Why do we have $\operatorname{inv.}\lim_n A/I^{(n)}\cong \operatorname{inv.}\lim_n A/(I^{(1)})^n$?

Question

In (b), why is $I_2/I_2^2=I_2$ an $A_2/I_2\cong A_1$-module.

Why do we have $\operatorname{inv.}\lim\limits_n A/I^{(n)}\cong \operatorname{inv.}\lim\limits_n A/(I^{(1)})^n$?

Answer 1

Since $I_2\trianglelefteq A_2$ is an ideal, it is automatically an $A_2$-module. And $I_2^2$ is also an ideal, and a subset of $I_2$, so it is an $A_2$-submodule of $I_2$. Then we may form the quotient $A_2$-module $I_2/I_2^2$. But, if we multiply anything in $I_2$ by another thing in $I_2$ we get something in $I_2^2$, so if we multiply something in $I_2/I_2^2$ by $I_2$ we get zero. Thus, $I_2$ is in the kernel of the module structure $A_2\to\mathrm{End}(I_2/I_2^2)$, and so by the first isomorphism theorem for rings there is a module structure $A_2/I_2\to\mathrm{End}(I_2/I_2^2)$, in other words $I_2/I_2^2$ is an $A_2/I_2$-module. The action is

$$ (a+I_2)(i+I_2^2)=ai+I_2^2. $$

You can verify it is well-defined. That is, if $a+I_2=a'+I_2$ and $i+I_2^2=i'+I_2^2$ then we must also have $ai+I_2^2=a'i'+I_2^2$.

The inverse limits $\lim A/I^{(n)}$ and $\lim A/(I^{(1)})^n$ match because $I^{(n)}=(I^{(1)})^n$.