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It can be shown that if $R$ is a commutative unital ring and $x$ is nilpotent, then $1-x$ is a unit of $R$. But is the converse true? I'm not sure how to approach this problem. That is, is it true that if $R$ is a commutative unital ring and $1-x$ is a unit, then $x$ is nilpotent?

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    Very easy example to show that this is not true: take $x \in \Bbb{Z}$ to be $x=2$. Then $1-x=-1$ is a unit, but $2$ is not nilpotent. – 2017-01-20
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    @sequence -- If you don't see how to prove something, you should consider that maybe it's false -- maybe it's even _usually_ false, and you should at least test it against the simplest examples. – 2017-01-20
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    @quasi true, but I for some reason was trying to find a general counterexample... Needless to say I couldn't. – 2017-01-20
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    As pointed out by another comment, this is not true in general. Such units are called **unipotent**. In a ring of matrices for example, those correspond to matrices with only $1$ as eigenvalue (whereas any invertible matrix is a unit of this ring). – 2017-01-20

2 Answers 2

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Seems not.

Take the field (thus commutative ring) of real numbers, and take $x$ to be any real number other than 1. Then $1-x$ is certainly a unit, but $x$ is not nilpotent.

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To get an instant counterexample, take any commutative unital ring $R$ such that

  • $R$ has no nonzero nilpotent elements.

  • $R$ has at least two units.

Now let $x = 1 - u$, where $u$ is a unit other than $1$.

Then $1 - x$ is a unit, but since $u \ne 1$, $x$ is nonzero, hence by choice of $R$, $x$ is not nilpotent.

For example, let $R$ be any integral domain such that $-1 \ne 1$ (i.e., $R$ does not have characteristic $2$), and let $x = 2$.