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Let $R$ be a commutative ring and $J(R)$ its Jacobson radical. It is easy to check that $J(R)$ consists of exactly those $f$ for which $1 + gf \in R^\times$ for all $g$.

Let $J'(R)$ be those $f$ for which $1 + uf \in R^\times$ for all units $u$. We might conjecture that in general $J(R) = J'(R)$. I assume this is false, since I can't find it anywhere, but I've been unable to come up with a counterexample - what are some counterexamples? Is there a nice characterization of when $J(R) = J'(R)$ holds?

user26857
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forget this
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1 Answers1

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Let $R=\mathbb{Z}_p[x]$. Then $p\in J'(R)$, since the units are just the constants that are not divisible by $p$, but $p\not\in J(R)$ since $1+px$ is not a unit.

More generally, if $S$ is any ring whose Jacobson radical is not equal to its nilradical, then you could take $R=S[x]$. Then for any $p\in S$ that is in the Jacobson radical but not the nilradical, $p\in J'(R)$ (because every unit of $R$ is a unit of $S$ plus a nilpotent element) but $p\not\in J(R)$ since $1+px$ is not a unit.

Eric Wofsey
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