Units in Formal Power Series Rings

Detailed Solution

In the polynomial ring $R[x]$ (with $R$ a field), the units are exactly the nonzero constant polynomials, since the degree of a product of nonzero polynomials adds and a unit must have degree $0$. In the power series ring $R[[x]]$, however, a series is a unit if and only if its constant term is nonzero. This is because if $$ f(x)=1-x, $$ we can check directly that $$ f(x)\cdot(1+x+x^2+\cdots)=(1-x)(1+x+x^2+\cdots)=1, $$ using the fact that the series multiplication is well-defined term by term and the terms telescope. So $f$ has a genuine inverse inside $R[[x]]$, even though $1-x$ is certainly not invertible as a polynomial in $R[x]$ (its only polynomial "inverse" would have to be an infinite series, which isn't a polynomial). This shows units in $R[[x]]$ are far more plentiful than units in $R[x]$: any series with nonzero constant term is invertible, not just the nonzero constants. Hence option A is false, C is false since $f$ is invertible in $R[[x]]$, and D is false since $R[[x]]$ is in fact an integral domain when $R$ is a field (no zero divisors at all), while this has nothing to do with matching $R[x]$'s zero divisor structure trivially — the real distinguishing feature here is the units, not zero divisors. Thus the correct option is $\boxed{\mathrm{B}}$.