Geometric meaning of localization at $(1+I)$?

Question

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 $A/I$?

Answer 1

Question: "What is the geometric interpretation of localization at this submonoid?"

Answer: If $k$ is any field and $\mathfrak{m}:=(x_1-a_1,..,x_n-a_n)\subseteq A:=k[x_1,..,x_n]$ the following holds: for any polynomial $g(x_1,..,x_n)$ it follows

$$g(x_1,..,x_n)=g(a_1+(x_1-a_1),..,a_n+(x_n-a_n))=g(a_1,..,a_n)+ y$$

with $y\in \mathfrak{m}$. If $S:=1+\mathfrak{m} \subseteq A-\mathfrak{m}$ it follows there is a canonical injective map

$$S^{-1}A \rightarrow A_{\mathfrak{m}}.$$

And any element $f(x)/g(x)\in A_{\mathfrak{m}}$ has $g(a_1,..,a_n):=a \neq 0$. Hence

$$\frac{f(x)}{g(x)}=\frac{f(x)}{a+y}= \frac{\frac{1}{a}f}{1+z}$$

with $z\in \mathfrak{m}$. Hence the canonical map $S^{-1}A \rightarrow A_{\mathfrak{m}}$ is an isomorphism. A similar argument holds for any algebraically closed field $k$ and $A$ any finitely generated $k$-algebra that is an integral domain.

You may wonder what happens if $I:=\mathfrak{m}_1^{l_1}\cdots \mathfrak{m}_d^{l_d}$ and you consider $(1+I)^{-1}A$. If you consider $I:=\mathfrak{m}^l$, I suspect the localization $S^{-1}A$ is related to "taking partial derivatives" at $\mathfrak{m}$ (vaguely). if $S_l:=(1+\mathfrak{m}^l)$ you get a sequence of subsets

$$ S_l \subseteq S_{l-1} \subseteq \dots \subseteq S_1$$

with canonical maps

$$S_l^{-1}A \rightarrow \cdots \rightarrow S_1^{-1}A \cong A_{\mathfrak{m}}.$$

Note: If $A$ is a domain it follows for any maximal ideal $\mathfrak{m}$, that $ (1+\mathfrak{m})^{-1}A \rightarrow A_{\mathfrak{m}} \subseteq K(A)$ are subrings of the quotient field $K(A)$. The quotient $A/\mathfrak{m}$ is not contained in the quotient field in general.