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Let $I\vartriangleleft A$ be an ideal of a commutative ring. Consider the submonoid $1+I\subset A$. What is the geometric interpretation of localization at this submonoid? How does it relate to the quotient $A/I$?

I asked this on MSE and received some interesting comments, but I would still like to know the big picture and perhaps some examples. Taking e.g $\langle x\rangle\vartriangleleft A[x]$, the localization $A[x]_{1+\langle x\rangle}$ seems like it takes $x$ to something "very infinitesimal". I don't recognize this ring as the functions on some familiar "space" as e.g the dual numbers are the functions on an infinitesimal line segment. What is the idea here?

Added. I thought about viewing the localization as a subring of the ring of formal power series comprised of the polynomials along with "some" power series, namely those of the form $\sum_k x^k f^k=(1-xf)^{-1}$. Don't really know where to go from here though. I also received the observation that $A[x]_{1+\langle x\rangle}$ inverts the polynomials whose value at zero is invertible, which remains true in the multivariate case. Still I wonder how to interpret the case of a general ideal.

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    $\begingroup$ It is the global section of the inverse image of $\mathcal O_{\operatorname{Spec}(A)}$ along the inclusion $V(I)\to\operatorname{Spec}(A)$. I don't know whether it is geometric enough. By the way, could you please include the link to the MSE post? $\endgroup$
    – Z. M
    Commented Sep 24, 2021 at 19:52
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    $\begingroup$ Unless I'm mistaken, the scheme you obtain is the intersection of all open subsets containing $V(I)$ (since $f$ is invertible in the localization iff $(f)+I=1$ iff $D(f)\supseteq V(I)$). I think this is as geometric as it gets $\endgroup$
    – Denis Nardin
    Commented Sep 24, 2021 at 20:22

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