The following is (part of) Exercise 3.11 from Atiyah-Macdonald's Commutative Algebra
Let $A$ be a commutative ring with $1$. Every prime ideal $A$ is maximal iff $A/\text{Nilrad}(A)$ is absolutely flat.
I have been trying to prove this without using the characterization of Exercise 2.17 that says a ring is absolutely flat iff every principal ideal is idempotent. Instead, I want to use the characterization from the previous exercise:
$A$ is absolutely flat iff the localization $A_{\mathfrak{m}}$ is a field for each maximal ideal $\mathfrak m$.
The positioning and both statements seem so close together that it would seem most natural to use one to prove the other, however, I had tried to no avail. I have seen some solution online, but it implicitly uses that $A$ is an integral domain.
What I have tried is: Assume the first statement. Clearly every prime/maximal contains $\mathfrak R:=\text{Nilrad}(A)$ so we don't "lose" any primes/maximals passing to the quotient $A/\mathfrak R$. Now, there is a bijective correspondence between the $\{$primes of $A/\mathfrak R$ contained in $\mathfrak m / \mathfrak R \}\longleftrightarrow \{$prime ideals of $(A / \mathfrak R)_{\mathfrak m / \mathfrak R} \}$, but the left hand set only has one element by assumption, that is, $\mathfrak m / \mathfrak R$, and it corresponds precisely to the unique maximal ideal of $(A / \mathfrak R)_{\mathfrak m / \mathfrak R}$. But now, this unique maximal ideal is, in particular, $(\mathfrak m / \mathfrak R)_{\mathfrak m / \mathfrak R}$, and since taking quotients commutes with localization, then taking the quotient of the former by the latter is isomorphic to $((A/\mathfrak R )/(\mathfrak m /\mathfrak R))_{\mathfrak m / \mathfrak R} \cong (A / \mathfrak m)_{\mathfrak m / \mathfrak R}$, but $A / \mathfrak m $ is a field, hence $(A / \mathfrak R)_{\mathfrak m / \mathfrak R}$ is a field, hence $A$ is absolutely flat.
Then the argument is completely reversible proving the reverse implication. Is this correct?