Localization of maximal ideal is again a maximal ideal and the residue fields are isomorphic

Question

Following is Lemma III.3.6 of Lorenzini's "An Invitation to Arithmetic Geometry":

Let $A$ be a commutative ring and $m\subset A$ be a maximal ideal. If $D\subset A\backslash m$ is a multiplicative set, then the fields $A/m$ and $D^{-1}A/D^{-1}m$ are isomorphic.

After spending one week, trying to figure out what the author wants to say in the proof, I think following are the key steps needed:

1. Show that $D^{-1}m$ is a maximal ideal of $D^{-1}A$.
2. Using the following composition of ring homomorphisms (where $f$ is defined as $f(x) = x/1$ and need not be injective) $$A\stackrel{f}{\rightarrow} D^{-1}A \stackrel{\pi}{\rightarrow} (D^{-1}A)/(D^{-1}m)$$ and the universal property of localization, need to conclude the desired isomorphism.

I am now tired of trying to understand the proof and hence will be grateful if somebody could shed some light on how to prove this lemma.

In case you are interested, following is the proof given by Lorenzini (taken from Google books):

Answer 1

Clearly, the author proves (I use his own notations) that the map: \begin{align} h:A/P &\longrightarrow S^{-1}A/S^{-1}P\\ a\bmod P& \longmapsto \frac{a}1 \bmod S^{-1}P \end{align} is surjective (which ensures it is bijective, hence an isomorphism since a homomorphism from a field is necessarily injective).

For this, he takes an element $\dfrac as\bmod S^{-1}P \in S^{-1}A/S^{-1}P$ and uses that $s$ is a unit mod. $P$ since $s\not\in P$ and $P$ is maximal, so that $\:\langle\, s, P\,\rangle=A$. This means there exists $t\in A$, $p\in P$ such that $ts+p=1$, which is equivalent to $ts-1=-p\in P$.

Then a small computation: $$\frac as=\frac{ta}{ts}\equiv\frac{ta}1\mod S^{-1}P= h(ta\bmod P)$$ shows that $\dfrac as\bmod S^{-1}P$ is in the image of $h$.

Answer 2

Though this is not the proof given by Lorenzini, the lemma can be seen as a consequence of the fact that localization is exact.

Consider the exact sequence of $A$-modules:
$0 \to m \to A \to A/m \to 0$

Localization gives an exact sequence:
$0 \to D^{-1}m \to D^{-1}A \to D^{-1}(A/m) \to 0$

This shows that $D^{-1}(A/m) \cong (D^{-1}A)/(D^{-1}m)$ where the isomorphism (of $D^{-1}A$-modules) is given by $\frac{\overline{a}}{d} \mapsto \overline{\frac{a}{d}}$. One checks that this is also a ring homomorphism.

Since we assume that $D \subset A\setminus m$ and $A/m$ is a field, every element of $D$ is invertible in $A/m$. Thus $D^{-1}(A/m) \cong A/m$.