Problem description
In the exercise in question we are supposed to prove that the natural map
$$A_f\to\mathcal O_{\text{Spec}(A)}(D(f))$$
is an isomorphism. Here $A_f$ is the localization of the ring $A$ at the set $\left\lbrace 1,f,f^2,\dots\right\rbrace$ and $\mathcal O_{\text{Spec}(A)}(D(f))$ is another localization of $A$, but this time at
$$S=\lbrace g\in A:\ g\ne 0\text{ outside of }V(f)\rbrace.$$
We can rephrase the condition to be $V(g)\subseteq V(f)$ or $D(f)\subseteq D(g)$, where $D(f)$ is "doesn't vanish $f$".
Regarding the natural part, Daniel McLaury addresses it here in this question.
My attempt
In the process of proving that this is an isomorphism, there's a hint which tells us to use exercise 3.5.E. This is the following equivalences:
$$D(f)\subseteq D(g)\iff \exists n(f^n\in\langle g\rangle)\iff g\text{ is invertible in }A_f.$$
By the first equivalence we can describe $S$ as
$$S=\lbrace g\in A:\ \exists n(f^n\in\langle g\rangle)\rbrace$$
which means that when localizing at $S$, we are localizing at some powers of $f$.
For example I might have one $g$ such that $f^n=ag$ and thus $f^{2n},f^{3n},\dots$ are now invertible, but I might also have an $h$ such that $f^{m}=a'h$ and so on.
My problem is that I'd like to localize at $S$ and have already localized at all the powers of $f$.If this was the case, then localizing at $S$ would already include the localization $A_f$, so the map $\iota:\ A_f\hookrightarrow \mathcal O_{\text{Spec}(A)}(D(f))$ becomes an inclusion.
The rest of the problem is based on showing that the function is surjective, but this is the case because $g$'s such that $D(f)\subseteq D(g)$ are already invertible in $A_f$ by the equivalence of exercise 3.5.E.
In summary
Is the idea correct? Can I have all the powers of $f$ be invertible when localizing by $S$? I can't see why having one power $f$ invertible would imply that all powers of $f$ are invertible. Or if this isn't the case, I'd appreciate any pointers in the right direction.
Thank you for taking the time to read, I really appreciate any help that you may be able to give me.
Best Answer
This is a general property of commutative unital rings. For any commutative ring $A$ and $f \in A$, if $f^n$ is invertible, say with inverse $g$, then $f^{n + 1}$ is invertible with inverse $f^{n - 1}g^2$. Then $f = f^{n + 1}g$ is the product of two units, so it is itself a unit, and also all powers of $f$ are units.