If $A/\mathfrak a$ is flat over $A$ then $V(\mathfrak a)$ is open. Why?

Question

I am trying to understand the following statement.

Let $A$ be a noetherian commutative ring and $\mathfrak a\subset A$ is an ideal. Suppose that the ring $A/\mathfrak a$ is flat over $A$, then $V(\mathfrak a)$ is open in $\operatorname{Spec} (A)$. How to prove this (just using the standard definition of flatness)?

Answer 1

In general, if $A$ is any ring (not necessarily Noetherian), then $A/I$ is flat over $A$ iff $\text{Supp}(I) \cap \text{Supp}(A/I) = \emptyset$ iff $\text{Spec}(A) = \text{Supp}(I) \sqcup \text{Supp}(A/I)$. This follows from the fact that a flat local map is faithfully flat, hence injective, so if $p \in \text{Supp}(A/I) = V(I)$, then $A_p \to A_p/I_p$ is injective, so $I_p = 0$. Thus, $V(I)$ is open iff $\text{Supp}(I)$ is a closed set, which happens e.g. if $I$ is finitely generated.

Answer 2

Since $A/\mathfrak{a}$ is $A$-flat, the ideal $\mathfrak{a}$ is generated by a single idempotent element $e$ (so $e^2=e$). This is a consequence of Nakayama's lemma and uses that $\mathfrak{a}$ is finitely generated (a consequence of the assumption that $A$ is Noetherian). (I'm pretty sure this is proved somewhere on the site.) You can now check that $\mathrm{Spec}(A)=D(e)\cup D(e-1)$ is a partition. Indeed, from $e^2=e$ we get $e(e-1)=0$, so for all prime ideals $\mathfrak{p}$, either $e\in\mathfrak{p}$ or $e-1\in\mathfrak{p}$, and we cannot have both (since $1\notin\mathfrak{p}$). So $D(e-1)=\mathrm{Spec}(A)\setminus D(e)=V(\mathfrak{a})$ is open.