1

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It is false, e.g. this answer shows the linear polynomial $\,ab+x\,$ is reducible in $R[[x]]$ if $\,a,b\neq 0\,$ are comaximal nonunits, giving an explicit factorization via a Catalan series $f(x)\in\Bbb Z[[x]],\,$ e.g.

$$\begin{eqnarray}{} 6+ x\, &=&\ (3 - x\:\!f(x))\ \ (2 + x\:\!f(x)) \\[0.5em] &=&\ \dfrac{5\!-\! \sqrt{1\!-\!4\:\!x}}{2}\,\ \dfrac{5\! +\! \sqrt{1\!-\!4\:\!x}}{2}\\ \end{eqnarray}\qquad\qquad$$

It is not difficult to construct algorithms for irreducibility testing and factorization in the UFD $R[[x]]$ over a PID $R,\,$ e.g. see J. Elliot: Factoring formal power series over principal ideal domains.

Bill Dubuque
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  • I assume here $R = \mathbb{Z}$? If so, the Catalan series have rational coefficients, right? So is this actually a decomposition in $\mathbb{Z}[[X]]$? – Anakhand Dec 06 '23 at 21:17
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    @Anakhand The Catalan series $,f(x)\in \Bbb Z[[x]]$ has integer coef's since it is the generating function of the Catalan integer sequence $,c_n = \frac{1}{n+1}{2n\choose n}.,$ Thus $,f(6x)\in\Bbb Z[[x]],,$ so both factors are in $\Bbb Z[[x]]$ – Bill Dubuque Dec 06 '23 at 21:31