Let $f:A\to B$ be a commutative ring morphism. Let $I\vartriangleleft A$ be an ideal. If $f(I)\subset \mathrm J(B)$ is contained in the Jacobson radical then $f(1+i)=1+f(i)\in B^\times$ is a unit so $1+I$ is inverted.

On the other hand, if $1+f(i)\in B^\times$ it does not seem to follow that $f(i)\in \mathrm J(B)$ since perhaps $1+bf(i)\notin B^\times$ for some $b\in B$.

Question. What are some examples where this implication fails i.e $f(1+I)\subset B^\times$ but $f(I)\nsubseteq\mathrm J(B)$?


There are some very basic examples. Let $f:\Bbb Z\to\Bbb Q$ be the inclusion map, and let $I$ be any non-trivial ideal, but for argument's sake say $I=3\Bbb Z$. Then $f(1+I)\subseteq\Bbb Q^\times$ but of course $J(\Bbb Q)=0$. Taking $I=3\Bbb Z$ we have another property: $1$ is the only invertible element of $1+I$ over $\Bbb Z$.

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  • $\begingroup$ Just came up with that example myself, but maybe you could help me clarify the source of my confusion. Say $f$ radicalizes $I$ if $f(I)\subset \mathrm J(B)$. The first paragraph of the question shows that if $f$ radicalizes $I$ then it inverts $1+I$. In this case, the universal property of localization gives a unique factorization of $f$ through $A[(1+I)^{-1}]$. A calculation shows this localization also radicalizes $I$, so it seems the localization represents the functor of radicalizations of $I$. What am I missing? $\endgroup$ – Arrow May 31 at 9:10
  • $\begingroup$ Ah, I think I see my error - radicalizations of $I$ are not functorial in the target ring. Thanks anyway! $\endgroup$ – Arrow May 31 at 9:26

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