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Let $R$ be a commutative ring with $1$. Let $a\in R^\times$ and $b\in\text{Nil}(R)$. Show that $a+b\in R^\times$.

Attempts:

Suppose $b^n=0$. I've tried to figure out a "simple" case as $n=2$;

I've tried to develop the expressions $(a+b)^2=a(a+2b)$ and $a^{-1}(a+b)=1+a^{-1}b$ but nothing came up.

Also $\exists x,y\in R$ s.t. $1=xa+yb$ but it doesn't help because we get $x=a^{-1},y=0$.

Another direction i've failed is to show $\langle 0\rangle=R/\langle a+b\rangle$.

J. Doe
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    https://math.stackexchange.com/q/119904/29335 and about a dozen of its duplicates should be more than sufficient. Please search first next time. – rschwieb Aug 26 '19 at 16:22

1 Answers1

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Hint the inverse of $1-r$ is $1+...+r^{n-1}$ if $r^n=0$.

If $a$ is invertible, $a+r=a(1+a^{-1}r)$ and $a^{-1}r$ is nilpotent. The product of invertibles elements is invertible.

Tsemo Aristide
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    If you're going to tailor a post that fits a user's attempt uniquely somehow, I can respect that. But this is just blind duplication of one of the most duplicated question-answer pairs in this site's history. I'm fairly sure you know the site and its contents well enough to know better. Please consider "doing the right thing" next time and either contributing genuinely new content or selecting a duplicate candidate. – rschwieb Aug 26 '19 at 16:19