Sum of a Unit and a Commuting Nilpotent Element

Detailed Solution

We are given that $u$ is a unit, $n$ is nilpotent, and $$ un=nu. $$ Since $u$ is a unit, there exists $u^{-1}\in R$ such that $$ uu^{-1}=u^{-1}u=1. $$ Factor $u+n$ as $$ u+n=u(1+u^{-1}n). $$ We will show that both factors on the right are units. Step 1: Show that $u^{-1}n$ is nilpotent. Since $un=nu$, the elements $u^{-1}$ and $n$ also commute. Therefore, $$ (u^{-1}n)^k=u^{-k}n^k=0. $$ Thus $u^{-1}n$ is nilpotent. Step 2: Show that $1+u^{-1}n$ is a unit. Let $a=u^{-1}n$. Since $a^k=0$, the finite geometric series gives $$ (1+a)^{-1}=1-a+a^2-\cdots+(-1)^{k-1}a^{k-1}. $$ Indeed, $$ (1+a)(1-a+a^2-\cdots+(-1)^{k-1}a^{k-1})=1, $$ because all intermediate terms cancel and $a^k=0$. The same product in the reverse order is also $1$. Hence $1+a=1+u^{-1}n$ is a unit. Step 3: Conclude that $u+n$ is a unit. We have expressed $u+n$ as a product of two units: $$ u+n=u(1+u^{-1}n). $$ Therefore, $u+n$ is a unit. An explicit inverse is $$ \boxed{ (u+n)^{-1} =\left(\sum_{j=0}^{k-1}(-u^{-1}n)^j\right)u^{-1}. } $$ Thus option A is correct. Why the other options are incorrect: Option B: A ring need not be free of zero divisors for this result to hold. The proof uses only the fact that $u$ is a unit, $n$ is nilpotent, and $u$ commutes with $n$. Option C: The sum need not be nilpotent. For example, in $\mathbb Z$, take $u=1$ and $n=0$. Then $u+n=1$, which is not nilpotent. Option D: False under the stated hypotheses. The sum $u+n$ is necessarily a unit. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$