Can a Unit Also be Nilpotent? Proving the Contradiction

[Bashyboy]

Homework Statement

First Claim: If $u \in R$ is a unit, then it cannot be nilpotent

Second Claim: If $u \in R$ is nilpotent, then it cannot be a unit

Homework Equations

The Attempt at a Solution

I realize these are simple problems, but I have no one to verify my work and I want to be certain I am doing things correctly. Here is a proof of the first claim:

Suppose the contrary, that $u$ is a unit and also nilpotent. This implies there exists an element $v$ that acts as a multiplicative inverse and a natural number $n$ such that $u^n = 0$. By the well ordering property, we can take $n$ to be the smallest natural number for which $u^n=0$. Then

$u^n = 0$

$u u^{n-1} = 0$

$vu u^{n-1} = v0$

$u^{n-1} = 0$,

contradicting the minimality of $n$.

Does this seem right? If I am not mistaken, then proof of the second claim is identical.

 

[fresh_42]

That's correct, although I wouldn't have used ordering and minimality, which looks kind of artificial to me, like a little stone in the shoe.
You could have simply multiplied by $v^n$ instead.