Is a ring of formal power series \(R[[x]]\) a unique factorization domain (UFD)?
Solution
Not in complete generality.
If \(R=k\) is a field, then \[ k[[x]] \] is a discrete valuation ring: every nonzero series has a unique form \[ f=x^m u, \] where \(u\) is a unit. Hence every ideal is \((x^m)\), so \(k[[x]]\) is a PID and therefore a UFD.
Likewise, for a field \(k\), each finite-variable formal power series ring \[ k[[x_1,\dots,x_n]] \] is a regular local ring and hence a UFD.
However, the implication \[ R\text{ UFD}\Longrightarrow R[[x]]\text{ UFD} \] is false in general. Samuel constructed UFDs \(R\) for which \(R[[x]]\) is not a UFD; see P. Samuel, On unique factorization domains, Illinois J. Math. 5 (1961), 1–17. Thus the polynomial-ring theorem of Gauss has no unrestricted formal-power-series analogue.
So the answer depends on hypotheses on \(R\): over a field the answer is yes, but for a general UFD the answer can be no.