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I’m stuck in this problem: Let $R$ be a ring, $N$ its subset of nilpotent elements. For $r ∈ R^x$ and
$x ∈ N$ show that $r + x$ is a unit.

I’ve tried a few things like $r(1+xr^-1) = r + x$ but idk which method to used in order to prove it

rschwieb
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SelfProclaimedDev
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    I don’t understand your notation — specifically, what do you mean by $r \in R^x, x \in N$? Please use MathJax to format your post. – Robert Shore Jan 04 '23 at 17:19
  • So $x$ is a nilpotent element of the ring $R$. What does $R^x$ mean? – FShrike Jan 04 '23 at 17:21
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    R^x stands for $R^\times$ i.e. the units of $R$ and $x\in N$ just means that $x$ is nilpotent. At least this is what I thought. – nilsw Jan 04 '23 at 17:39
  • @nilsw If that is the case (and I think you're right) this question is a duplicate many times over. – rschwieb Jan 04 '23 at 17:41
  • @selfproclaimeddev please read this tutorial before posting more questions. You'll have a lot less trouble with people not understanding/misunderstanding your notation then. – rschwieb Jan 04 '23 at 17:43

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We have for any $y\in R$ $$ y\in R^\times \iff (\forall \mathfrak{m}\subset R \ \text{maximal ideal}:y\notin \mathfrak{m}).$$ Let $r\in R^\times$ and $x\in R$ be nilpotent. Then there is some $n\in \mathbb{N}$ s.t. $x^n = 0$. Let $\mathfrak{m}\subset R$ be any maximal ideal. In particular $\mathfrak{m}$ is a prime ideal and $x^n = 0\in \mathfrak{m}$ so $x\in \mathfrak{m}$. Suppose we would have $r+x\in \mathfrak{m}$ then clearly $r = r+x - x$ and the latter two terms are in $\mathfrak{m}$, so $r\in \mathfrak{m}$ but $r$ is a unit, hence $\mathfrak{m}=R$ but this contradicts $\mathfrak{m}$ being maximal. So we get $r+x\notin \mathfrak{m}$. As $\mathfrak{m}\subset R$ was an arbitrary maximal ideal we get $r+x\in R^\times$.

nilsw
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