# Examples where $1+I$ is inverted but $I$ is not mapped into the Jacobson radical

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 $$\Bbb Z$$.
• 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? – Arrow May 31 at 9:10
• Ah, I think I see my error - radicalizations of $I$ are not functorial in the target ring. Thanks anyway! – Arrow May 31 at 9:26