**Lemma.** Let be a commutative ring with 1. If is nilpotent and is a unit, then is a unit.

*Proof*. So for some integer and for some Let

and see that

**Problem.** Let be a commutative ring with 1. Let be an element of the polynomial ring Prove that is a unit if and only if is a unit and all are nilpotent.

**First Solution.** () Suppose that are nilpotent and is a unit. Then clearly is nilpotent and thus is a unit, by the lemma.

() We’ll use induction on the degree of It’s clear for So suppose that the claim is true for any polynomial which is a unit and has degree less than Let be a unit. So there exists some such that Then and so is a unit. We also have

So where

Thus Therefore and hence because is a unit. Thus and so is nilpotent. So is a unit, by the lemma. Finally, since we can apply the induction hypothesis to finish the proof.

**Second Solution.** () This part is the same as the first solution.

() Let be a unit of and let be such that Then and so is a unit. To prove that is nilpotent for all we consider two cases:

*Case 1* . *is an integral domain*. Suppose that Then from we get which is impossible because both and are non-zero and is an integral domain. So and we are done.

*Case 2* . *is arbitrary*. Let be any prime ideal of and let For every let Let

Then clearly in and thus, since is an integral domain, for all by case 1. Hence for all So is in every prime ideal of and thus is nilpotent.

For the lemma : given $b.c=1$ for some $c$ then , if suppose there is no element such that $(a+b)u \not =1$ then by multiplying by $a^{n-1} \not =$ we get a contradiction that $a^{n-1} \not = a^{n-1} $ . does that sound ok ?

Hi, I just saw this. Do you know what can be said in the case where is non-commutative?

Thank you for posting this proof. There’s an insignificant mistake: since is a matrix.

You’re right, thank you. I fixed it.

Using the concept of matrix in the converse of the first one really made it simpler. I was trying to solve the problem for last few days. Thanks for the post.