ra.rings and algebras - When are the units of R[x] exactly the units of R?

If $R$ is a commutative ring, then by the following result, the answer is "if and only if $R$ is reduced."

If $R$ is a commutative ring, then $a_0+a_1x+cdots + a_nx^nin R[x]$ is a unit if and only if $a_0$ a unit in $R$ and $a_i$ is nilpotent for $i>0$.

Proof. One direction is easy. Any polynomial of the given form is a unit because the sum of a unit and a nilpotent element is always a unit.

The other direction isn't too hard if $R$ is a domain (the product of non-zero elements is always non-zero). If $g=b_0+cdots b_mx^m$ (with $b_mneq 0$) is the inverse of $f=a_0+cdots+a_nx^n$ (with $a_nneq 0$), then the highest order term of $1=fcdot g$ is $a_nb_mx^{n+m}$, so we must have $n=m=0$ and $a_0$ invertible (with inverse $b_0$)

For the general case, suppose $a_0+cdots +a_nx^n$ is a unit. Reducing modulo $x$, we must get a unit in $R[x]/(x)cong R$, so $a_0$ must be a unit. Reducing modulo any prime $mathfrak psubseteq R$, we get a unit in $(R/mathfrak p)[x]$. Since $R/mathfrak p$ is a domain, the previous paragraph shows that $a_iin mathfrak p$ for all $i>0$ and all primes $mathfrak p$. Since the intersection of all primes is the nilradical, each $a_i$ must be nilpotent.

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A more "bare hands" elementary proof is given in Ex. 1.32 of Lam's Exercises in Classical Ring Theory. He also gives counterexamples to both implications if $R$ is not assumed commutative and mentions a really interesting related question. If $Isubseteq R$ is an ideal all of whose elements are nilpotent and $a_iin I$, then does it follow that $1+a_1x+cdots +a_nx^n$ is a unit in $R[x]$? If you can prove that it does, it would imply the Köthe conjecture, a famous problem in ring theory.